What You Need To Know
Latent Heat


Introduction

In order to reduce the temperature of an object you must extract heat. However, extracting heat does not necessarily lead to a reduction in temperature. It could instead lead to a change of phase. The two commonest changes of phase are In either case the transition from one phase occurs at constant temperature. Only once the phase transition is complete can the temperature be changed again.

Freezing and Melting

These two processes are the reverse of each other. Freezing requires the extraction of heat, whereas melting requires the addition of heat. The most common example is the freezing of liquid water to make ice, which occurs at 273 K (0o C or 32o F), although a wide range of materials can be made to undergo the same process. The amount of energy which must be extracted (or added) depends only on Writing this as a simple formula, the heat Q is given by

Q = m Lf

where m is the mass of the sample, and Lf is known as the latent heat of fusion. It is a parameter which contains all the information about the material and phase of the sample.

Boiling and Condensing

The process for boiling and condensing is much the same as for freezing and condensing, and we will write

Q = m Lv

where m is the mass of the sample, and Lv is known as the latent heat of vaporization. It is a parameter which contains all the information about the material and phase of the sample.
Material Latent Heat 
of Fusion 
(J/kg)
Latent Heat 
of Vaporization 
(J/kg)
Water 333,000 2,600,000
Freon - 12 165,500

Values of the latent heat

For any pure substance each of the latent heats is fixed. You can look up their values in a book. Two values are listed in the table to the right.

Note the units for the specific heat. The unit for the thermal energy Q must be Joules; and for the mass, kilograms. That makes the unit for the specific heat Joules/Kelvin or J/K.

Calculations

Q. What is the amount of heat which must be extracted to freeze 0.3 kg of water.
A. Looking up the latent heat of fusion for water from the table, we get Q = m Lf = 0.3 * 333,000 = 100,000 J
Q. How much heat would we have to supply to boil 0.3 kg of water
A. Looking up the latent heat of vaporization for water from the table, we get Q = m Lv = 0.3 * 2,600,000 = 780,000 J