1. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 5 Energy and States of Matter 5.6 Melting and Freezing 5.7 Boiling and Condensation

2. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 2 Changes of State

3. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 3 A substance is melting while it changes from a solid to a liquid. A substance is freezing while it changes from a liquid to a solid. The freezing (melting) point of water is 0°C. Melting and Freezing

4. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 4 The heat of fusion is the amount of heat released when 1 gram of liquid freezes at its freezing point. The heat of fusion is the amount of heat needed to melt 1 gram of a solid at its melting point. For water the heat of fusion (at 0°C) is 80. cal 1 g water Calculations Using Heat of Fusion

5. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 5 The heat involved in the freezing (or melting) a specific mass of water (or ice) is calculated using the heat of fusion. Heat = g water x 80. cal g water Problem: How much heat in calories is needed to melt 15.0 g of water? 15.0 g water x 80. cal = 1200 cal 1 g water Calculation Using Heat of Fusion

6. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 6 A. How many calories are needed to melt 5.0 g of ice of 0°C? 1) 80. cal2) 400 cal 3) 0 cal B. How many calories are released when 25 g of water at 0°C freezes? 1) 80. cal2) 0 cal 3) 2000 cal Learning Check

7. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 7 A. How many calories are needed to melt 5.0 g of ice of 0°C? 2) 400 cal5.0 g x 80. cal 1 g B. How many calories are released when 25 g of water at 0°C freezes? 3) 2000 cal 25 g x 80. cal 1 g Solution

8. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 8 Boiling Water evaporates when molecules on the surface gain enough energy to form a gas. At boiling, all the water molecules acquire enough energy to form a gas.

9. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 9 The heat of vaporization Is the amount of heat needed to change 1 g of liquid to gas at the boiling point. Is the amount of heat released when 1 g of a gas changes to liquid at the boiling point. Boiling (Condensing) Point of Water = 100°C Heat of Vaporization (water) = 540 cal 1 g water Heat of Vaporization

10. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 10 Learning Check How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C? 1) 27 kcal 2) 540 kcal 3) 27 000 kcal

11. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 11 Solution How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C? 1) 27 kcal 50.0 g steam x 540 cal x 1 kcal = 27 kcal 1 g steam 1000 cal

12. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 12 Chapter 5 Energy and States of Matter 5.8 Heating and Cooling Curves

13. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 13 Heating Curve A heating curve illustrates the changes of state as a solid is heated. Sloped lines indicate an increase in temperature. Plateaus (flat lines) indicate a change of state.

14. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 14 A. A flat 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 Learning Check

15. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 15 A. A flat line on a heating curve represents 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change Solution

16. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 16 Cooling Curve  A cooling curve illustrates the changes of state as a gas is cooled.  Sloped lines indicate a decrease in temperature.  This cooling curve for water begins at 140°C and ends at -30°C.

17. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 17 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0°C2) 50°C3) 100°C B. At a temperature of 0°C, water 1) freezes2) melts3) changes to a gas C. At 40 °C, water is a 1) solid 2) liquid3) gas D. When water freezes, heat is 1) removed2) added Learning Check

18. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 18 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 3) 100°C B. At a temperature of 0°C, water 1) freezes C. At 40 °C, water is a 2) liquid D. When water freezes, heat is 1) removed Solution

19. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 19 To reduce a fever, an infant is packed in 250 g of ice. If the ice at 0°C melts and warms to body temperature (37.0°C), how many calories are removed from the body? Step 1: Diagram the changes. 37°C  T = 37.0°C – 0°C = 37.0°C 0°C S L Combined Heat Calculations

20. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 20 Combined Heat Calculations (continued) Step 2: Calculate the heat to melt ice (fusion) 250 g ice x 80. cal= 20 000 cal 1 g ice Step 3: Calculate the heat to warm the water from 0°C to 37.0°C 250 g x 37.0°C x 1.00 cal = 9250 cal g °C Total: Step 2 + Step 3 = 29 250 cal Final answer= 29 000 cal

21. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 21 Learning Check When a volcano erupts, 150 g of steam at 100°C is released. How many kilocalories are lost when the steam condenses and cools to 15°C? 1) 81 kcal 2) 13 kcal 3) 94 kcal

22. Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 22 Solution 3) 94 kcal Condense: 150 g x 540 cal x 1 kcal = 81 kcal 1 g 1000 cal Cool: 150 g x 85°C x 1 cal x 1 kcal = 13 kcal g °C 1000 cal Total: 81 kcal + 13 kcal = 94 kcal