1. Conservation of Energy
Like mass, energy cannot be created or destroyed, just transferred from one place to another within the system
The most common energy transfer is in the form of heat
Transfer of heat is measured as a temperature change
Temperature is a measure of the average kinetic energy of the particles in the substance

2. Measuring Heat SI unit for heat is Joules Specific heat
4.184 Joules = 1 calorie
Calorie is the amount of heat it takes to raise 1 gram of water by 1 degree Celsius
Specific heat
The amount of heat necessary to raise one gram of a substance by one degree Celsius
Unique for each substance
The specific heat of water is 1 calorie/gºC
Also J/ gºC

3. Kinetic Theory As the energy of the particles in a substance increases
Particles move faster
Density decreases
As the energy of the particles in a substance decreases
Particles move more slowly
Density increases

4. Calculating Heat Heat is calculated using the heat equation
Q = (m) (ΔT) (cp)
Q = heat
M = mass
ΔT = change in temperature
Tf – Ti (final temp – initial temp)
Cp = specific heat

5. Example Problem
How much heat is lost when a solid aluminum ingot with a mass of 4110 g cools from C to 25.0 C?
q=(m) (T) (Cp)
Mass is 4110g
Initial temp is C
Final temp is 25.0 C
Specific heat of aluminum is J/gC
Q = (4110) ( 660C – 25.0 C ) ( J/gC)
Q = (4110) ( 635C) ( J/gC)
Q = 2.35 x 106

6. Energy and Change of State
If the change in kinetic energy is large enough this will result in a change of state
As the amount of energy changes, the temperature will increase or decrease until the boiling (heat increase) or freezing (heat loss) point is reached
At the freezing or boiling point two phases of matter can exist at the same temperature

7. To make the change from one phase to another more energy will be absorbed (boiling or melting) or lost (condensing or freezing) without a corresponding change in temperature
This is because this energy is used merely to overcome the intermolecular forces maintaining one state and move to the new state

8. Heat of vaporization Hv
Heat necessary to change from liquid to gas or from gas to liquid at the boiling/condensation point
Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and gas

9. Heat of fusion Hf
Heat necessary to change from liquid to solid or from solid to liquid at the freezing/melting point
Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and solid

10. Phase change graph

11. Slanted sections of the graph represent change in kinetic energy M ΔTCp
Flat portions represent the change in potential energy necessary to change from one phase or state of matter to another – Hf or Hv; look at it again

12. Phase change graph

13. If one were to begin with a cold solid, and add heat, each step as heat is added must be calculated
The solid warms up to the melting point
M ΔTCp (using the specific heat for the solid)
At the melting point, some liquid will begin to form
the heat necessary to melt it completely is calculated by mass of the solid times the heat of fusion

14. The liquid increases temperature until reaching the boiling point
calculated by M ΔTCp (using the specific heat for the liquid)
At the boiling point gas will begin to form
the heat necessary to vaporize it completely is calculated by mass of the solid times the heat of vaporization

15. Vapor may continue to increase in temperature
calculated by M ΔTCp (using the specific heat for the gas)
The sum of all of these heats (from initial temperature to final temperature) is used to calculate the heat gained or lost by the substance

16. If a 26. 5g block of ice at –15. 0 ºC is heated to water at 20
If a 26.5g block of ice at –15.0 ºC is heated to water at 20.0 ºC, how much heat is used?
Ist heat the ice to its melting point
M ΔTCp
(26.5g)(0.0C – (-15.0C))(0.53cal/gºC)
210.7 cal

17. Now melt the ice Now warm the water M(Hf) 26.5g (80 cal/g) 2120 cal
M ΔTCp
26.5g (20 ºC - 0 ºC)(1.00cal/g ºC)
530 cal

18. Now add up the calories 210.7 cal + 2120 cal + 530 cal 2860.7 cal
2860 cal (2 sig fig.)
Now try one yourself

19. How much heat is needed to heat 117g of water at 42 ºC to steam at 136 ºC

20. 117g (100 ºC - 42 ºC)(1.00 cal/g ºC)
6786 cal
117g (540 cal/g)
63180 cal
117(136 ºC ºC)(0.480 cal/g ºC)
cal
6786 cal cal cal
71988 cal
7.20 x 104 cal

21. How much heat does it take to raise the temp of 112g of ice at –4
How much heat does it take to raise the temp of 112g of ice at –4.0 ºC to 120 ºC?