1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 10 Structures of Solids and Liquids 10.5 Changes of State

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Melting and Freezing A substance is melting while it changes from a solid to a liquid is freezing while it changes from a liquid to a solid such as water has a freezing (melting) point of 0 °C

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Calculations Using Heat of Fusion The heat of fusion is the amount of heat released when 1 g of liquid freezes (at its freezing point) is the amount of heat needed to melt 1 g of solid (at its melting point) for water (at 0 °C) is 334 J or 80. cal 1 g H 2 O1 g H 2 O

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Calculations Using Heat of Fusion 4

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 How much heat in joules is needed to melt 15.0 g of ice (H 2 O)? STEP 1 List the grams of substance and change of state. Given 15.0 g of H 2 O(s) Need number of joules to melt ice to H 2 O(l) STEP 2 Write the plan to convert grams to heat. grams of ice H 2 O(s) joules (to melt) Example of a Calculation Using Heat of Fusion

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 STEP 3 Write the heat conversion factors and metric factors if needed. 1 g of H 2 O(s l) = 334 J 334 J and 1 g H 2 O g of H 2 O 334 J STEP 4 Set up the problem with factors. Heat to melt ice at 0 °C 15.0 g ice x 334 J = 5010 J 1 g ice Example of a Calculation Using Heat of Fusion (continued)

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 How many kilojoules are released when 25.0 g of water at 0 °C freezes? 1) 0.335 kJ 2) 0 kJ 3) 8.35 kJ Learning Check

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 STEP 1 List the grams of substance and change of state. Given 25.0 g of H 2 O(l) Need number of kilojoules to freeze to H 2 O(s) STEP 2 Write the plan to convert grams to heat. grams of H 2 O(l) kilojoules (to freeze) Solution

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 STEP 3 Write the heat conversion factors and metric factors if needed. 1 g of H 2 O(l s) = 334 J 334 J and 1 g H 2 O 1 g H 2 O 334 J 1 kJ = 1000 J 1 kJ and 1000 J 1000 J 1 kJ Solution (continued)

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 STEP 4 Set up the problem with factors. Heat to freeze water at 0 °C 25.0 g H 2 O x 334 J x 1 kJ = 8.35 kJ (3) 1 g H 2 O 1000 J Solution (continued)

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Sublimation occurs when a solid changes directly to a gas is typical of dry ice, which sublimes at -78  C takes place in frost-free refrigerators is used to prepare freeze- dried foods for long-term storage

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Evaporation and Condensation Water evaporates when molecules of water on the surface gain sufficient energy to form a gas condenses when water molecules in a gas lose energy and form a liquid

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Boiling of Water At boiling, all the water molecules acquire the energy to form a gas (vaporize) bubbles of water vapor appear throughout the liquid

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 Heat of Vaporization The heat of vaporization is the amount of heat absorbed to change 1 g of liquid to gas at the boiling point released when 1 g of gas changes to liquid at the boiling point boiling point of H 2 O = 100 °C heat of vaporization (water) 2260 J or 540 cal 1 g H 2 O

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Heats of Vaporization and Fusion of Polar and Nonpolar Compounds

16. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 Heats of Fusion and Vaporization

17. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Calculations Using Heat of Vaporization 17

18. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 18 Learning Check How many kilojoules (kJ) are released when 50.0 g of steam from a volcano condenses at 100 °C? 1) 113 kJ 2) 2260 kJ 3) 113 000 kJ

19. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 19 Solution STEP 1 List the grams of substance and change of state. Given 50.0 g of H 2 O(g) Need number of kilojoules to condense to H 2 O(l) STEP 2 Write the plan to convert grams to heat. grams of H 2 O(g) kilojoules (to condense)

20. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 20 Solution (continued) STEP 3 Write the heat conversion factors and metric factors if needed. 1 g of H 2 O(g l) = 2260 J 2260 J and 1 g H 2 O 1 g H 2 O 2260 J 1 kJ = 1000 J 1 kJ and 1000 J 1000 J 1 kJ

21. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 Solution (continued) STEP 4 Set up the problem with factors. Heat to condense H 2 O at 100 °C 50.0 g H 2 O x 2260 J x 1 kJ = 113 kJ (1) 1 g H 2 O 1000 J

22. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 22 Summary of Changes of State

23. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 23 Heating Curve A heating curve illustrates the changes of state as a solid is heated uses sloped lines to show an increase in temperature uses plateaus (horizontal lines) to indicate a change of state

24. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 24 A. A plateau (horizontal 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

25. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 25 A. A plateau (horizontal 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

26. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 26 Cooling Curve A cooling curve illustrates the changes of state as a gas is cooled uses sloped lines to indicate a decrease in temperature uses plateaus (horizontal lines) to indicate a change of state

27. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 27 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, liquid 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

28. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 28 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, liquid water 1) freezes C. At 40 °C, water is a 2) liquid D. When water freezes, heat is 1) removed Solution

29. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 29 To reduce a fever, an infant is packed in 250. g of ice H 2 O(s). If the ice (at 0 °C) melts and warms to body temperature (37.0 °C), how many joules are removed? Example of Combining Heat Calculations

30. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 30 STEP 1 1 List the grams of substance and change of state. Given 250. g of ice H 2 O(s); H 2 O(l) water at 37.0 °C Need joules to melt H 2 O(s) at 0 °C and warm to 37.0 °C Example of Combining Heat Calculations (continued)

31. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 31 STEP 2 Write the plan to convert grams to heat. Total heat = joules to melt ice at 0 °C and to warm liquid water from 0 °C to 37 °C. For several changes, we can draw a heating diagram. 37.0 °C temperature increase S L melting 0 °C Example of Combining Heat Calculations (continued)

32. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 32 Example of Combining Heat Calculations (continued) STEP 3 Write heat conversion factors and metric factors if needed. 1 g of H 2 O(s l) = 334 J 334 J and 1 g H 2 O 1 g H 2 O 334 J SH of H 2 O = 4.184 J/g °C 4.184 J and g °C g °C 4.184 J

33. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 33 Example of Combining Heat Calculations (continued) STEP 4 Set up problem with factors.  T = 37.0 °C – 0 °C = 37.0 °C Heat to melt the water at 0 °C 250. g H 2 O x 334 J = 83 500 J 1 g H 2 O Heat to warm the water from 0 °C to 37 °C 250. g x 37.0 °C x 4.184 J = 38 700 J g °C Total: 83 500 J + 38 700 J = 122 200 J

34. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 34 Learning Check When a volcano erupts, 175 g of steam at 100 °C is released. How many kilojoules are lost when the steam condenses, cools, freezes, and cools to -5 °C (SH of H 2 O(s) = 2.03 J/g °C)? 1) 396 kJ 2) 529 kJ 3) 133 kJ

35. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 35 Solution STEP 1 List the grams of substance and change of state. Given 175. g of steam H 2 O(g); to H 2 O(s) at -5 °C Need kilojoules to condense H 2 O(g) at 100 °C; cool to 0  C, freeze at 0 °C, and cool to -5 °C

36. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 36 Solution (continued) STEP 2 Write the plan to convert grams to heat. Total heat = joules to condense steam at 100 °C, cool water to 0 °C, freeze water at 0 °C, cool ice to -5 °C. For several changes, we can draw a cooling diagram. 100 °C (condensing) 100 °C to 0 °C (cooling) 0 °C (freezing) -5 °C (cooling)

37. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 3 Write heat conversion factors and metric factors if needed. 1 g of H 2 O(g l) = 2260 J 2260 J and 1 g H 2 O 1 g H 2 O 2260 J SH of H 2 O(l) = 4.184 J/g °C 4.184 J and g °C g °C 4.184 J 1 kJ = 1000 J 1 kJ and 1000 J 1000 J 1 kJ 37

38. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution (continued) STEP 4 Write heat conversion factors and metric factors if needed. 1 g of H 2 O(l s) = 334 J 334 J and 1 g H 2 O 1 g H 2 O 334 J SH of H 2 O(s) = 2.03 J/g °C 2.03 J and g °C g °C 2.03 J 38

39. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 39 Solution (continued) STEP 5 Set up problem with factors. Steam condenses at 100  C 175 g x 2260 J x 1 kJ = 396 kJ 1 g 1000 J Water cools from 100  C to 0  C 175 g x 100 °C x 4.184 J x 1 kJ = 73.2 kJ g °C 1000 J Water freezes to ice at 0  C: 175 g x 334 J x 1 kJ = 58.5 kJ g 1000 J Ice cools from 0  C to -5  C 175 g x 5 °C x 2.03 J x 1 kJ = 1.78 kJ g °C 1000 J 529 kJ (2)