chapter 10 structures of solids and liquids
DESCRIPTION
Chapter 10 Structures of Solids and Liquids. 10.5 Changes of State. 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. - PowerPoint PPT PresentationTRANSCRIPT
Basic Chemistry Copyright © 2011 Pearson Education, Inc.1
Chapter 10 Structures of Solids and Liquids
10.5 Changes of State
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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 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
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Calculations Using Heat of Fusion
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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 2 Write the plan to convert grams to heat.
grams of ice H2O(s) joules (to melt)
Example of a Calculation Using Heat of Fusion
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STEP 3 Write 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 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)
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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
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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 2 Write the plan to convert grams to heat.
grams of H2O(l) kilojoules (to freeze)
Solution
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STEP 3 Write 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
Solution (continued)
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STEP 4 Set 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
Solution (continued)
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SublimationSublimation• 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
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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
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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
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Heat of VaporizationThe 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 H2O = 100 °Cheat of vaporization (water)
2260 J or 540 cal1 g H2O 1 g H2O
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Heats of Vaporization and Fusion of Polar and Nonpolar Compounds
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Heats of Fusion and Vaporization
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Calculations Using Heat of Vaporization
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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
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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 2 Write the plan to convert grams to heat.
grams of H2O(g) kilojoules (to condense)
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Solution (continued)
STEP 3 Write 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
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Solution (continued)
STEP 4 Set 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
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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 (horizontal lines) to indicate a change of state
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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
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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
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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 (horizontal lines) to indicate a change of state
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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 °CB. At a temperature of 0 °C, liquid water
1) freezesC. At 40 °C, water is a
2) liquidD. 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 H2O(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
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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
Example of Combining Heat Calculations (continued)
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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)
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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
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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 °C250. g H2O x 334 J = 83 500 J
1 g H2O
Heat to warm the water from 0 °C to 37 °C250. g x 37.0 °C x 4.184 J = 38 700 J
g °C Total: 83 500 J + 38 700 J = 122 200 J
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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
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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
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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)
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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
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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
38
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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 JWater 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 C175 g x 5 °C x 2.03 J x 1 kJ = 1.78 kJ
g °C 1000 J 529 kJ (2)