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Energy and Matter Unit 2

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Page 1: Energy and Matter Unit 2

wwwunit5orgchemistry

Energy and Matter

Unit 2

Chemical ChangeChemical Change

Any change involving a rearrangement of atomsAny change involving a

rearrangement of atoms

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

Physical and Chemical PropertiesExamples of Physical Properties

Boiling point Color Slipperiness Electrical conductivity

Melting point Taste Odor Dissolves in water

Shininess (luster) Softness Ductility Viscosity (resistance to flow)

Volatility Hardness Malleability Density (mass volume ratio)

Examples of Chemical Properties

Burns in air Reacts with certain acids Decomposes when heated

Explodes Reacts with certain metals Reacts with certain nonmetals

Tarnishes Reacts with water Is toxic

Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction

Three Possible Types of Bonds

+ -

d+ d-

Covalent eg H2

Polar Covalent eg HCl

Ionic eg NaCl

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 2: Energy and Matter Unit 2

Chemical ChangeChemical Change

Any change involving a rearrangement of atomsAny change involving a

rearrangement of atoms

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

Physical and Chemical PropertiesExamples of Physical Properties

Boiling point Color Slipperiness Electrical conductivity

Melting point Taste Odor Dissolves in water

Shininess (luster) Softness Ductility Viscosity (resistance to flow)

Volatility Hardness Malleability Density (mass volume ratio)

Examples of Chemical Properties

Burns in air Reacts with certain acids Decomposes when heated

Explodes Reacts with certain metals Reacts with certain nonmetals

Tarnishes Reacts with water Is toxic

Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction

Three Possible Types of Bonds

+ -

d+ d-

Covalent eg H2

Polar Covalent eg HCl

Ionic eg NaCl

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 3: Energy and Matter Unit 2

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

During aldquochemical reactionrdquonew materials are

formed by a change in the way atoms are bonded together

Physical and Chemical PropertiesExamples of Physical Properties

Boiling point Color Slipperiness Electrical conductivity

Melting point Taste Odor Dissolves in water

Shininess (luster) Softness Ductility Viscosity (resistance to flow)

Volatility Hardness Malleability Density (mass volume ratio)

Examples of Chemical Properties

Burns in air Reacts with certain acids Decomposes when heated

Explodes Reacts with certain metals Reacts with certain nonmetals

Tarnishes Reacts with water Is toxic

Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction

Three Possible Types of Bonds

+ -

d+ d-

Covalent eg H2

Polar Covalent eg HCl

Ionic eg NaCl

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 4: Energy and Matter Unit 2

Physical and Chemical PropertiesExamples of Physical Properties

Boiling point Color Slipperiness Electrical conductivity

Melting point Taste Odor Dissolves in water

Shininess (luster) Softness Ductility Viscosity (resistance to flow)

Volatility Hardness Malleability Density (mass volume ratio)

Examples of Chemical Properties

Burns in air Reacts with certain acids Decomposes when heated

Explodes Reacts with certain metals Reacts with certain nonmetals

Tarnishes Reacts with water Is toxic

Ralph A Burns Fundamentals of Chemistry 1999 page 23Chemical properties can ONLY be observed during a chemical reaction

Three Possible Types of Bonds

+ -

d+ d-

Covalent eg H2

Polar Covalent eg HCl

Ionic eg NaCl

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 5: Energy and Matter Unit 2

Three Possible Types of Bonds

+ -

d+ d-

Covalent eg H2

Polar Covalent eg HCl

Ionic eg NaCl

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 6: Energy and Matter Unit 2

Shattering an Ionic Crystal Bending a Metal

Bailar Jr Moeller Kleinberg Guss Castellion Metz Chemistry 1984 page 248

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+ +

+++ ++

++++

+ +++

+

+

+++ ++

++++

+ +++

++ + + + ++ +

+ ++

+

+ + + ++ + ++

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

+-+ --- +

+- -+

- ++-+ -

-+

+ -+-- - -++

-+

+ - + - -+ + -

An ionic crystal

A metal

No electrostatic forces of repulsion ndash metal is deformed (malleable)

Electrostatic forces

of repulsion

Force

Force

broken crystal

>

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 7: Energy and Matter Unit 2

Chemical Bonds

Increasing ionic character

Nonpolar covalent

Electrons are sharedequally

Cl Cl

Polar covalent

Electrons are sharedunequally

ClH

Ionic bonding

Electrons are transferred

Cl1-Na1+

Ralph A Burns Fundamentals of Chemistry 1999 page 229

bull between two identical nonmetal atoms are nonpolar covalentbull between two different nonmetal atoms are polar covalentbull between nonmetals and reactive metals are primarily ionic

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 8: Energy and Matter Unit 2

Covalent vs Ionic

Covalent

Transferelectrons

(ions formed)

+ -

BetweenMetal andNonmetal

StrongBonds

(high melting point)

Shareelectrons

(polar vs nonpolar)

BetweenTwo

Nonmetals

Weak Bonds

(low melting point)

Alike Different

Electronsare

involved

ChemicalBonds

Ionic

Different

Topic Topic

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 9: Energy and Matter Unit 2

Temperature Scales

Fahrenheit

212 oF

180 oF

32 oF

Celcius

100 oC

100 oC

0 oC

Kelvin

373 K

100 K

273 K

Boiling point of water

Freezing point of water

Notice that 1 kelvin degree = 1 degree Celcius

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 10: Energy and Matter Unit 2

Heat versus Temperature

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

TOTALKinetic ENERGY

= Heat

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 11: Energy and Matter Unit 2

Molecular Velocities

speed

Fra

ctio

ns o

f pa

rtic

les many different molecular speeds

molecules sorted by speed

the Maxwell speed distribution

httpantoinefrostburgeduchemsenese101gasesslidessld016htm

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 12: Energy and Matter Unit 2

Temperature vs Heat

Measuredwith a

Calorimeter

TotalKineticEnergy

Joules(calories)

Measuredwith a

Thermometer

AverageKineticEnergy

oCelcius(or Kelvin)

Alike Different

A Propertyof

Matter

HaveKineticEnergy

Heat

Different

Topic Topic

Temperature

thermometercalorimeter

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 13: Energy and Matter Unit 2

Conservation of Matter

Reactants yield Products

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 14: Energy and Matter Unit 2

Densitybull

Density is an INTENSIVE property of matter

- does NOT depend on quantity of matter - color melting point boiling point odor density

bull Contrast with

EXTENSIVE - depends on quantity of matter- mass volume heat content (calories)

Styrofoam Brick

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 15: Energy and Matter Unit 2

Properties of Matter

httpantoinefrostburgeduchemsenese101matterslidessld001htm

Pyrex Pyrex

ExtensiveProperties

IntensiveProperties

volumemass

densitytemperature

100 mL999347 g

0999 gmL20oC

15 mL149902 g

0999 gmL20oC

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 16: Energy and Matter Unit 2

Volume and DensityRelationship Between Volume and Density for Identical Masses of Common Substances

Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (gcm3)

Lithium

Water

Aluminum

Lead

10 19 053

10 10 10

10 37 27

10 058 114

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 17: Energy and Matter Unit 2

Density

D

M

Vensity

ass

olume

D = M V

M = D x V

V =M D

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 18: Energy and Matter Unit 2

Two ways of viewing density

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 71

Equal volumeshellip

hellipbut unequal masses

The more massive object(the gold cube) has thegreater density

aluminum gold

(A)

Equal masseshelliphellipbut unequal volumes

(B)

gold

aluminumThe object with the larger volume (aluminum cube) has the smaller density

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 19: Energy and Matter Unit 2

Specific Gravity

Jaffe New World of Chemistry 1955 page 66

09025

water 10

ice

cork

aluminum

27

>
>
>

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 20: Energy and Matter Unit 2

Archimedes Principle

Vfinal = 985 cm3

- Vinitial = 445 cm3

Vfishing sinker = 540 cm3

Before immersion

Water

445 cm3

After immersion

Fishing sinker

985 cm3

Thread

>

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 21: Energy and Matter Unit 2

Dissolving of Salt in Water

NaCl(s) + H2O Na+(aq) + Cl-(aq)

Cl-

ions

Na+

ions Water molecules

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 22: Energy and Matter Unit 2

Liquids

The two key properties we need to describe areEVAPORATION and its opposite CONDENSATION

add energy and break intermolecular bonds

EVAPORATION

release energy and form intermolecular bonds

CONDENSATION

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 23: Energy and Matter Unit 2

States of Matter

Solid Liquid Gas

Holds Shape

Fixed Volume

Shape of Container

Free Surface

Fixed Volume

Shape of Container

Volume of Container

heat heat

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 24: Energy and Matter Unit 2

Some Properties of Solids Liquids and Gases

Property Solid Liquid Gas

Shape Has definite shape Takes the shape of Takes the shape the container of its container

Volume Has a definite volume Has a definite volume Fills the volume of the container

Arrangement of Fixed very close Random close Random far apartParticles

Interactions between Very strong Strong Essentially noneparticles

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 25: Energy and Matter Unit 2

bull To evaporate molecules must have sufficient energy to break IM forces

bull Molecules at the surface break away and become gas

bull Only those with enough KE escapebull Breaking IM forces requires energy The

process of evaporation is endothermicbull Evaporation is a cooling processbull It requires heat

Evaporation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 26: Energy and Matter Unit 2

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system

What is a closed system

A closed system means matter canrsquot go in or out (put a cork in it)

What the heck is a ldquodynamic equilibriumrdquo

Condensation

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 27: Energy and Matter Unit 2

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid - some condense back to a liquid

The rate at which the molecules evaporate and condense are equal

Dynamic Equilibrium

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 28: Energy and Matter Unit 2

As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense

Equilibrium is reached when

Rate of Vaporization = Rate of Condensation

Molecules are constantly changing phase ldquodynamicrdquo

The total amount of liquid and vapor remains constant ldquoequilibriumrdquo

Dynamic Equilibrium

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 29: Energy and Matter Unit 2

bull Vaporization is an endothermic process - it requires heat

bull Energy is required to overcome intermolecular forces

bull Responsible for cool earthbull Why we sweat

Vaporization

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 30: Energy and Matter Unit 2

Energy Changes Accompanying Phase Changes

Solid

Liquid

Gas

Melting Freezing

Deposition

CondensationVaporization

Sublimation

Ene

rgy

of s

yste

m

Brown LeMay Bursten Chemistry 2000 page 405

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 31: Energy and Matter Unit 2

solid

liquid

gas

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 32: Energy and Matter Unit 2

solid

liquid

gas

vaporization

condensation

melting

freezing

Heat added

Tem

pera

ture

(oC

)

A

B

C

DE

Heating Curve for Water

0

100

LeMay Jr Beall Robblee Brower Chemistry Connections to Our Changing World 1996 page 487

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 33: Energy and Matter Unit 2

Latent Heat

bull Take 1 kg of water from ndash10 oC up to 150 oC we can plot temperature rise against absorbed heat

water

steam(water vapor)

-10 C

0 C

100 C

ice

Lf = 80 calg Lv = 540 calg

Lf is the latent heat of fusionLv is the latent heat of vaporization

Q heat absorbed

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 34: Energy and Matter Unit 2

MATTER

Can it be physically separated

Homogeneous Mixture

(solution)

Heterogeneous Mixture Compound Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed

noyesIs the composition uniform

noyes

Colloids Suspensions

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 35: Energy and Matter Unit 2

Elements

only one kindof atom atomsare bonded ifthe element

is diatomic orpolyatomic

Compounds

two ormore kindsof atomsthat arebonded

substancewith

definitemakeup

andproperties

Mixtures

two or moresubstances

that arephysically

mixed

two ormore

kinds ofand

Both elements and compounds have a definite makeup and definite properties

Packard Jacobs Marshall Chemistry Pearson AGS Globe page (Figure 241)

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 36: Energy and Matter Unit 2

Matter Flowchart

Examples

ndash graphite

ndash pepper

ndash sugar (sucrose)

ndash paint

ndash soda

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

element

hetero mixture

compound

solution homo mixture

hetero mixture

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 37: Energy and Matter Unit 2

Pure Substances

Elementndash composed of identical atomsndash examples copper wire aluminum foil

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 38: Energy and Matter Unit 2

Pure Substances

Compoundndash composed of 2 or more elements

in a fixed ratio

ndash properties differ from those of individual elements

ndash EX table salt (NaCl)

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 39: Energy and Matter Unit 2

Pure Substances

Law of Definite Compositionndash A given compound always contains the same

fixed ratio of elements

Law of Multiple Proportionsndash Elements can combine in different ratios to

form different compounds

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 40: Energy and Matter Unit 2

Mixtures

Variable combination of two or more pure substances

Heterogeneous Homogeneous

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 41: Energy and Matter Unit 2

Mixtures

Solutionndash homogeneousndash very small particlesndash no Tyndall effect Tyndall Effect

ndash particles donrsquot settlendash EX rubbing alcohol

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 42: Energy and Matter Unit 2

Mixtures

Colloidndash heterogeneousndash medium-sized particlesndash Tyndall effectndash particles donrsquot settlendash EX milk

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 43: Energy and Matter Unit 2

Mixtures

Suspensionndash heterogeneousndash large particlesndash Tyndall effectndash particles settlendash EX fresh-squeezed

lemonade

Courtesy Christy Johannesson wwwnisdnetcommunicationsartspageschem

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 44: Energy and Matter Unit 2

Classification of Matter

Materials

HomogeneousHeterogeneous

Heterogeneousmixture

Homogeneousmixture

Substance

Element Compound Solution Mixture

Speci

fic

Gen

eral

Order Disorder

Smoot Smith Price Chemistry A Modern Course 1990 page 43

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 45: Energy and Matter Unit 2

Classification of Matter

MATTER(gas Liquid

solid plasma)

PURESUBSTANCES MIXTURES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURESELEMENTSCOMPOUNDS

Separated by

physical means into

Separated by

chemical means into

Kotz amp Treichel Chemistry amp Chemical Reactivity 3rd Edition 1996 page 31

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 46: Energy and Matter Unit 2

Classification of Matter

uniformproperties

fixedcomposition

chemicallydecomposable

no

no

no

yes

hetero-geneousmixture

solution

element

compound

httpantoinefrostburgeduchemsenese101matterslidessld003htm

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 47: Energy and Matter Unit 2

Elements Compounds and Mixtures

(a)an element(hydrogen)

(b)a compound(water)

(c)a mixture(hydrogen and oxygen)

(d)a mixture(hydrogenand oxygen)

Dorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 68

hydrogenatoms hydrogen

atoms

oxygen atoms

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 48: Energy and Matter Unit 2

Mixture vs Compound

Mixture

FixedComposition

Bonds between

components

Can ONLY beseparated by

chemical means

VariableComposition

No bondsbetween

components

Can beseparated by

physical means

Alike Different

Contain two or more

elements

Can beseparated

intoelements

Involvesubstances

Compound

Different

Topic Topic

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 49: Energy and Matter Unit 2

Compounds vs Mixtures

bull Compounds have properties that are uniquely different from the elements from which they are made ndash A formula can always be written for a compoundndash eg NaCl Na + Cl2

bull Mixtures retain their individual propertiesndash eg Salt water is salty and wet

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 50: Energy and Matter Unit 2

Diatomic Elements 1 and 7H2

N2 O2 F2

Cl2

Br2

I2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 51: Energy and Matter Unit 2

MatterMatter

SubstanceDefinite composition

(homogeneous)

SubstanceDefinite composition

(homogeneous)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Element(Examples iron sulfur

carbon hydrogenoxygen silver)

Mixture ofSubstances

Variable composition

Mixture ofSubstances

Variable composition

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Compound(Examples water

iron (II) sulfide methaneAluminum silicate)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Homogeneous mixtureUniform throughoutalso called a solution

(Examples air tap watergold alloy)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Heterogeneous mixtureNonuniform

distinct phases(Examples soup concrete granite)

Chemicallyseparable

Physicallyseparable

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 52: Energy and Matter Unit 2

The Organization of Matter

MATTER

PURESUBSTANCES

HETEROGENEOUSMIXTURE

HOMOGENEOUSMIXTURES

ELEMENTS COMPOUNDS

Physical methods

Chemical methods

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 41

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 53: Energy and Matter Unit 2

Phosphorous (P4)

TWO ALLOTROPIC FORMS

White phosphorousspontaneously ignites

Red phosphorousused for matches

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 457

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 54: Energy and Matter Unit 2

Allotropes of Carbon

Graphite BuckminsterfullereneDiamond

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 27

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 55: Energy and Matter Unit 2

Gold

24 karat gold 18 karat gold 14 karat gold

Gold

Copper

Silver

1824 atoms Au2424 atoms Au 1424 atoms Au

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 56: Energy and Matter Unit 2

Solid Brass

An alloy is a mixture of metals

bull Brass = Copper + Zincbull Solid brass

bull homogeneous mixturebull a substitutional alloy

Copper

Zinc

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 57: Energy and Matter Unit 2

Brass Plated

bull Brass = Copper + Zincbull Brass plated

bull heterogeneous mixturebull Only brass on outside

Copper

Zinc

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 58: Energy and Matter Unit 2

Galvanized Nails and Screws

bull Zinc coating prevents rustndash Use deck screws for any outdoor project

bull Iron will rust if untreated ndash Weaken and break

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 59: Energy and Matter Unit 2

Methods of Separating Mixtures

bull Magnetbull Filterbull Decantbull Evaporationbull Centrifugebull Chromatographybull Distillation

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 60: Energy and Matter Unit 2

Chromatography

bull Tie-dye t-shirt

bull Black pen ink

bull DNA testingndash Tomb of Unknown Soldiersndash Crime scene ndash Paternity testing

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 61: Energy and Matter Unit 2

Paper Chromatographyof Water-Soluble Dyes

orange red yellow

Initial spots of dyes

Direction of Water(mobile phase)

movement

Filter paper(stationary phase)

Orange mixture ofredand

yellow

Suggested red dyeis not

homogeneous

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 62: Energy and Matter Unit 2

Separation by Chromatography

samplemixture

a chromatographic column

stationary phaseselectively absorbs

components

mobile phasesweeps sampledown column

detector

httpantoinefrostburgeduchemsenese101matterslidessld006htm

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 63: Energy and Matter Unit 2

Ion chromatogram of orange juice

time (minutes)

de

tec

tor

res

po

ns

e

0 5 10 15 20 25

Na+

K+

Mg2+ Fe3+

Ca2+

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 64: Energy and Matter Unit 2

A Distillation Apparatus

liquid with a soliddissolved in it

thermometer

condenser

tube

distillingflask

pure liquid

receiving flaskhose connected to

cold water faucetDorin Demmin Gabel Chemistry The Study of Matter 3rd Edition 1990 page 282

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 65: Energy and Matter Unit 2

Centrifugation

bull Spin sample very rapidly denser materials go to bottom (outside)

bull Separate blood into serum and plasmandash Serum (clear)ndash Plasma (contains red blood

cells lsquoRBCsrsquo)bull Check for anemia (lack of iron)

Blood

RBCrsquos

Serum

A B C

AFTER

Before

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 66: Energy and Matter Unit 2

The decomposition of two water molecules

2 H2O O2 + 2 H2

Electriccurrent

Watermolecules

Diatomic Diatomicoxygen molecule hydrogen molecules+

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 67: Energy and Matter Unit 2

Electrolysis

Must add acid catalyst to conduct electricity

H1+

water oxygen hydrogen

ldquoelectrordquo = electricity ldquolysisrdquo = to split

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 32

Water

Hydrogengas forms

Oxygengas forms

ElectrodeSource ofdirect current

H2O(l) O2 (g) + 2 H2 (g)

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 68: Energy and Matter Unit 2

Reviewing ConceptsPhysical Properties

bull List seven examples of physical properties

bull Describe three uses of physical propertiesbull Name two processes that are used to

separate mixturesbull When you describe a liquid as thick are

you saying that it has a high or low viscosity

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 69: Energy and Matter Unit 2

Reviewing ConceptsPhysical Properties

bull Explain why sharpening a pencil is an example of a physical change

bull What allows a mixture to be separated by distillation

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 70: Energy and Matter Unit 2

Reviewing ConceptsChemical Properties

bull Under what conditions can chemical properties be observed

bull List three common types of evidence for a chemical change

bull How do chemical changes differ from physical changes

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 71: Energy and Matter Unit 2

Reviewing ConceptsChemical Properties

bull Explain why the rusting of an iron bar decreases the strength of the bar

bull A pat of butter melts and then burns in a hot frying pan Which of these changes is physical and which is chemical

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 72: Energy and Matter Unit 2

ELEMENT

hydrogen molecule H2

ELEMENT

oxygen molecule O2

MIXTURE

a mixture ofhydrogen and oxygen molecules

CHEMICAL REACTION

if molecules collide with enoughforce to break them into atoms a can take place

COMPOUND

water H2O

>

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 73: Energy and Matter Unit 2

2 H22 H2 O2

O2 2 H2O2 H2O++ ++ EE

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 74: Energy and Matter Unit 2

Specific Heats of Some Substances

Specific Heat

Substance (cal g oC) (Jg oC)

Water 100 418Alcohol 058 24Wood 042 18Aluminum 022 090Sand 019 079Iron 011 046Copper 0093 039Silver 0057 024Gold 0031 013

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 75: Energy and Matter Unit 2

Copyright copy 2007 Pearson Benjamin Cummings All rights reserved

(a) Radiant energy (b) Thermal energy

(c) Chemical energy (d) Nuclear energy (e) Electrical energy

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 76: Energy and Matter Unit 2

The energy something possesses due to its motion depending on mass and velocity

Potential energy

Energy in Energy out

kinetic energy kinetic energy

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 77: Energy and Matter Unit 2

Energy

Kinetic Energy ndash energy of motion

KE = frac12 m v 2

Potential Energy ndash stored energy

Batteries (chemical potential energy)

Spring in a watch (mechanical potential energy)

Water trapped above a dam (gravitational potential energy)

mass velocity (speed)

B

AC

>

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 78: Energy and Matter Unit 2

School Bus or Bullet

Which has more kinetic energy a slow moving school bus or a fast moving bullet

Recall KE = frac12 m v 2

KE = frac12 m v 2 KE = frac12 m v

2

BUS BULLET

KE(bus) = frac12 (10000 lbs) (05 mph)2 KE(bullet) = frac12 (0002 lbs) (240 mph)2

Either may have more KE it depends on the mass of the bus and the velocity of the bullet

Which is a more important factor mass or velocity Why (Velocity)2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 79: Energy and Matter Unit 2

Kinetic Energy and Reaction Rate

Kinetic energy

Fra

ctio

ns o

f pa

rtic

les

lower temperature

higher temperature

minimum energyfor reaction

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 80: Energy and Matter Unit 2

Hot vs Cold Tea

Kinetic energy

Many molecules have anintermediate kinetic energy

Few molecules have avery high kinetic energy

Low temperature(iced tea)

High temperature(hot tea)

Perc

ent o

f mol

ecul

es

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 81: Energy and Matter Unit 2

Decomposition of Nitrogen Triiodide

2 NI3(s) N2(g) + 3 I2(g)

NI3 I2

N2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 82: Energy and Matter Unit 2

Exothermic ReactionReactants Products + Energy 10 energy = 8 energy + 2 energy

Reactants

Products

-DH

Ene

rgy

Energy of reactants

Energy of products

Reaction Progress

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 83: Energy and Matter Unit 2

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 84: Energy and Matter Unit 2

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reactionWhat is a catalyst What does it do during a chemical reaction

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 85: Energy and Matter Unit 2

Burning of a Match

Zumdahl Zumdahl DeCoste World of Chemistry 2002 page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

D(PE)

Pot

entia

l ene

rgy

(Products)

>

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 86: Energy and Matter Unit 2

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example the energy of the reactants and products increases while the energy of the surroundings decreases

In every case however the total energy does not change

Myers Oldham Tocci Chemistry 2004 page 41

Endothermic Reaction

Reactant + Energy Product

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 87: Energy and Matter Unit 2

Direction of Heat Flow

Surroundings

ENDOthermicqsys gt 0

EXOthermicqsys lt 0

System

Kotz Purcell Chemistry amp Chemical Reactivity 1991 page 207

System

H2O(s) + heat H2O(l)

melting

H2O(l) H2O(s) + heat

freezing

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 88: Energy and Matter Unit 2

Caloric Values

Food joulesgrams caloriesgram Caloriesgram

Protein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4

Smoot Smith Price Chemistry A Modern Course 1990 page 51

1000 calories = 1 Calorie

science food

1calories = 4184 joules

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 89: Energy and Matter Unit 2

Units of energy

Most common units of energy

1 S unit of energy is the joule (J) defined as 1 (kilogrambullmeter2)second2 energy is also

expressed in kilojoules (1 kJ = 103J)

2 Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed

to raise the temperature of 1 g of water by 1degC

One cal = 4184 J or 1J = 02390 cal

Units of energy are the same regardless of the form of energy

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 90: Energy and Matter Unit 2

Typical apparatus used in this activity include a boiler (such as large glass beaker) a heat source (Bunsen burner or hot plate) a stand or tripod for the boiler a calorimeter thermometers samples (typically samples of copper aluminum zinc tin or lead) tongs (or forceps or string) to handle samples and a balance

Experimental Determination of Specific Heat of a Metal

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 91: Energy and Matter Unit 2

A Bomb Calorimeter

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 92: Energy and Matter Unit 2

Heating CurvesTe

mp

erat

ure

(oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 93: Energy and Matter Unit 2

Calculating Energy Changes - Heating Curve for Water

Tem

per

atu

re (

oC

)

40

20

0

-20

-40

-60

-80

-100

120

100

80

60

140

Time

DH = mol x Cfus

DH = mol x Cvap

DH = mass x DT x Cp liquid

DH = mass x DT x Cp gas

DH = mass x DT x Cp solid

Cp gas = 187 JgoC

Cp liquid = 4184 JgoC

Cp solid = 2077 JgoC

Cf water = 333 Jg

Cv water = 2256 Jg

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 94: Energy and Matter Unit 2

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 95: Energy and Matter Unit 2

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C333

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system

a) 60oC b) 30oC c) 20oC d)

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 96: Energy and Matter Unit 2

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

Assume NO heat energy is ldquolostrdquo to the surroundings from the system

20 g (40oC) 10 g (20oC) 333oC

C726

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 267oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 97: Energy and Matter Unit 2

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 300oC

Block ldquoArdquo Block ldquoBrdquoFinal

Temperature

20 g (40oC) 10 g (20oC) 333oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 267oC

AgH2O

Real Final Temperature = 266oC

Why

Wersquove been assuming ALL materialstransfer heat equally well

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 98: Energy and Matter Unit 2

Specific Heat

bull Water and silver do not transfer heat equally well Water has a specific heat Cp = 4184 JgoC Silver has a specific heat Cp = 0235 JgoC

bull What does that mean It requires 4184 Joules of energy to heat 1 gram of water 1oC and only 0235 Joules of energy to heat 1 gram of silver 1oC

bull Law of Conservation of Energyhellip In our situation (silver is ldquohotrdquo and water is ldquocoldrdquo)hellip this means water heats up slowly and requires a lot of energy

whereas silver will cool off quickly and not release much energy

bull Lets look at the math

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 99: Energy and Matter Unit 2

ldquolosesrdquo heat

Calorimetry

C266 x

3208x 8550

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 266oC

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 100: Energy and Matter Unit 2

Calorimetry

C266 x

8550 3208x

7845 3138x x 057 705

algebra the solve and units Drop

C25 -x g 75CgJ 1844 C100 -x g 30CgJ 2350

equation into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 101: Energy and Matter Unit 2

1 Calorie = 1000 calories

ldquofoodrdquo = ldquosciencerdquo

Candy bar300 Calories = 300000 calories

English

Metric = _______Joules

1 calorie - amount of heat needed to raise 1 gram of water 1oC

1 calorie = 4184 Joules

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 102: Energy and Matter Unit 2

Cp(ice) = 2077 Jg oC

It takes 2077 Joules to raise 1 gram ice 1oC

X Joules to raise 10 gram ice 1oC

(10 g)(2077 Jg oC) = 2077 Joules

X Joules to raise 10 gram ice 10oC

(10oC)(10 g)(2077 Jg oC) = 2077 Joules

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DTTe

mpe

ratu

re (

o C)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 103: Energy and Matter Unit 2

Heat = (specific heat) (mass) (change in temperature)

q = Cp m DT

T m C q p(ice)

initialfinalp(ice) TT m C q

C)30(C20- g 10 C g

J 2077 q oo

o

Given Ti = -30oC

Tf = -20oC

q = 2077 Joules

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 104: Energy and Matter Unit 2

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC) When thermal equilibrium is reached the system has a temperature of 42oC Find the mass of the iron

Calorimetry Problems 2 question 5

FeT = 500oCmass = grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(CpFe) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(04495 JgoC) (X g) (42oC - 500oC)] = (4184 JgoC) (240 g) (42oC - 20oC)

Drop Units - [(04495) (X) (-458)] = (4184) (240 g) (22)

2059 X = 22091

X = 1073 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 105: Energy and Matter Unit 2

A 97 g sample of gold at 785oC is dropped into 323 g of water which has an initial temperature of 15oC If gold has a specific heat of 0129 JgoC what is the final temperature of the mixture Assume that the gold experiences no change in state of matter

Calorimetry Problems 2 question 8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(CpAu) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0129 JgoC) (97 g) (Tf - 785oC)] = (4184 JgoC) (323 g) (Tf - 15oC) Drop Units

- [(125) (Tf - 785oC)] = (135x 103) (Tf - 15oC)

-125 Tf + 982 x 103 = 135 x 103 Tf - 202 x 104

3 x 104 = 136 x 103 Tf

Tf = 221oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 106: Energy and Matter Unit 2

If 59 g of water at 13oC are mixed with 87 g of water at 72oC find the final temperature of the system

Calorimetry Problems 2 question 9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(4184 JgoC) (59 g) (Tf - 13oC)] = (4184 JgoC) (87 g) (Tf - 72oC) Drop Units

- [(2468) (Tf - 13oC)] = (3640) (Tf - 72oC)

-2468 Tf + 3208 = 364 Tf - 26208

29416 = 6108 Tf

Tf = 482oC

T = 72oC

mass = 87 g

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 107: Energy and Matter Unit 2

A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC Find the systems final temperature

Calorimetry Problems 2 question 10

iceT = -11oCmass = 38 g

T = 56oC

mass = 214 g

LOSE heat = GAIN heat-

- [(CpH2O) (mass) (DT)] = (CpH2O) (mass) (DT) + (Cf) (mass) + (CpH2O) (mass) (DT)

- [(4184 JgoC)(214 g)(Tf - 56oC)] = (2077 JgoC)(38 g)(11oC) + (333 Jg)(38 g) + (4184 JgoC)(38 g)(Tf - 0oC)

- [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)

- 895 Tf + 50141 = 868 + 12654 + 159 Tf

- 895 Tf + 50141 = 13522 + 159 Tf

Tf = 347oC

36619 = 1054 Tf

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

warm icemelt ice

warm water

water cools

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 108: Energy and Matter Unit 2

25 g of 116oC steam are bubbled into 02384 kg of water at 8oC Find the final temperature of the system

Calorimetry Problems 2 question 11

- [(CpH2O) (mass) (DT)] + (CvH2O) (mass) + (CpH2O) (mass) (DT) = [(CpH2O) (mass) (DT)]

- [ - 8168 - 56400 + 1045Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(CpH2O) (mass) (DT)]

qA = [(2042 JgoC) (25 g) (100o - 116oC)]

qA = - 8168 J

qB = (CvH2O) (mass)

qB = (2256 Jg) (25 g)

qB = - 56400 J

qC = [(CpH2O) (mass) (DT)]

qC = [(4184 JgoC) (25 g) (Tf - 100oC)]

qC = 1045Tf - 10450

qD = (4184 JgoC) (2384 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

8168 + 56400 - 1045Tf + 10450 = 997Tf - 7972

67667 - 1045Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 686oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp liquid

Heat = mass x Dt x Cp gas

Heat = mass x Dt x Cp solid

A

B

C

D

(1000 g = 1 kg)

2384 g

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 109: Energy and Matter Unit 2

A 322 g sample of lead (specific heat = 0138 JgoC) is placed into 264 g of water at 25oCIf the systems final temperature is 46oC what was the initial temperature of the lead

Calorimetry Problems 2 question 12

PbT = oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(CpPb) (mass) (DT)] = (CpH2O) (mass) (DT)

- [(0138 JgoC) (322 g) (46oC - Ti)] = (4184 JgoC) (264 g) (46oC- 25oC) Drop Units

- [(4444) (46oC - Ti)] = (11046) (21oC)

- 2044 + 4444 Ti = 23197

4444 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 110: Energy and Matter Unit 2

A sample of ice at ndash12oC is placed into 68 g of water at 85oC If the final temperature of the system is 24oC what was the mass of the ice

Calorimetry Problems 2 question 13

H2OT = -12oCmass = g

Ti = 85oC

mass = 68 g

GAIN heat = - LOSE heat

[ qA + qB + qC ] = - [(CpH2O) (mass) (DT)]

4582 m = - 17339

m = 378 g

iceTf = 24oC

qA = [(CpH2O) (mass) (DT)]

qC = [(CpH2O) (mass) (DT)]

qB = (CfH2O) (mass)

qA = [(2077 JgoC) (mass) (12oC)]

qB = (333 Jg) (mass)

qC = [(4184 JgoC) (mass) (24oC)]

[ qA + qB + qC ] = - [(4184 JgoC) (68 g) (-61oC)]

249 m

333 m

1003 m

4582 mqTotal = qA + qB + qC

4582 4582

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 111: Energy and Matter Unit 2

Endothermic Reaction

Energy + Reactants Products

+DH Endothermic

Reaction progress

Ene

rgy

Reactants

ProductsActivation Energy

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 112: Energy and Matter Unit 2

O

Catalytic Converter

C O

N O

CO

OCO

NN

One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas

CO

N

NN

OO

OC

OCO

2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 113: Energy and Matter Unit 2

Enthalpy Diagram

H2O(g)H2O(g)

H2O(l)H2O(l)

H2(g) + frac12 O2(g)

- 44 kJExothermic

+44 kJEndothermic

DH = +242 kJEndothermic

-242 kJExothermic

-286 kJEndothermic

DH = -286 kJExothermic

Ene

rgy

H2(g) + 12O2(g) H2O(g) + 242 kJ H = -242 kJKotz Purcell Chemistry amp Chemical Reactivity 1991 page 211

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 114: Energy and Matter Unit 2

Hessrsquos LawCalculate the enthalpy of formation of carbon dioxide from its elements

C(g) + 2O(g) CO2(g)

Use the following data2O(g) O2(g) H = - 250 kJC(s) C(g) H = +720 kJCO2(g) C(s) + O2(g) H = +390 kJ

Smith Smoot Himes pg 141

2O(g) O2(g) H = - 250 kJ

C(g) + 2O(g) CO2(g) H = -1360 kJ

C(g) C(s) H = - 720 kJC(s) + O2(g) CO2(g) H = - 390 kJ

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 115: Energy and Matter Unit 2

Fission vs Fusion

Fuse small atoms2H2 He

NO Radioactive

waste

Very HighTemperatures~5000000 oC

(SUN)

Split large atoms

U-235

Radioactive waste

(long half-life)

NuclearPowerPlants

Alike Different

Create Large Amounts

of EnergyE = mc2

Transmutationof Elements

Occurs

Change Nucleus

of Atoms

Fusion

Different

Topic Topic

Fission

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 116: Energy and Matter Unit 2

bull Use fear and selective facts

to promote an agenda

bull Eating animalsbull Radiation = Bad

Look who is funding research it may bias the results

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 117: Energy and Matter Unit 2

Shielding Radiation

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 118: Energy and Matter Unit 2

Nuclear Fission

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 119: Energy and Matter Unit 2

Nuclear Fission

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 120: Energy and Matter Unit 2

Nuclear Power Plants

map Nuclear Energy Institute

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 121: Energy and Matter Unit 2

Nuclear Fusion

Sun

+ +

Fourhydrogen

nuclei(protons)

Two betaparticles

(electrons)

Oneheliumnucleus

He e2 H4 4

2

0

1-

1

1 + Energy

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 122: Energy and Matter Unit 2

Conservation of Masshellipmass is converted into energy

Hydrogen (H2) H = 1008 amuHelium (He) He = 4004 amu

FUSION

2 H2 1 He + ENERGY

1008 amux 440032 amu = 4004 amu + 0028 amu

This relationship was discovered by Albert EinsteinE = mc2

Energy= (mass) (speed of light)2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 123: Energy and Matter Unit 2

Tokamak Reactor

bull Fusion reactorbull 10000000 o Celciusbull Russian for torroidial

(doughnut shaped) ring

bull Magnetic field contains plasma

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 124: Energy and Matter Unit 2

Cold Fusion

bull Fraudbull Experiments must

be repeatable to

be valid

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 125: Energy and Matter Unit 2

0 1 2 3 4Number of half-lives

Rad

iois

otop

e re

mai

ning

(

)

100

50

25

125

Half-life of Radiation

Initial amountof radioisotope

t12

t12

t12

After 1 half-life

After 2 half-lives

After 3 half-lives

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 126: Energy and Matter Unit 2

Objectives - Matter

bull Explain why mass is used as a measure of the quantity of matter

bull Describe the characteristics of elements compounds and mixtures

bull Solve density problems by applying an understanding of the concepts of density

bull Distinguish between physical and chemical properties and physical and chemical changes

bull Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 127: Energy and Matter Unit 2

Objectives - Energy

bull Identify various forms of energybull Describe changes in energy that take place

during a chemical reactionbull Distinguish between heat and temperaturebull Solve calorimetry problemsbull Describe the interactions that occur between

electrostatic charges

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 128: Energy and Matter Unit 2

Law of Conservation of EnergyEafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 129: Energy and Matter Unit 2

Law of Conservation of Energy

ENERGY

CO2 + H2OC2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2 H2 + O2 2 H2O + energy

+ + WOOF

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 130: Energy and Matter Unit 2

Law of Conservation of Energy

ENERGY

C2H2 + O2C2H2 + O2

PEreactants

PEproducts

KEstopper

heat light sound

Eafter = Ebefore

2C2H2 + 5O2 4 CO2 + 2H2O + energy

Energy Changes

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures
Page 131: Energy and Matter Unit 2

First experimental image showing internal atomic structures

copy 2005 University of Augsburg Experimental Physics VI httpwwwphysikuni-augs

  • Energy and Matter
  • Slide 2
  • Slide 3
  • Physical and Chemical Properties
  • Three Possible Types of Bonds
  • Shattering an Ionic Crystal Bending a Metal
  • Chemical Bonds
  • Covalent vs Ionic
  • Temperature Scales
  • Heat versus Temperature
  • Molecular Velocities
  • Temperature vs Heat
  • Conservation of Matter
  • Density
  • Properties of Matter
  • Volume and Density
  • Density (2)
  • Two ways of viewing density
  • Specific Gravity
  • Archimedes Principle
  • Dissolving of Salt in Water
  • Liquids
  • States of Matter
  • Some Properties of Solids Liquids and Gases
  • Slide 25
  • Condensation
  • Dynamic Equilibrium
  • Dynamic Equilibrium (2)
  • Vaporization
  • Energy Changes Accompanying Phase Changes
  • Heating Curve for Water
  • Heating Curve for Water (2)
  • Latent Heat
  • Slide 34
  • Slide 35
  • Matter Flowchart
  • Pure Substances
  • Pure Substances (2)
  • Pure Substances (3)
  • Mixtures
  • Mixtures (2)
  • Mixtures (3)
  • Mixtures (4)
  • Classification of Matter
  • Classification of Matter (2)
  • Classification of Matter (3)
  • Elements Compounds and Mixtures
  • Mixture vs Compound
  • Compounds vs Mixtures
  • Diatomic Elements 1 and 7
  • Slide 51
  • The Organization of Matter
  • Phosphorous (P4)
  • Allotropes of Carbon
  • Gold
  • Solid Brass
  • Slide 57
  • Galvanized Nails and Screws
  • Methods of Separating Mixtures
  • Chromatography
  • Paper Chromatography of Water-Soluble Dyes
  • Separation by Chromatography
  • Ion chromatogram of orange juice
  • A Distillation Apparatus
  • Centrifugation
  • The decomposition of two water molecules
  • Electrolysis
  • Reviewing Concepts Physical Properties
  • Reviewing Concepts Physical Properties (2)
  • Reviewing Concepts Chemical Properties
  • Reviewing Concepts Chemical Properties (2)
  • Slide 72
  • Slide 73
  • Specific Heats of Some Substances
  • Slide 75
  • Slide 76
  • Energy
  • School Bus or Bullet
  • Kinetic Energy and Reaction Rate
  • Hot vs Cold Tea
  • Decomposition of Nitrogen Triiodide
  • Exothermic Reaction
  • Endothermic Reaction
  • Effect of Catalyst on Reaction Rate
  • Burning of a Match
  • Conservation of Energy in a Chemical Reaction
  • Direction of Heat Flow
  • Slide 88
  • Units of energy
  • Slide 90
  • A Bomb Calorimeter
  • Heating Curves
  • Calculating Energy Changes - Heating Curve for Water
  • Heat Transfer
  • Heat Transfer (2)
  • Heat Transfer (3)
  • Heat Transfer (4)
  • Specific Heat
  • Calorimetry
  • Calorimetry (2)
  • Slide 101
  • Slide 102
  • Slide 103
  • Slide 104
  • Slide 105
  • Slide 106
  • Slide 107
  • Slide 108
  • Slide 109
  • Slide 110
  • Endothermic Reaction (2)
  • Catalytic Converter
  • Enthalpy Diagram
  • Hessrsquos Law
  • Fission vs Fusion
  • Irradiated Spam
  • Shielding Radiation
  • Nuclear Fission
  • Nuclear Fission (2)
  • Nuclear Power Plants
  • Nuclear Fusion
  • Conservation of Mass
  • Tokamak Reactor
  • Cold Fusion
  • Half-life of Radiation
  • Objectives - Matter
  • Objectives - Energy
  • Law of Conservation of Energy
  • Law of Conservation of Energy (2)
  • Law of Conservation of Energy (3)
  • First experimental image showing internal atomic structures