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Chapter 2 Energy and Matter
2.6 Changes of State
Copyright © 2009 by Pearson Education, Inc.
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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.
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Calculations Using 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
80. cal 1 g water
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The heat needed to freeze (or melt) a specific mass of water (or ice) is calculated using the heat of fusion.
Heat = g water (ice) x 80. cal 1 g water
(ice)
Example: How much heat in cal is needed to melt 15. g of ice?
15. g ice x 80. cal = 1200 cal
1 g ice
Calculation Using Heat of Fusion
Study Tip: Using Heat of F Fusion
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A. How many calories are needed to melt 5.0 g of ice at 0 °C?
1) 80. cal 2) 4.0 x 102 cal 3) 0 cal
B. How many calories are released when 25 g of water at 0 °C freezes?
1) 80. cal 2) 0 cal 3) 2.0 x 103 cal
Learning Check
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A. How many calories are needed to melt 5.00 g of ice at 0 °C?
2) 4.0 x 102 cal 5.0 g x 80. cal 1 g
B. How many calories are released when 25.0 g of water at 0 °C freezes?
3) 2.0 x 103 cal 25 g x 80. cal 1 g
Solution
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Sublimation
Sublimation• occurs when particles
change directly from solid 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.
Copyright © 2009 by Pearson Education, Inc.
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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.
Copyright © 2009 by Pearson Education, Inc.
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Boiling
At boiling,
• all the water molecules acquire enough energy to form a gas.
• bubbles appear throughout the liquid.
Copyright © 2009 by Pearson Education, Inc.
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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) = 540 cal
1 g water
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Learning Check
How many kilocalories (kcal) are released when 50.0 g of steam from a volcano condenses at 100 °C?
1) 27 kcal
2) 540 kcal
3) 2700 kcal
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Solution
How many kilocalories (kcal) are released when 50.0 gof 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
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Summary of Changes of State
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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 (flat lines) to indicate a change of state.
Copyright © 2009 by Pearson Education, Inc.
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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
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A. A flat line on a heating curve represents 2) a constant temperature. 3) a change of state.
(note that this is happening as energy is being put into the system. The energy is being used to change state instead of increase the temperature.)
B. A sloped line on a heating curve represents 1) a temperature change.
Solution
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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 (flat lines) to indicate a change of state.
Copyright © 2009 by Pearson Education, Inc.
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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.
Learning Check
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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
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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
temperature increase
0 °C solid liquid
melting
Combined Heat Calculations
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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 = 9 250 cal g °C
Total: Step 2 + Step 3 = 29 000 cal(rounded to 2 SF)
