1. Changes in State
Pix:

2. Warming Curve Consider what happens to an ice cube which is heated.
boiling
Time ->
Temperature ->
warm gas
warm solid
warm liquid
melting
Pix: dclips.fundraw.com

3. Within One State
warm solid
warm liquid
warm gas
Time ->
Temperature ->
Rising temperature indicates a change in kinetic energy: molecules are moving faster.
This is measured by the specific heat (heat capacity): the calories needed to raise the temperature of 1 gram of a substance by 1°C.

4. Changing States
melt
boil
Time ->
Temperature ->
A temperature plateau indicates a change in potential energy: bonds between molecules are changing, and energy can be stored in them.
This is measured by the heat of vaporization (gas/liquid) or heat of fusion (solid/liquid): the calories needed to change the state of 1 gram.

5. Time ->
Temperature -> 1 2 3 4 5
Cooling Curve
Can you identify what’s happening at each stage of this graph?
1= cooling the gas; 2 = condensing; 3 = cooling the liquid; 4 = freezing; 5 = cooling the solid
1 = gas; 2 = gas + liquid; 3 = liquid; 4 = liquid + solid; 5 = solid
KE is affected in stages 1,3,5; PE is affected in stages 2,4
Can you identify what states are present at each stage?
When is kinetic energy affected? When is potential energy affected? Answers are in the notes.

6. Entropy crystalline solid highly ordered minimum entropy liquid
Ice:
Water:
Steam:
crystalline solid
highly ordered
minimum entropy
liquid
some order
some entropy
gas
very random
maximum entropy

7. Calculations with Specific Heat
1. How many calories are needed to warm up 3.45 g of water from 18.5°C to 60.0°C?
The specific heat of water is 1 cal/g°C.
Heat = (1 cal/g°C)(3.45 g)(41.5°C) = 143 cal
2. How many calories are removed (released) when kg of H2O are cooled from 98.5°C to 20.0°C?
Heat = (1 cal/g°C)(1,290 g)(78.5°C) = 101,000 cal
Technically, this is negative because heat is removed.

8. More Calculations with Specific Heat
3. What is the final temperature when 20.0 cal of heat energy are added to g of iron (specific heat cal/g°C) at 19°C?
(sp. heat) *(T) *(mass) = heat
( cal/g°C)*(T)*(0.235 g) = 20.0 cal
T = 667 °C
Heat is being added, so Tfinal = 19°C °C
Tfinal = 686 °C

9. Heats of Vaporization & Fusion
1. How much heat must be released to freeze 25.0 g of water? The heat of fusion of water is 1.14 kcal/mole.
25.0 g x (1 mole/18.0 g) x (1.14 kcal/1 mole)
= 1.77 kcal
2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 24.5 cal/g.
125.0 g x x (24.5 cal/1 g) = 3060 cal

10. Putting it together How much heat energy must be removed from
Heat of fusion H2O = 1.14 kcal/mol
Sp. Heat of Ice = 0.50 cal/g°C
Putting it together
How much heat energy must be removed from
35.0 g of water at 80.°C to turn it into ice at -25 °C?
FIRST: Identify what parts of the heating or cooling diagram you need to work with.
Time ->
Temperature ->
 starts at 80 °C; liquid
 ends at -25 °C; solid
Second: This tells you there are 3 stages to consider: (a) cooling the liquid, (b) freezing, (c) cooling the solid.

11. Putting it together How much heat energy must be removed from
Heat of fusion H2O = 1.14 kcal/mol
Sp. Heat of Ice = 0.50 cal/g°C
Putting it together
How much heat energy must be removed from
35.0 g of water at 80.°C to turn it into ice at -25 °C?
First, cool the water to its freezing point.
Heat = (1 cal/g°C)(35.0 g)( °C) = cal
Second, freeze the water into ice.
(35.0 g)(1 mole/18.0 g)(1.14 kcal/1 mole) = 2.22 kcal
Third, cool the ice to its final temperature.
Heat = (0.50 cal/g°C)(35.0 g)(0-[-25°C]) = cal
Last, add all the heats together. Make sure units agree!
2800 cal cal cal = 5460 cal