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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 conČtents, 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
 
 
 
© Keith Gibbs 2010