CH104 CH104 Chapter 2: Energy & Matter Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves

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CH104 CH104 Chapter 2: Energy & MatterChapter 2: Energy & Matter

TemperatureTemperature

Energy in ReactionsEnergy in Reactions

Specific HeatSpecific Heat

Energy from FoodEnergy from Food

States of MatterStates of Matter

Heating & Cooling CurvesHeating & Cooling Curves

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Chapter 2 Energy and MatterChapter 2 Energy and Matter

2.2Temperature

2

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Metric

SI

Common

ConversionsLengthVolumeMassTemperature

Units of MeasurementUnits of Measurement

meter (m) 1 m = 1.09 yd

liter (L) 1 L = 1.06 qt

gram (g) 1 kg = 2.2 lb

Celsius (oC) C = (F-32)/1.8

Kelvin (K) K = C + 273

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180180 oo100100 oo


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Learning CheckLearning Check

A. What is the temperature of freezing water?

1) 0 °F 2) 0 °C 3) 0 K

B. What is the temperature of boiling water?

1) 100 °F 2) 32 °F 3) 373 K

C. How many Celsius units are between the boiling and freezing points of water?

1) 100 2) 180 3) 273

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SolutionSolution

A. What is the temperature of freezing water?

1) 0 °F 2) 0 °C 3) 0 K

B. What is the temperature of boiling water?

1) 100 °F 2) 32 °F 3) 373 K

C. How many Celsius units are between the boiling and freezing points of water?

1) 100 2) 180 3) 273

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• At what temperature does At what temperature does ooF = F = ooC?C?

Trivial Pursuit QuestionTrivial Pursuit Question

- 40 - 40 ooF = - 40 F = - 40 ooCC

- 40 - 40 oo

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Temperature conversionTemperature conversion

• Common scales used• Fahrenheit,Fahrenheit, CelsiusCelsius andand KelvinKelvin..

ooFF = 1.8 = 1.8 ooC + 32 C + 32

ooCC = = ((ooF - 32)F - 32)1.81.8

KK = = ooC + 273C + 273 SI unitSI unit

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Solving a Temperature ProblemSolving a Temperature ProblemA person with hypothermia has abody temperature of 34.8 °C. WhatIs that temperature in °F?

F = 1.8(C) + 32 °

F = (1.8)(34.8 °C) + 32 ° exact tenth’s exact

= 62.6 ° + 32 ° = 94.6 °F tenth’s

9

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Practice: Temp ConversionPractice: Temp Conversion

• What is 75.0 º F in ºC?• ºC = (75.0 º F -32 º) = 23.9 ºC

1.8

• What is -12 º F in ºC?• º F = 1.8 (-12) + 32 º F = 10 º C

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Learning CheckLearning Check The normal temperature of a chickadee is 105.8

°F. What is that temperature on the Celsius scale?

1) 73.8 °C 2) 58.8 °C 3) 41.0 °C

11

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SolutionSolution

TC = TF – 32 °

1.8

= (105.8 – 32 °)

1.8

= 73.8 °F = 41.0 °C

1.8 ° tenth’s place

12

The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale?

1) 73.8 °C 2) 58.8 °C 3) 41.0 °C

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Learning Check Learning Check A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale?

1) 423 °C 2) 235 °C 3) 221 °C

13

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Solution Solution

TF – 32 ° = TC

1.8

(455 – 32 °) = 235 °C

1.8 one’s place

14

A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale?

1) 423 °C 2) 235 °C 3) 221 °C

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Learning Check Learning Check On a cold winter day, the temperature is –15 °C.What is that temperature in °F?1) 19 °F 2) 59 °F 3) 5 °F

15

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Solution Solution

TF = 1.8TC + 32 °

TF = 1.8(–15 °C) + 32 °

= – 27 + 32 °

= 5 °F one’s place

16

On a cold winter day, the temperature is –15 °C.What is that temperature in °F?1) 19 °F 2) 59 °F 3) 5 °F

Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign.

1.8 x 15 +/ – = –27

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TemperaturesTemperatures

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What is normal body temperature of 37 °C in kelvins?

1) 236 K 2) 310 K 3) 342 K

18

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Solution Solution

TK = TC + 273

= 37 °C + 273

= 310. K

one’s place

19

What is normal body temperature of 37 °C in kelvins?

1) 236 K 2) 310 K 3) 342 K

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2.1Energy

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EnergyEnergy

EnergyEnergy = The capacity to cause change

HeatLightWind

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Potential Energy Potential Energy = stored Energystored Energy (Has potential for motion)

Kinetic Energy Kinetic Energy = Energy in motionEnergy in motion (Fulfilling its potential)

X

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Identify each of the following as potential energy or kinetic energy.

A. roller blading

B. a peanut butter and jelly sandwich

C. mowing the lawn

D. gasoline in the gas tank

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SolutionSolution

Identify each of the following as potential energy or kinetic energy.

A. roller blading kinetic kinetic

B. a peanut butter and jelly sandwich potential potential

C. mowing the lawn kinetic kinetic

D. gasoline in the gas tank potential potential

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Kinetic EnergyKinetic EnergyKEKE = 1 mv2

2 Which has more E?

•Truck moving at 5 mph

•Bicycle moving at 5 mph

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MetricMetric

SISI

CommonCommon

ConversionsConversions

LengthLength

VolumeVolume

MassMass


EnergyEnergy

Units of MeasurementUnits of Measurement

meter (m) 1 m = 1.09 ydmeter (m) 1 m = 1.09 yd

liter (L) 1 L = 1.06 qtliter (L) 1 L = 1.06 qt

gram (g) 1 kg = 2.2 lbgram (g) 1 kg = 2.2 lb

calorie (cal) 1Kcal = 1000 cal = 1Calcalorie (cal) 1Kcal = 1000 cal = 1CalJoule (J) 1 cal = 4.18 JJoule (J) 1 cal = 4.18 J

Celsius (oC) C = (F-32)/1.8

Kelvin (K) K = C + 273

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Examples of Energy Values in JoulesExamples of Energy Values in Joules

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Learning CheckLearning CheckHow many calories are obtained from a pat

of butter if it provides 150 J of energy when metabolized?

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SolutionSolutionHow many calories are obtained from a pat

of butter if it provides 150 J of energy when metabolized?

Given

150 J

Need

= cal

1 cal and 4.184 J4.184 J 1 cal

1 cal4.184 J

36

Conversion Factors:

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2.3Specific Heat

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EnergyEnergyUnits of Energy =Units of Energy =

caloriecalorie calcal

kilocaloriekilocalorie kcalkcal 1000 cal = 1 kcal1000 cal = 1 kcal

CalorieCalorie CalCal Cal = kcalCal = kcal

joulejoule JJ 4.18 J = 1 cal4.18 J = 1 cal

British British

Thermal UnitThermal UnitBTUBTU

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EnergyEnergycalorie: calorie:

•E E toto raise 1 g H raise 1 g H220 0 byby 1 1 ooCC

H2O 11 cal

1g 1oC

11 cal

1g 1oC


1oC

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Specific HeatSpecific HeatE to raise Temp of 1g substance by 1 E to raise Temp of 1g substance by 1 ooCC

Fe Fe

0.110.11

Cu

0.093H2O

1.00

Ag

0.057

Au

0.031Sand

0.19

Al

0.22

0oC = start Add 1 cal

cal

g oC

cal

g oC

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Fe Fe

0.110.11

Cu

0.093

Ag

0.057

Au

0.031

Sand

0.19

Al

0.22

0oC = start

H2O

1.00

E to raise Temp of 1g substance by 1 E to raise Temp of 1g substance by 1 ooCC

cal

g oC

cal

g oC

1o

10o

30o

20o

Add 1 cal

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High SpHt; Resists changeHigh SpHt; Resists change


Fe Fe

0.110.11

Cu

0.093

Ag

0.057

Au

0.031

Sand

0.19

Al

0.22H2O

1.00

1o

10o

30o

20o

Low SpHt; Heats quickly

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Sand

0.19H2O

1.00

Resists change

Stays cold

Heats quickly

Gets hot

Hydrated person

Resists change in body temp

Dehydrated person

Body temp rises quickly

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Learning CheckLearning CheckA. For the same amount of heat added, a

substance with a large specific heat

1) has a smaller increase in temperature

2) has a greater increase in temperature

B. When ocean water cools, the surrounding air

1) cools 2) warms 3) stays the same

C. Sand in the desert is hot in the day and cool at night. Sand must have a

1) high specific heat 2) low specific heat

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SolutionSolutionA. For the same amount of heat added, a

substance with a large specific heat

1) has a smaller increase in temperature

2) has a greater increase in temperature

B. When ocean water cools, the surrounding air

1) cools 2) warms 3) stays the same

C. Sand in the desert is hot in the day and cool at night. Sand must have a

1) high specific heat 2) low specific heat

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What is the specific heat of a metal if 24.8 g absorbs 65.7 cal 65.7 cal of energy and the temp rises from 20.2 C to 24.5 C?

20.2oC

Specific HeatSpecific HeatSample Problem:Sample Problem:

24.5oCT = 4.3 CoT = 4.3 Co

cal

g oC

cal

g oC

m =m =

SpHt =SpHt =SpHt =SpHt =

24.8g

= cal

24.8g 4.34.3oC

= cal

24.8g 4.34.3oC

65.765.7 = cal

1g 1oC

= cal

1g 1oC

0.620.62

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75oC


How much energy does is take to heat 50 g’s of water from 75oC to 87oC?How much energy does is take to heat 50 g’s of water from 75oC to 87oC?

87oCT = 12 CoT = 12 Co

E = m E = m T SpHtT SpHt

11 cal

1g 1oC

11 cal

1g 1oC

m =m =


50g H2O

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50g H2O


How much energy does is take to heat 50 g’s of water from 75oC to 87oC?How much energy does is take to heat 50 g’s of water from 75oC to 87oC?

12 Co 12 Co


11 cal

1g 1oC

11 cal

1g 1oC

mm SpHtSpHtTT

= = cal calto heat waterto heat water

600600

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37oC


A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles?


65oCT = 28 CoT = 28 Co


11 cal

1g 1oC

11 cal

1g 1oC

m =m =


750g H2O

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750g H2O


28 Co 28 Co


11 cal

1g 1oC

11 cal

1g 1oC

mm SpHtSpHtTT

= = cal calfrom cool waterfrom cool water

2100021000



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Learning CheckLearning CheckHow many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to

77.0 °C?

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SolutionSolutionHow many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to

77.0 °C?

15oC

77oC


T = 62 CoT = 62 Co

325g H2O 62 Co 62 Co 4.1844.184 J

1g 1oC

4.1844.184 J

1g 1oC

mm SpHtSpHtTT

== KJ KJ84.384.311 KJ

1000 J

11 KJ

1000 J

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2.4Energy and Nutrition

1 Cal = 1000 calories

1 Cal = 1 kcal

1 Cal = 4184 J

1 Cal = 4.184 kJ

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EnergyEnergyUnits of Energy =Units of Energy =

1000 cal = 1 kcal1000 cal = 1 kcal

1 Cal1 Cal

1 Cal1 Cal

4184 J = 4.184 kJ4184 J = 4.184 kJ

1 cal = 4.18 J 1 cal = 4.18 J

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CalorimetersCalorimetersA calorimeter • A reaction chamber

& thermometer in H2O used to measure heat transfer

• indicates the heat lost by a sample

• indicates the heat gained by water

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Energy from FoodEnergy from Food

CarbohydrateCarbohydrate

FatFat

ProteinProtein

4 kcal

g

4 kcal

g

9 kcal

g

9 kcal

g

4 kcal

g

4 kcal

g

17 kJ

g

17 kJ

g

38 kcal

g

38 kcal

g

17 kcal

g

17 kcal

g

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= 196 Cal= 196 Cal

Energy from FoodEnergy from FoodSample Problem:Sample Problem:How much energy in kcal (= Cal) would be obtained from 2 Tbl peanut butter containing 6 g carb, 16 g fat, & 7 g protein?

How much energy in kcal (= Cal) would be obtained from 2 Tbl peanut butter containing 6 g carb, 16 g fat, & 7 g protein?

6 g carb6 g carb

16 g fat16 g fat

4 kcal

1 g carb

4 kcal

1 g carb

9 kcal

1 g fat

9 kcal

1 g fat

4 kcal

1 g protein

4 kcal

1 g protein7 g protein7 g protein

= 24 kcal= 24 kcal

= 144 kcal= 144 kcal

= 28 kcal= 28 kcal


= 200 Cal= 200 Cal

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A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 9.0 g of protein. How many kcal (Cal) does a cup of milk contain? (Round to the nearest 10 kcal.)


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A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 5.0 g of protein. How many kcal (Cal) does a cup of milk contain? (Round to the nearest 10 kcal.)

SolutionSolution

= 149 Cal= 149 Cal

12 g carb12 g carb

9 g fat9 g fat

4 kcal

1 g carb

4 kcal

1 g carb

9 kcal

1 g fat

9 kcal

1 g fat

4 kcal

1 g protein

4 kcal

1 g protein5 g protein5 g protein

= 48 kcal= 48 kcal

= 81 kcal= 81 kcal

= 20 kcal= 20 kcal


= 150 Cal= 150 Cal

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2.5Classification of Matter

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The stuffstuff things are made of.

Has MassMass and takes up space.

(Air, water, rocks, etc..)

• MatterMatter

The amountamount of stuff (in g’s) (Bowling Ball > Balloon)

WeightWeight on earth.

Pull of Gravity on matter.

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Classification of matterClassification of matter

MatterMatter

Pure Substance Mixture

Element Compound

FeFe FeSFeS

MgMg MgOMgO Mg + OMg + O22

Fe + SFe + S

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Pizza

MixtureMixture

HomogeneousHomogeneous

(Solution)(Solution)HeterogeneousHeterogeneous

Fe + SFe + S

MixturesMixtures

NNoonn--uunniiffoorrmm ccoommppoossiittiioonnUniform compositionUniform composition

Air

UrineGasoline

Sand

Tea w/ice

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Physical Separation of A MixturePhysical Separation of A MixtureMixtures can be separated• involves only physical changes– Like Filtering & distilling

Like when pasta and water are separated with a strainer

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Identify each of the following as a pure substance or a mixture.

A. pasta and tomato sauce

B. aluminum foil

C. helium

D. Air


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Identify each of the following as a pure substance or a mixture.

A. pasta and tomato sauce mixture

B. aluminum foil pure substance

C. helium pure substance

D. Air mixture

SolutionSolution

CH104 64

Identify each of the following as a homogeneous or heterogeneous mixture.

A. hot fudge sundae

B. shampoo

C. sugar water

D. peach pie


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Identify each of the following as a homogeneous or heterogeneous mixture.

A. hot fudge sundae heterogeneous

B. Shampoo homogeneous

C. sugar water homogeneous

D. peach pie heterogeneous

SolutionSolution

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2.6States and Properties

of Matter

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Elemental states at 25Elemental states at 25ooCC

He

Rn

XeI

KrBrSe

ArClS

NeFO

P

NC

H

Li

Na

Cs

Rb

K

TlHgAuHfLsBa

Fr

PtIrOsReWTa PoBiPb

Be

Mg

Sr

Ca

CdAgZrY PdRhRuTcMoNb

AcRa

ZnCuTiSc NiCoFeMnCrV

In SbSn

Ga Ge

Al

Gd

Cm

Tb

Bk

Sm

Pu

Eu

Am

Nd

U

Pm

Np

Ce

Th

Pr

Pa

Yb

No

Lu

Lr

Er

Fm

Tm

Md

Dy

Cf

Ho

Es

At

Te

As

Si

B

6 - 2

SolidLiquid

Gas

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DensityDensity

ShapeShape

CompressibilityCompressibility

ThermalThermalexpansionexpansion

Properties of matterProperties of matter

PropertyPhysical StatePhysical State

SolidSolid LiquidLiquid Gas Gas

High Highlike solids

Low

Fixed Shape ofcontainer

expandsto fillcontainer

Small Small Large

Verysmall

Small Moderate

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Three States of WaterThree States of Water

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•

Frozen HFrozen H22O:O: Slow moving moleculesSlow moving moleculesH-BondH-Bond in patterns in patterns

Hydrogen Bonding of Water Hydrogen Bonding of Water

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Learning CheckLearning CheckIdentify each as a S) solid, L) liquid, or G) gas.__ A. It has a definite volume but takes the

shape of the container.__ B. Its particles are moving very rapidly.__ C. It fills the volume of a container.__ D. It has particles in a fixed arrangement. __ E. It has particles that are close together

and are mobile.

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SolutionSolutionIdentify each as a S) solid, L) liquid, or GG) gas.L A. It has a definite volume but takes the

shape of the container.GG B. Its particles are moving very rapidly.G G C. It fills the volume of a container.S D. It has particles in a fixed arrangement. L E. It has particles that are close together

and are mobile.

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Physical PropertiesPhysical PropertiesPhysical properties• are observed or measured without changing the

identity of a substance• include shape and color• include melting point and boiling point

Physical Properties of CopperPhysical Properties of Copper

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Physical ChangePhysical ChangeIn a physical change,

• the identity and composition of the substance do not change

• the state can change or the material can be torn into smaller pieces

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Examples of Physical ChangeExamples of Physical Change

77

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Chemical PropertiesChemical PropertiesChemical properties

• describe the ability of a substance to change into a new substance

78

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Chemical ChangeChemical ChangeDuring a chemical

change, • reacting substances

form new substances with different compositions and properties

• a chemical reaction takes place

Iron

Fe

Iron (III) oxide

Fe2O3

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Examples of Chemical ChangeExamples of Chemical Change

80

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SummarySummary

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Classify each of the following as a 1) physical change or 2) chemical change.

A. ____ burning a candle

B. ____ ice melting on the street

C. ____ toasting a marshmallow

D. ____ cutting a pizza

E. ____ polishing a silver bowl


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Classify each of the following as a

1) physical change or 2) chemical change.

A. ____ burning a candle 2) chemical

B. ____ ice melting on the street 1) physical

C. ____ toasting a marshmallow 2) chemical

D. ____ cutting a pizza 1) physical

E. ____ polishing a silver bowl 2) chemical

SolutionSolution

CH104 84


2.7Changes of State

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Changes of StateChanges of State

Melting Pt = Melting Pt = Freezing PtFreezing Pt

Boiling PtBoiling Pt

SolidSolid

LiquidLiquid

VaporVapor

CondenseCondenseCondenseCondense

FreezeFreezeFreezeFreezeMeltMeltMeltMelt

VaporizeVaporizeVaporizeVaporize

Slow, close,Slow, close,Fixed Fixed

arrangementarrangement

Moderate, close,Moderate, close,Random Random

arrangementarrangement

Fast, far apart,Fast, far apart,RandomRandom

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SolidSolid

LiquidLiquid

VaporVapor

Changes of StateChanges of State

FrostFrostDepositDepositDepositDeposit

SublimeSublimeSublimeSublimeFreeze DryFreeze Dry

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Heating CurveHeating Curve

Specific Heat of IceSpecific Heat of Ice0.50 cal0.50 cal to heat ice to heat ice g g o o CC

Specific Heat of IceSpecific Heat of Ice0.50 cal0.50 cal to heat ice to heat ice g g o o CC

Heat of FusionHeat of Fusion80 cal80 cal to melt ice to melt ice gg

Heat of FusionHeat of Fusion80 cal80 cal to melt ice to melt ice gg

Specific Heat of HSpecific Heat of H22OO

1.00 cal1.00 cal to heat water to heat water g g o o CC

Specific Heat of HSpecific Heat of H22OO


Heat of VaporizationHeat of Vaporization540 cal540 cal to vaporize water to vaporize water gg

Heat of VaporizationHeat of Vaporization540 cal540 cal to vaporize water to vaporize water gg

Specific Heat of SteamSpecific Heat of Steam0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC

Specific Heat of SteamSpecific Heat of Steam0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC

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Heating CurveHeating Curve

0.50 cal0.50 cal to heat ice to heat ice g g o o CC

80 cal80 cal to melt ice to melt ice gg


540 cal540 cal to vaporize water to vaporize water gg

0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC

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Example:Example: Calculate the total amount of heat needed to change 500. g of ice at –10 oC into 500. g of steam at 120oC.

500g500g 10 10 o o CC 0.50 cal 0.50 cal g g o o CC

500g 500g 80 cal80 cal gg


500g500g 540 cal540 cal gg


== 2500 cal 2500 cal to heat iceto heat ice

== 40,000 cal 40,000 cal to melt iceto melt ice

= = 50,000 cal50,000 calto heat waterto heat water

= = 270,000 cal270,000 calto vaporize waterto vaporize water

= = 4800 cal4800 calto heat vaporto heat vapor

367,300 cal = 3.67 x 10367,300 cal = 3.67 x 1055 cal cal

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Learning CheckLearning CheckA. A plateau (horizontal line) on a heating

curve represents 1) a temperature change 2) a constant temperature 3) a change of state

B. A sloped line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state

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SolutionSolutionA. A plateau (horizontal line) on a heating

curve represents 1) a temperature change 2) a constant temperature 3) a change of state

B. A sloped line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state

CH104 92

Learning CheckLearning CheckUse the cooling curve for water to answer each.

A. Water condenses at a temperature of

1) 0 °C 2) 50 °C 3) 100 °C

B. At a temperature of 0 °C, liquid water

1) freezes 2) melts 3) changes to a gas

C. At 40 °C, water is a

1) solid 2) liquid 3) gas

D. When water freezes, heat is

1) removed 2) added

CH104 93

SolutionSolutionUse the cooling curve for water to answer each.

A. Water condenses at a temperature of

1) 0 °C 2) 50 °C 3) 100 °C

B. At a temperature of 0 °C, liquid water

1) freezes 2) melts 3) changes to a gas

C. At 40 °C, water is a

1) solid 2) liquid 3) gas

D. When water freezes, heat is

1) removed 2) added

CH104 94

Learning CheckLearning CheckTo reduce a fever, an infant is packed in 250 g of ice. If the ice (at 0 °C)

melts and warms to body temp (37.0 °C), how many calories are removed?

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SolutionSolutionTo reduce a fever, an infant is packed in 250 g of ice. If the ice (at 0 °C)

melts and warms to body temp (37.0 °C), how many calories are removed?

STEP 1

37.0 °C

0.0 °C

STEP 2250g 250g 80 cal80 cal gg


== 20,000 cal 20,000 cal to melt iceto melt ice

= = 9,250 cal9,250 calto heat waterto heat water

29,000 cal = 2.9 x 1029,000 cal = 2.9 x 1044 cal cal

Documents

CH104 CH104 Chapter 2: Energy & Matter Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves