Chemistry » Liquids and Solids » Phase Transitions

# Enthalpy of Vaporization

## Enthalpy of Vaporization

Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, ΔHvap. For example, the vaporization of water at standard temperature is represented by:

$${\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(g)\phantom{\rule{5em}{0ex}}\text{Δ}{H}_{\text{vap}}=\text{44.01 kJ/mol}$$

As described in the tutorial on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:

$${\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{5em}{0ex}}\text{Δ}{H}_{\text{con}}=\text{−Δ}{H}_{\text{vap}}=-44.01\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$$

## Example

### Using Enthalpy of Vaporization

One way our body is cooled is by evaporation of the water in sweat (see the figure below). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); ΔHvap = 43.46 kJ/mol at 37 °C.

Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr)

### Solution

We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

$$1.5\phantom{\rule{0.2em}{0ex}} \require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1000\phantom{\rule{0.2em}{0ex}} \require{cancel}\cancel{\text{g}}}{1 \require{cancel}\cancel{\text{L}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}} \require{cancel}\cancel{\text{mol}}}{18\phantom{\rule{0.2em}{0ex}} \require{cancel}\cancel{\text{g}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{43.46\phantom{\rule{0.2em}{0ex}}\text{kJ}}{1\phantom{\rule{0.2em}{0ex}} \require{cancel}\cancel{\text{mol}}}\phantom{\rule{0.2em}{0ex}}=3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.

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