Chapter 10 Structures of Solids and Liquids

1. Chapter 10 Structures of Solids and Liquids 10.5 Changes of State

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. 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 H2O 1 g H2O

4. Calculations Using Heat of Fusion

5. Example of a Calculation Using Heat of Fusion How much heat in joules is needed to melt 15.0 g of ice (H2O)? STEP 1 List the grams of substance and change of state. Given 15.0 g of H2O(s) Need number of joules to melt ice to H2O(l) STEP 2Write the plan to convert grams to heat. grams of ice H2O(s) joules (to melt)

6. Example of a Calculation Using Heat of Fusion (continued) STEP 3Write the heat conversion factors and metric factors if needed. 1 g of H2O(s l) = 334 J 334 J and 1 g H2O g of H2O 334 J STEP 4Set up the problem with factors. Heat to melt ice at 0 °C 15.0 g ice x 334 J = 5010 J 1 g ice

7. Learning Check 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

8. Solution STEP 1 List the grams of substance and change of state. Given 25.0 g of H2O(l) Need number of kilojoules to freeze to H2O(s) STEP 2Write the plan to convert grams to heat. grams of H2O(l) kilojoules (to freeze)

9. Solution (continued) STEP 3Write the heat conversion factors and metric factors if needed. 1 g of H2O(l s) = 334 J 334 J and 1 g H2O 1 g H2O 334 J 1 kJ = 1000 J 1 kJ and 1000 J 1000 J 1 kJ

10. Solution (continued) STEP 4Set up the problem with factors. Heat to freeze water at 0 °C 25.0 g H2O x 334 J x 1 kJ = 8.35 kJ (3) 1 g H2O 1000 J

11. Sublimation 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. Evaporation and Condensation Water • evaporateswhen molecules of water on the surface gain sufficient energy to form a gas • condenseswhen water molecules in a gas lose energy and form a liquid

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. Heat of Vaporization The heat of vaporizationis 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 H2O = 100 °C heat of vaporization (water) 2260 J or 540 cal 1 g H2O 1 g H2O

15. Heats of Vaporization and Fusion of Polar and Nonpolar Compounds

16. Heats of Fusion and Vaporization

17. Calculations Using Heat of Vaporization

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. Solution STEP 1 List the grams of substance and change of state. Given 50.0 g of H2O(g) Need number of kilojoules to condense to H2O(l) STEP 2Write the plan to convert grams to heat. grams of H2O(g) kilojoules (to condense)

20. Solution (continued) STEP 3Write the heat conversion factors and metric factors if needed. 1 g of H2O(g l) = 2260 J 2260 J and 1 g H2O 1 g H2O 2260 J 1 kJ = 1000 J 1 kJ and 1000 J 1000 J 1 kJ

21. Solution (continued) STEP 4Set up the problem with factors. Heat to condense H2O at 100 °C 50.0 g H2O x 2260 J x 1 kJ = 113 kJ (1) 1 g H2O 1000 J

22. Summary of Changes of State

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. Learning Check 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

25. Solution 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

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. Learning Check Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0 °C 2) 50 °C 3) 100 °C B. At a temperature of 0 °C, liquid water 1) freezes 2) melts 3) changes to a gas C. At 40 °C, water is a 1) solid 2) liquid 3) gas D. When water freezes, heat is 1) removed 2) added

28. Solution 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

29. Example of Combining Heat Calculations To reduce a fever, an infant is packed in 250. g of ice H2O(s). If the ice (at 0 °C) melts and warms to body temperature (37.0 °C), how many joules are removed?

30. Example of Combining Heat Calculations (continued) STEP 1 1 List the grams of substance and change of state. Given 250. g of ice H2O(s); H2O(l) water at 37.0 °C Need joules to melt H2O(s) at 0 °C and warm to 37.0 °C

31. Example of Combining Heat Calculations (continued) STEP 2Write 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

32. Example of Combining Heat Calculations (continued) STEP 3 Write heat conversion factors and metric factors if needed. 1 g of H2O(s l) = 334 J 334 J and 1 g H2O 1 g H2O 334 J SH of H2O = 4.184 J/g °C 4.184 J and g °C g °C 4.184 J

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 H2O x 334 J = 83 500 J 1 g H2O 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. 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 H2O(s) = 2.03 J/g °C)? 1) 396 kJ 2) 529 kJ 3) 133 kJ

35. Solution STEP 1 List the grams of substance and change of state. Given 175. g of steam H2O(g); to H2O(s) at -5 °C Need kilojoules to condense H2O(g) at 100 °C; cool to 0 C, freeze at 0 °C, and cool to -5 °C

36. Solution (continued) STEP 2Write 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. Solution (continued) STEP 3 Write heat conversion factors and metric factors if needed. 1 g of H2O(g l) = 2260 J 2260 J and 1 g H2O 1 g H2O 2260 J SH of H2O(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

38. Solution (continued) STEP 4 Write heat conversion factors and metric factors if needed. 1 g of H2O(l s) = 334 J 334 J and 1 g H2O 1 g H2O 334 J SH of H2O(s) = 2.03 J/g °C 2.03 J and g °C g °C 2.03 J

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)