Change of state and latent heat
When a substance changes from one
state to another energy is either absorbed or liberated. This heat energy is called the latent
heat, and part of it is the energy used to overcome the forces of attraction between the
molecules.
It is clearly useful to know the energy required to change the state of unit
mass of the substance. This is known as the specific latent heat and is defined as follows:
The specific latent heat is the energy required to change the state of 1 kg of the substance
Any material has two specific
latent heats:
The specific latent heat fusion is the heat energy needed to change 1 kg of the material in its solid state at its melting point to 1 kg of the material in its liquid state, and that released when 1 kg of the liquid changes to 1 kg of solid
The specific latent heat of vaporisation of a liquid is the heat energy needed to change 1 kg of the material in its liquid state at its boiling point to 1 kg of the material in its gaseous state, and that released when 1 kg of vapour changes to 1 kg of liquid
It is
important to realise that no temperature change occurs during the change of state. The
temperature will only rise or fall when the entire specimen has changed from one state to the
other.
Some examples of specific latent heats are given in the following table:
| Material |
Specific latent heat of fusion (Jkg-1) |
Material |
Specific latent heat of vaporisation (Jkg-1) |
| Aluminium |
390000 |
Benzene |
400000 |
| Copper |
210000 |
Ethanol |
850000 |
| Iron |
270000 |
Ether |
350000 |
| Lead |
2600 |
Turpentine |
270000 |
| Mercury |
1300 |
Water |
2260000 |
| Naphthalene |
150000 |
- |
- |
| Solder |
70000 |
- |
- |
| Water |
330000 |
- |
- |
Generally the specific latent heats of
vaporisation are greater than the specific latent heats of fusion. The change of state from a
liquid to a gas results in a large increase of volume and therefore a large amount of work has to
be done against the surrounding atmosphere.
In general energy is needed to
(a)
change the state of the material at a constant temperature (and pressure), and
(b) to do
external work if there is a change of volume during the change of state.
This external
work is usually positive, although there are exceptions. Ice contracts when it melts, the volume
of a sample of water being a minimum at 4 oC, and therefore the external work done on
melting is negative.
If a volatile liquid is allowed to evaporate from the surface of an
object then its latent heat of vaporisation may be used to cool the object: the heat energy
needed to evaporate the liquid is drawn from the object itself and so its temperature falls. You
can hand will cool.
Figure 1(a) shows how the temperature of a specimen might alter
with time due to a steady heat input - heat losses to the exterior have been ignored here. Figure
1(b) shows how the molecular arrangements within the material change as the heat energy is
supplied
Student investigation
A shiny electric kettle (Figure 2) is used in a simple determination of the specific latent heat of vaporisation of water. The kettle and its contents are weighed and the kettle is then boiled for a given time and the loss of weight is found.
Assuming the value given above for the specific latent heat capacity of water, make measurements to find the heat lost by the kettle during this time and thence estimate the accuracy of a measurement made by this method.
Example problems
1. A double-walled flask containing water of total mass 1 kg is heated with a 16W heater and it is found that it takes 30 minutes for the temperature to rise from 20 oC to 100 oC.
(a) Estimate an upper limit for the value of the mean specific heat capacity of the inner flask and its contents.
(b)Calculate the mass of water that would be vaporised after 30 minutes of steady heating when the power is supplied at a rate of 60 W.
Take the specific latent heat of vaporisation of water to be 2.26x 106J kg-1.
Initial energy input = 16 x 30 x 60 = 28 800 J.
Therefore (assuming no heat losses) the upper limit for the specific heat capacity of the inner flask and contents would be 28 800/80 = 360 J kg-1K-1.
New energy input during input at 60 W = 60 x 30 x 60 = 108000 J.
Of this, 28 000 J is required to heat the flask and con¬tents, and therefore a further 79 200 J is available to vaporise the water.
Mass of water vaporised = 79 200/ 2.26x 106 = 3.5 x10-2 kg = 35 g
This result assumes that there are no heat losses from the flask during the boiling phase and therefore no energy is needed to keep the flask at 100 oC.
2. Calculate the amount of ice that would be melted by a 65 W heater in five minutes at 0oC if all other heat energy exchanges are ignored.
Specific latent heat of fusion of ice = 330 000 Jkg-1
Electrical energy input = 65x5x60 = 19500 J
Mass of ice melted = 19500/330 000 = 0.059 kg = 59 g
