Chapter 5
Energy and States of Matter
Changes of State
5.6 Melting and Freezing
5.7 Boiling and Condensation
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Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Melting and Freezing
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Calculations Using Heat of Fusion
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.
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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
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Calculation Using Heat of Fusion
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Learning Check
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
A. How many calories are needed to melt 5.0 g of ice
of 0°C?
1) 80. cal
2) 400 cal
3) 0 cal
80. cal
g water
Problem: How much heat in calories is
needed to melt 15.0 g of water?
15.0 g water x
x
B. How many calories are released when 25 g of
water at 0°C freezes?
1) 80. cal
2) 0 cal
3) 2000 cal
80. cal
= 1200 cal
1 g water
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Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Solution
Boiling
A. How many calories are needed to melt 5.0 g of ice
of 0°C?
2) 400 cal
5.0 g x 80. cal
1g
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B. How many calories are released when 25 g of
water at 0°C freezes?
3) 2000 cal
25 g x 80. cal
1g
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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.
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Heat of Vaporization
Learning Check
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.
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
Boiling (Condensing) Point of Water = 100°C
Heat of Vaporization (water) =
540 cal
1 g water
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Chapter 5
Energy and States of Matter
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
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Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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5.8 Heating and Cooling Curves
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Heating Curve
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Learning Check
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.
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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
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Solution
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Cooling Curve
A. A flat line on a heating curve represents
2) a constant temperature
3) a change of state
 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.
B. A sloped line on a heating curve represents
1) a temperature change
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Learning Check
Solution
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, 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
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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, water
1) freezes
C. At 40 °C, water is a
2) liquid
D. When water freezes, heat is
1) removed
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Combined Heat Calculations
(continued)
Combined Heat Calculations
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
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
ΔT = 37.0°C – 0°C =
37.0°C
0°C
S
L
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Learning Check
Solution
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
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3) 94 kcal
Condense: 150 g x 540 cal x 1 kcal = 81 kcal
1g
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
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