The Foundations of Chemistry

Condensation is the reverse of evaporation. The amount of heat that must be removed

from a vapor to condense it (without change in temperature) is called the heat of conden-

sation.

evaporation

liquid
heat3:::::::::4vapor

condensation

The heat of condensation of a liquid is equal in magnitude to the heat of vaporization. It

is released by the vapor during condensation.

Because 2.26 kJ must be absorbed to vaporize one gram of water at 100°C, that same

amount of heat must be released to the environment when one gram of steam at 100°C

condenses to form liquid water at 100°C. In steam-heated radiators, steam condenses and

releases 2.26 kJ of heat per gram as its molecules collide with the cooler radiator walls

and condense there. The metallic walls conduct heat well. They transfer the heat to the

air in contact with the outside walls of the radiator. The heats of condensation and vapor-

ization of non–hydrogen-bonded liquids, such as benzene, have smaller magnitudes than

those of hydrogen-bonded liquids (see Table 13-5). They are therefore much less effec-

tive as heating and cooling agents.

EXAMPLE 13-1 Heat of Vaporization

Calculate the amount of heat, in joules, required to convert 180. grams of water at 10.0°C to

steam at 105.0°C.

Plan

The total amount of heat absorbed is the sum of the amounts required to (1) warm the liquid

water from 10.0°C to 100.0°C, (2) convert the liquid water to steam at 100.0°C, and (3) warm

the steam from 100.0°C to 105.0°C.

Because of the large amount of heat
released by steam as it condenses,
burns caused by steam at 100°C are
much more severe than burns caused
by liquid water at 100°C.

500 CHAPTER 13: Liquids and Solids

180. g H 2 O(
     )
     at 10.0 C
     180. g H 2 O(
     )
     at 100.0 C

Step 1:

warm the liquid

(temp. change)

Step 2:

boil the liquid

(phase change)

180. g H 2 O(g)
     at 100.0 C
     180. g H 2 O(g)
     at 105.0 C

Step 3:

warm the steam

(temp. change)

Steps 1 and 3 involve the specific heats of water and steam, 4.18 J/g °C and 2.03 J/g °C, respec-

tively (Appendix E), whereas step 2 involves the heat of vaporization of water (2.26 103 J/g).

Solution

1._?_J180. g

4

g

.

1

°

8

C

J

(100.0°C10.0°C)6.77 104 J0.677 105 J

2._?_J180. g

2.26

g

103 J

 4.07^105 J

3._?_J180. g

2

g

.

0

°

3

C

J

(105.0°C100.0°C)1.8 103 J0.018 105 J

Total amount of heat absorbed 4.76 105 J

You should now work Exercises 50 and 51.

Distillation is not an economical way to purify large quantities of water for public water

supplies. The high heat of vaporization of water makes it too expensive to vaporize large

volumes of water.

Steps 1 and 3 of this example involve
warming with nophase change. Such
calculations were introduced in Section
1-13.

Step 1: Temperature change only

Step 2: Phase change only

Step 3: Temperature change only