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
-
changing from a solid to a liquid (melting) or from
a liquid to a solid (freezing)
-
changing from a liquid to a gas (boiling) or
from a gas to a liquid (condensing)
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
-
the number of atoms, that is the mass of our sample
-
what the sample is made from, and whether it is in
the solid, liquid, or gas phase.
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
