Temperature and States of Matter Help (page 3)

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By — McGraw-Hill Professional
Updated on Sep 5, 2011

Heat Of Vaporization

It takes a certain amount of energy to change a sample of liquid to its gaseous state, assuming that the matter is of the sort that can exist in either of these two states. In the case of water, it takes 540 cal to convert 1 g of liquid at +100°C to 1 g of pure water vapor at +100°C. This quantity varies for different substances and is called the heat of vaporization for the substance.

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

Heat of vaporization, like heat of fusion, is sometimes expressed in calories per mole (cal/mol) rather than in cal/g. However, this is not the usual case.

If the heat of vaporization (in calories per gram) is symbolized 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 following formula holds:

h v = h / m

This is the same as the formula for heat of fusion, except that h v has been substituted for h f .

Heat of Vaporization Practice Problem


Suppose that a certain substance boils and condenses at +500°C. Imagine a beaker of this material whose mass is 67.5 g, and it is entirely liquid at +500°C. It heat of vaporization is specified as 845 cal/g. How much heat, in calories and in kilocalories, is required to completely boil away this liquid?


Our units are already in agreement: grams for m and calories per gram for h v . We must manipulate the preceding formula so that it expresses the heat h (in calories) in terms of the other given quantities. This can be done by multiplying both sides by m , giving us this formula:

h = h v m

Now it is simply a matter of plugging in the numbers:

h = 845 × 67.5

= 5.70 × 10 4 cal = 57.0 kcal

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

Practice problems of these concepts can be found at: Temperature, Pressure, And Changes Of State Practice Problem

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