The Foundations of Chemistry

EXAMPLE 13-2 Heat of Vaporization

Compare the amount of “cooling” experienced by an individual who drinks 400. mL of ice
water (0.0°C) with the amount of “cooling” experienced by an individual who “sweats out”

400. mL of water. Assume that the sweat is essentially pure water and that all of it evaporates.
     The density of water is very nearly 1.00 g/mL at both 0.0°C and 37.0°C, average body temper-
     ature. The heat of vaporization of water is 2.41 kJ/g at 37.0°C.

Plan

In the case of drinking ice water, the body is cooled by the amount of heat required to raise
the temperature of 400. mL (400. g) of water from 0.0°C to 37.0°C. The amount of heat lost
by perspiration is equal to the amount of heat required to vaporize 400. g of water at 37.0°C.

Solution

Raising the temperature of 400. g of water from 0.0°C to 37.0°C requires

_?_J(400. g)(4.18 J/g °C)(37.0°C)6.19 104 J, or 61.9 kJ

Evaporating (i.e., “sweating out”) 400. mL of water at 37°C requires

_?_J(400. g)(2.41 103 J/g)9.64 105 J, or 964 kJ

Thus, we see that “sweating out” 400. mL of water removes 964 kJ of heat from one’s body,
whereas drinking 400. mL of ice water cools it by only 61.9 kJ. Stated differently, sweating
removes (964/61.9)15.6 times more heat than drinking ice water!

You should now work Exercise 55.

For health reasons, it is important

to replace the water lost by

perspiration.

Problem-Solving Tip:Temperature Change or Phase Change?

A problem such as Example 13-1 can be broken down into steps so that each involves

eithera temperature change ora phase change, but not both. A temperature change calcu-

lation uses the specific heat of the substance (steps 1 and 3 of Example 13-1); remember

that each different phase has its own specific heat. A phase change always takes place

with no changein temperature, so that calculation does not involve temperature (step 2

of Example 13-1).

The Clausius–Clapeyron Equation

We have seen (Figure 13-13) that vapor pressure increases with increasing temperature. Let
us now discuss the quantitative expression of this relationship.
When the temperature of a liquid is changed from T 1 to T 2 , the vapor pressure of the
liquid changes from P 1 to P 2. These changes are related to the molar heat of vaporization,
Hvap, for the liquid by the Clausius–Clapeyron equation.

ln    

Although Hvapchanges somewhat with temperature, it is usually adequate to use the value
tabulated at the normal boiling point of the liquid (Appendix E) unless more precise values
are available. The units of Rmust be consistent with those of Hvap.

1



T 2

1



T 1

Hvap



R

P 2



P 1

E

nrichment

The Clausius–Clapeyron equation is

used for three types of calculations:

(1) to predict the vapor pressure of a

liquid at a specified temperature, as in

Example 13-3; (2) to determine the

temperature at which a liquid has a

specified vapor pressure; and (3) to

calculate Hvapfrom measurement

of vapor pressures at different

temperatures.