Energy and Transformation
Aim 7– What are some of the different types of energy we encounter?
Energy in the US
• Energy is the capacity for doing work
• It exists in many forms, and sources of energy in the US come from specific substances and technologies
• The chart at the right represents a breakdown of energy sources in the United States as of 2010
• What percentage of the total resources are finite (will run out)?
• Energy may exist in various forms
• Kinetic energy
– Energy associated with motion
– Moving objects have kinetic energy
– There are different subforms of kinetic energy…
• Mechanical energy– Moving machine parts
transfers kinetic energy within the machine
– Example: a car’s engine and drive train
• Electrical Energy– Moving electrons
carry energy through electrical devices
– Example: power running through an iPhone or light bulb
• Electromagnetic radiation
– Energy carried by photons
• the elementary particles of radio waves, light, X-rays, gamma rays, etc.
– Longer the wavelength
– the lower the energy
• Potential energy– Energy associated with position– Often referred to as stored energy– There are different subforms of potential
energy
• Gravitational Potential Energy– Energy stored in
an object due to its position
– Example: – A rollercoaster
• Chemical Energy– Energy stored within a
chemical substance– released when a
substance goes through a chemical reaction
– Examples: – energy stored in:
• a candle’s wax• the chemical bonds of
dynamite• the chemical
reactions of a battery
• Nuclear Energy– Energy stored within
the nucleus of an atom
• Examples: • Nuclear weapons
– split atoms in uncontrolled reactions
• Fission power plants – split atoms in
controlled reactions
Examples of Energy Transformations
State what energies are being converted:
• Coal is burned to run a generator
chemical energy to electrical energy
• A rock is dropped off a building and hits a car
gravitational potential energy to kinetic energy
• A cell phone is plugged in to a socket and charged
electrical energy to chemical energy
• A cheeseburger is digested and gives a runner energy
chemical energy to mechanical (muscles/bones)
• Light waves allow a plant to do photosynthesis
electromagnetic energy to chemical energy (sugar)
• Mr. Foley set off a nuclear bomb in Hauppauge
nuclear energy to heat and light energy
Aim 8 – What is heat flow?
• Thermal energy – Thermal energy is a result of the kinetic
energy of the molecules’ motion – The total kinetic energy of all the molecules
combined is called thermal energy– The amount of thermal energy contained in the
particles depends on • how fast they are moving and • how many particles there are
– Thermal energy may be transferred from one system to another
• Heat – is a transfer of thermal energy – flows from a body of higher temperature to a body
of lower temperature
• Units of Heat flow • Joule (J)
– a base unit of metric heat measure
– 4.18 joules heat will raise one gram of water 1oC
• calorie (cal) – another base unit of heat measure– 1 calorie of heat will raise one gram of water 1oC
• Kilocalorie (kcal)– Is equal to 1,000 calories– 1 kilocalorie is equal to 1 nutritional Calorie(the
unit used to measure food intake)
• Thermal energy is NOT the same as temperature• Temperature is NOT a form of energy• Temperature
– (def) a measure of the average kinetic energy of the particles in an object and a system
– Measuring temperature allows us to calculate:• the heat in a system
or object;• The heat being
transferred between systems
• Temperature Scales • Three scales are used to
measure temperature– Fahrenheit (rarely
used in chem) and Celsius are based on the freezing and boiling points of water
– Kelvin is based on absolute zero, the absence of all particle motion
– therefore, to convert from Kelvin to Celsius we use:
Kelvin = oC +273
Practice ProblemsConvert each of the following using the
formula K = oC + 273
a) 273 K = _________ oC
b) 373 oC = _________ K
c) -100 K = _________ oC
d) 100 oC = _________ K
e) 5 K = _________ oC
f) 38 oC = _________ K
• Energy Conversions and the Law of Science– Theories that have been proven over and
over again to be valid– Examples: – Law of Gravity – all objects exert a gravitational
force in proportion to the sized of the object– Kepler’s Law of Planetary Motion – all planetary
bodies move in an elliptical orbit
• Chemistry has laws as well!• The first three are:
– The Law of Conservation of Energy– The Law of Conservation of Matter– The Law of Conservation of Energy and Matter
• The Law of Conservation of Energy– states that the total energy of an isolated
system cannot change– it is said to be conserved over time– Energy can be neither created nor destroyed,
but can change form– Example: – chemical energy
can be converted to kinetic energy in the explosion of a stick of dynamite
• The Law of Conservation of Matter– States that in chemical reactions, the total
mass of the reactants must equal the total mass of the reactants.
– Example 2 X + Y 2 X2Y
• In this reaction, 2 particles of X react with 1 particle of Y to make 2 particles of X2Y
• Nothing is lost when the reaction occurs• it is simply rearranged
– Matter cannot be neither created or destroyed, but may be changed from one form to another
J Deutsch 2003 19
The Law of Conservation of Matter and Energy
• states that neither matter or energy can neither be created nor destroyed, but may be converted into the other.
• This was explained in part by a very short but extremely famous formula, created by a gentleman by the name of Albert Einstein
• E = mc2, where “E” is energy, “m” is the mass equivalent, and “c2 “ is the speed of light squared
Phase Changes
Aim 9 – What happens when temperatures increase?
Phase Changes and Energy
• Changing energy causes changes in phase as attractive forces are broken or formed
• These energy changes and temperature changes can be represented on a graph
• This graph is called a phase change diagram
Phase Changes and Energy • Particles with temps above 0 Kelvin are
moving• Particles have
– kinetic energy as they move in place (solids) or away from each other (liquids and gases)
– potential energy due to attractive forces that hold solids and liquids together
• As temperature increases– The motion of the particles increases– Therefore kinetic energy increases– But the attractive forces are not being broken– So no phase change occurs
Phase Changes and Energy • Eventually, enough energy is added to
begin breaking the attractive forces• Temperature stops increasing
– Therefore kinetic energy stops going up• Eventually, the attractive forces holding the
particles to each other in the solid and liquid phases break– Potential energy increases– But particles don’t move any faster– So the average kinetic energy doesn’t
increase
• At the Melting Point
– the temperature a solid reaches the point where attractive forces break,
– melting it and forming a liquid state
• After the substance melts completely
– The temperature continues to go up till it reaches the
• Boiling Point
– the temperature a liquid reaches where all the remaining attractive forces are broken
– The liquid changes over to the gas state
Phase Diagrams - Heating Curves• As heat is added to a system• What types of changes occur?
– Particles move faster– Average kinetic energy increases– Temperature increases– Forces of attraction are broken as
• solids change to liquids
• liquids change to gases
• Because energy is added to break these forces of attraction– Stored energy increases– So potential energy increases
Heating Curve of Water
Phase Changes
Aim 10 – What happens when temperatures cool?
Phase Diagrams – Cooling Curve• As heat is REMOVED from a system• What types of changes occur?
– Particles start to slow down– Average kinetic energy decreases– Temperature goes down
• Forces of attraction are reformed as – gases change to liquids– liquids change to solids
• Forming attractions lowers energy – Stored energy decreases– So potential energy decreases
Phase Diagram - Cooling Curve
Energy changes during Phase changes
• Enthalpy – refers to heat being either lost or gained from any system
– Exothermic changes
• energy is lost from a system
• Example – a burning match
– Endothermic changes
• energy is gained from a system
• Example – a cold pack
• Entropy– Refers to the amount of disorder
particles have in a sample– The more disordered, the higher the entropy
• Solids – most organized, lowest entropy• Gases – least organized, greatest entropy
– As substances go thru phase changes, their entropy will either increase or decrease
• Question – you melt ice… what happens to the ice’s entropy?
J Deutsch 2003 32
Phases of Matter Changes:
Phase changeChange occurring (example)
Exothermic or endothermic?
Increase or decrease in enthalpy?
Fusion (melting)
Ice(s) water (l)
Endothermic Increase – less organized
Solidification (freezing)
Water (l) Ice (s)
Exothermic Decrease – more organized
Vaporization (boiling)
Water (l) Steam (g)
Endothermic Increase – less organized
CondensationSteam (g) Water (l)
Exothermic Decrease – more organized
SublimationDry Ice (s) Vapor (g)
Endothermic Increase – less organized
DepositionIodine (g) Iodine (s)
Exothermic Decrease – more organized
• Label the diagram with the following terms:boiling PE Solid onlymelting KE Liquid only
Gas only
Aim 11 – How much heat is absorbed or released during phase changes?
Calculating the Energy Changes during Phase Changes
• When a phase change occurs, heat is either – gained (heating occurs)– or lost (cooling occurs)
• In order to calculate the heat change in the sample or system, you need: – the mass of the sample undergoing the phase
change– and the change in heat constant for that
particular substance during a particular phase change
• A constant is a value that is the same for every situation
• There are two constants we use for phase changes– Heat of fusion – the amount of energy
gained or lost when a substance either melts or solidifies (freezes)
– Heat of vaporization – the amount of energy gained or lost when a substance either vaporizes (boils or evaporates) or condenses
• For example, for water– Heat of fusion (Hfusion ) = 334 joules/gram
– Heat of vaporization (Hvap ) = 2,260 joules/gram
– Joule – a unit of heat energy, similar to calories
• Knowing the mass of the sample, • and knowing what type of phase
change is occurring, • we can calculate the amount of heat (q)
gained or lost by the system using the formulas on Table T in the CRT
• Example: a 100.0 gram sample of ice is heated at 0oC and completely melted. How much heat (q) is used?
q = mass of the sample x Hfusion
q = 100.0 grams x 334 J/gq = 33,400 joules of heat energy
• Example 2: a student is distilling a 100.0 gram sample of water at 100oC. How much heat must be added to completely boil away the water?
q = mass of the sample x Hvaporization
q = 100.0 grams x 2260 J/g
q = 226,000 joules of heat energy
• Note – this is a huge number – how many kilojoules is this sample?
226,000 J x 1 kJ = 226 kJ
1,000 J
1. How much heat energy is required to freeze 5.00 grams of ice?
q = m x Hfusion
q = 5.00 grams x -334 J/gq = 1,670 J
2. How much heat energy is required to boil 20.00 g of water?
q = m x Hvaporization
q = 20.00 x 2260 J/gq = 45,200 J
3. How much ice will melt if 668,000 joules of energy are added to a block of ice?
q = m x Hfusion
668,000 J = m x 334 J/gm = 2,000 g
Aim 12 – how do we calculate the heat change in my cup of tea?
• When a phase changes occur– Heat is either gained or lost and the temperature
does not change– Particles are not moving faster or slower– Instead, attractive forces are being broken or formed– Potential energy changes, but not average kinetic energy
• When we are heating a solid, a liquid, or a gas and NOT causing a phase change– Heat is either gained or lost and the temperature IS
increasing or decreasing– The average kinetic energy is increasing or decreasing– Potential energy is staying the same (no attractive forces
formed or broken)– Heat is transferred to different materials at different rates
• Specific heat capacity (C) – the types of materials determine the rate at which heat will
be absorbed. – Each material has a specific heat capacity constant (C)
• Examples of specific heats– water = 4.18 J/g.oC– aluminum = 0.91 J/g.oC– iron = 0.45 J/g.oC– copper = 0.39 J/g.oC– J/g.oC is the number of joules of heat energy
needed per gram for a 1oC temperature change
• This is why when you touch the desk top, and the metal chair leg, they feel different temperatures!
• Demo – Ice Melting Blocks
• When temperature changes, the heat gained or lost (q) can be calculated if three things are known: The mass (m) of the sample The temperature change (T) of the
sample
(or, Tfinal – Tinitial) The specific heat capacity (C) of the
sample
• From Table T in the CRTs
q = m x T x C
1. A 100.0 gram sample of water has its temperature changed from 10.0oC to 30.0oC. If the specific heat of water (C) is 4.18 J/g.oC, what is the heat energy (q)?
q = m x T x C
q = 100.0 g x (30.0 – 10.0)oC x 4.18 J/g.oC
q = 100.0 g x 20.0oC x 4.18 J/g.oC
q = 8,360 J
Example 2 – A 20.0 gram sample of water cools from 40oC to 20oC. If the specific heat of water (C) is 4.18 J/g.oC, what is the heat energy (q) LOST by the water? How can you tell it was lost?
q = - 1,672 J
Example 3 – A 10.0 gram sample of aluminum has its temperature changed from 25oC to 30oC. If the specific heat of water (C) is aluminum is 0.91 J/g.oC, what is the heat energy (q) gained by the metal?
q = 45.5 J
Example – Mr. Foley’s cup of Earl Grey tea holds 200.0 g of tea (specific heat = 4.18 J/g.oC). First made, it has a temperature of 90oC. By lunch time, it is 30oC. What is the heat energy (q) lost by the tea?
q = - 50,160 J
• Calorimetry – is used to determine
the heat content or the heat transfer in a system
– the amount of temperature change that occurs in the water in the cup allows us to measure the heat
– note – it is a CALCULATED value
– At right – a simple calorimeter
• Calorimeter – is a heat measuring device. – is generally used to measure the amount
of heat energy– It then uses that to calculate the specific heat of
a substance or other heat related information
Example 4 – A macadamia nut burns in a calorimeter set up. A 50.0 gram sample of water has its temperature changed from 20oC to 30oC. If the specific heat of water (C) is 4.18 J/g.oC, what is the heat energy (q) gained by the water? How much heat was released by the macadamia nut?
q = m x T x Cq = 50.0 g x (30.0 – 20.0)oC x 4.18 J/g.oCq = 50.0 g x 20.0oC x 4.18 J/g.oCq = 4,180 J gained by the water, the same lost by the nut