1. Chapter 2 Energy and Matter
2.7 Changes of State

2. Melting and Freezing A substance
is melting when it changes from a solid to a liquid
is freezing when it changes from a liquid to a solid
such as water has a freezing (melting) point of 0 °C

3. Heat of Fusion The heat of fusion
is the amount of heat released when 1 gram of liquid freezes (at its freezing point)
is the amount of heat needed to melt 1 gram of a solid (at its melting point)
for water (at 0 °C) is
334 J or 80 cal
1 g of water

4. Guide to Calculations Using Heat of Fusion (or Vaporization)

5. Calculations Using Heat of Fusion
How much heat in calories is needed to melt 15.0 g of
ice at 0 °C ?
STEP 1 Given: 15.0 g of water(s)
change of state: melting at 0 °C
STEP 2 Plan: g of water(s) g of water(l)
STEP 3 Write conversion factors:
1 g of water = cal
1 g of water and cal
80 cal g of water

6. Calculations Using Heat of Fusion (continued)
STEP 4 Set up the problem to calculate calories:
15.0 g water x cal = cal
1 g water

7. Learning Check How many joules are released when 25.0 g of water
at 0 °C freezes?
1) 334 J
2) J
3) J

8. Solution 3) 8350 J STEP 1 Given: 25.0 g of water(l)
change of state: freezing at 0 °C
STEP 2 Plan: g of water g of water
STEP 3 Write conversion factors:
1 g of water = 334 J
1 g of water and __334 J___
334 cal 1 g of water

9. Solution (continued) STEP 4 Set up the problem to calculate joules:
25.0 g water x J = J (3)
1 g water

10. 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

11. Evaporation and Condensation
Water
evaporates when molecules on the surface gain sufficient energy to form a gas
condenses when gas molecules lose energy and form a liquid

12. Boiling At boiling, water molecules acquire the energy to form a gas
bubbles appear throughout the liquid

13. Heat of Vaporization The heat of vaporization is the amount of heat
absorbed to vaporize 1 g of a liquid to gas at the boiling point
released when 1 g of a gas condenses to liquid at the boiling point
Boiling Point of Water = 100 °C
Heat of Vaporization (water) = J or 540 cal
1 g of water

14. Comparing Heat of Fusion and Heat of Vaporization

15. Learning Check How many kilocalories (kcal) are released when 50.0 g
of water(g) as steam from a volcano condenses at
100 °C?
1) 27 kcal
2) 540 kcal
3) kcal

16. Solution 1) 27 kcal STEP 1 50.0 g of water(g) = 50.0 g of water(l)
Change of state: water condensing at 100 °C
STEP 2 Plan: g of water(g) g of water(l)
STEP 3 Write conversion factors:
1 g of water = 540 cal
1 g of water and cal
540 cal g of water
1 kcal = cal
1 kcal and cal
1000 cal kcal

17. Solution (continued)
STEP 4 Set up the problem to calculate kilocalories:
50.0 g water x cal x 1 kcal = 27 kcal
1 g water cal

18. Summary of Changes of State

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. Combined Heat Calculations
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?
STEP 1 Given: 250 g ice changes to water at 37.0 °C
Need: calories to melt at 0 °C and warm to 37.0 °C
STEP 2  T = 37.0 °C – 0 °C = 37.0 °C
37.0 °C
0 °C

26. Combined Heat Calculations (continued)
STEP 3 Conversion factors:
1 g of water (0 °C) = 80 cal
1 g of water (0 °C) and cal
80 cal g of water (0 °C)
1 g of water(l) = cal/g °C
1 g of water(l) and cal/g °C
1.00 cal/g °C g of water(l)

27. Combined Heat Calculations (continued)
STEP 4 Set up problem:
(1) Heat to melt ice (fusion)
250 g ice x cal = cal
1 g ice
(2) Heat the water(l) from 0 °C to 37.0 °C
250 g x °C x cal = cal g °C
Total: cal cal = cal
(rounded)