Heat Of Vaporization:

It acquires a certain quantity of energy to change a sample of liquid to its gaseous state, supposing that the matter is of the type that can exist in either of these two states. In the situation of water, it acquires 540 cal to convert 1 g of liquid at +100°C to 1 g of pure water vapor at +100°C. This amount differs for various substances and is termed as the heat of vaporization for the substance.

In the reverse situation, when 1 g of pure water vapor at +100°C condenses entirely and becomes liquid water at +100°C, it gives up 540 cal of heat. The heat of vaporization is expressed in similar units as heat of fusion, which is, in calories per gram (cal/g). It can also be expressed in kilocalories per kilogram (kcal/kg) and will result exactly the same numbers as the cal/g figures for all substances. Whenever the substance is something other than water, then the boiling/condensation point of that substance should be replaced for +100°C.

Heat of vaporization, such as heat of fusion, is at times expressed in calories per mole (cal/mol) instead of in cal/g. Though, this is not the common case.

When the heat of vaporization (in calories per gram) is represented by h_v, the heat added or given up by a sample of matter (in calories) is h, and the mass of the sample (in grams) is m, then the formula below holds:

h_v = h/m

This is similar as the formula for heat of fusion, except that h_v has been replaced for h_f.

PROBLEM:

Assume that a certain substance boils and condenses at +500°C. Envision a beaker of this material whose mass is 67.5 g, and it is wholly liquid at +500°C. When heat of vaporization is identified as 845 cal/g. Determine heat, in calories and in kilocalories, is needed to entirely boil away this liquid?

SOLUTION:

We should manipulate the formula so that it represents the heat h (in calories) in terms of the other given quantities. This can be completed by multiplying both sides by m, giving us the formula which is shown below:

h = h_vm

Now, it is merely a matter of plugging in the numbers:

h = 845 x 67.5

   = 5.70 x 10⁴ cal = 57.0 kcal

It has been rounded off to three significant figures, the extent of the correctness of our input data.