chapter 10 structures of solids and liquids

Basic Chemistry Copyright © 2011 Pearson Education, Inc.1

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


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

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Calculations Using Heat of Fusion

4


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


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)


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



Given 25.0 g of H2O(l)

Need number of kilojoules to freeze to H2O(s)


grams of H2O(l) kilojoules (to freeze)

Solution



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)



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



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


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


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


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


Heats of Vaporization and Fusion of Polar and Nonpolar Compounds


Heats of Fusion and Vaporization


Calculations Using Heat of Vaporization

17


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


Solution


Given 50.0 g of H2O(g)

Need number of kilojoules to condense to H2O(l)


grams of H2O(g) kilojoules (to condense)




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




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


Summary of Changes of State


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


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


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


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


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


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


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


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)


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)


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


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


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


Solution


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




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)


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

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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)

chapter 10 structures of solids and liquids

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