1. Changes in State

2. Warming Curve warming the solid melting boiling warming the liquid warming the gas Time -> Temperature -> Consider what happens to an ice cube which is heated.

3. Within One State warm solid warm liquid warm gas Time -> Temperature -> 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. Rising temperature indicates a change in kinetic energy: molecules are moving faster.

4. Changing States melt boil Time -> Temperature -> 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. A temperature plateau indicates a change in potential energy: bonds between molecules are changing, and energy can be stored in them.

5. SUMMARY warm solid melt boil warm liquid warm gas Time -> Temperature -> sp. heat solid sp. heat liquid sp. heat gas heat of fusion heat of vaporatization

6. Energy Amount cal or kcalJoules or kJgrams or kg moles You will see a variety of units for both energy and amount of material in these problems.

7.  W happening at each stage of this graph? Time -> Temperature -> 1 2 3 4 5  When is kinetic energy affected? When is potential energy affected? Answers are in the notes.  What states are present at each stage?  What property describes each stage?

8. 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 1.29 kg of H 2 O 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. We will not be using this convention, leaving all energy without signs and using works to decribe addition/loss of energy.

9. More Calculations with Specific Heat 3. What is the final temperature when 85.0 J of heat energy are added to 23.5 g of iron (specific heat 0.125 J/g°C) at 19°C? (0.125 J/g°C)* (23.5 g) * (  T) = 85.0 J  T = 28.9 °C Heat is being added, so T final = 19°C + 28.9°C T final = 47.9 °C Note: If the problem stated that heat was being removed the iron, then  T would still be 28.9 °C but the final temperature would be -9.9 °C. Do you see why?

10. 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. (1.14 kcal/1 mole) x [25.0 g x (1 mole/18.0 g)] = 1.77 kcal 2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 102.4 J/g. (102.4J/1 g) x 125 g = 12,800 J 1.14 kcal = x kcal alternative proportional solution 18.0 g 25.0 g x = 1.77 kcal 102.4 J = x J alternative proportional solution 1 g 125 g x = 12,800 J

11. How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? Heat of fusion H 2 O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C FIRST: Sketch a heating or cooling curve and mark the plateaus and your starting/ending points. This tells you there are 3 stages to consider: (a) cooling the liquid, (b) freezing, (c) cooling the solid. Time -> Temperature ->  starts at 80 °C; liquid  ends at -25 °C; solid 100 °C 0 °C

12. How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? Heat of fusion H 2 O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C Second: Find the properties that control each stage of the problem. (a)cooling the liquid = specific heat of liquid: 1 cal/g°C (b) freezing = heat of fusion: 1.14 kcal/mol (c) cooling the solid = specific heat of solid: 0.50 cal/g°C The units of each property show you how to do the math!

13. How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? 1. Cool the water to its freezing point. (1 cal/g°C)(35.0 g)(80.-0.0°C) = 2800 cal Heat of fusion H 2 O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C 2. Freeze the water into ice. (1.14 kcal/1 mole) (35.0 g)(1 mole/18.0 g) = 2.22 kcal 3. Cool the ice to its final temperature. (0.50 cal/g°C)(35.0 g)(0-[-25°C]) = 440 cal Last, add all the heats together. Make sure units agree! 2800 cal + 2220 cal + 440 cal = 5460 cal

14. crystalline solid highly ordered minimum entropy liquid some order some entropy gas very random maximum entropy

15. cal/g°C J/g°C cal/g, cal/mole kcal/g, kcal/mole J/g, J/mole