  # Heat capacity and specific heat capacity Why is it that you can put out a candle flame with moist fingers without hurting yourself although it is at 750 oC but it is very painful to put your fingers into a cup of hot water at 80 oC? The reason is because of the difference between heat energy and temperature.

There is much more heat energy "locked away" in the hot water than in the candle flame, partly because of the low mass of the flame and partly because of a property of the gas in the flame and the water known as their specific heat capacity.

This section will enable you to estimate the amount of heat energy transferred to your hand when you put it in water or the candle flame. For the candle flame this is about 1 J and for the hot water about 40 kJ!

The temperature of an object is a measure of the energy of each individual particle within it while the heat energy is a measure of the total energy of the object as a whole. In the candle flame the molecules are moving at very high speeds because of their high temperature but there are very few of them. In the cup of water although the speed of the molecules is much less there are very many more molecules and so the heat energy content is much greater.

The amount of heat needed to change the temperature of a body depends on
(a) the material of the body,
(b) the mass of the body, and
(c) the change in temperature (positive or negative).

For a given body we can define a quantity known as the thermal capacity as the heat energy needed to raise its temperature by 1 K. A rather more useful quantity is the specific heat capacity of a material defined as follows:

The specific heat capacity of a material is the heat energy needed to raise the temperature of 1 kg of the material by 1 K.

The units of thermal capacity are joules per kelvin (J K-1) and those of specific heat capacity joules per kilogram kelvin (Jkg- 1K-1). Therefore:

Heat energy = mass x specific heat capacity x temperature change

The values of some specific heat capacities are given in the following table.

 Material Specific heat capacityJkg-1K-1 Material Specific heat capacityJkg-1K-1 Water 4200 Copper 385 Ethanol 2500 Lead 126 Paraffin oil 2130 Aluminium 913 Turpentine 1760 Sodium 1240 Hydrogen 14 300 Iron 106 Air 993 Steel 420 Helium 5240 Concrete 3350 Oxygen 913 Polypropylene 900 Granite 820 Marble 2100 Beryllium 1970 Glass 600

It is interesting to note the large specific heat capacity of water compared with other liquids, which makes it useful as a coolant. Sodium too has a high specific beat capacity and is used in liquid form as a coolant in some nuclear reactors.

The specific heat capacity of air is also deceptive; it actually requires quite a large amount of energy to raise the temperature of a kilogram of air. Remember, however, that at normal atmospheric pressure this mass of air has a volume of about 1 m3

Example problem
A block of metal of mass 0.5 kg initially at a temperature of 100 oC is gently lowered into an insulated copper container of mass 0.05 kg containing 0.9 kg of water at 20 oC. If the final temperature of the mixture is 25 oC, calculate the specific heat capacity (c) of the metal of the block.
(Assume no loss of heat and that no water is vaporised.)

Heat lost by block = 0.5c(100 - 25) = 37.5c
Heat gained by water and container = (0.9 x 4200x 5) + (0.05x385x5) = 18996
Therefore 37.5c = 18996
Specific heat capacity (c) = 18996/37.5 = 506.6 J kg-1K-1 All specific heat capacities have been found to vary with temperature: the variation for three solids is shown in Figure 1. The exact reasons for this variation are complex and outside the scope of this text.

## Molar heat capacities

If the molar heat capacities (the heat required to raise the temperature of one mole of a substance by 1 K) of metals are considered, then it is found that they are all approximately the same and equal to about 25 J mol-1K-1, a fact first noticed by Dulong and Petit in 1819.

It follows that the heat required to raise the temperature of a sample of metal depends on how many particles the sample contains, and not on the mass of an individual molecule. The specific heat capacity is therefore directly related to the molecular structure of the material.

Student investigation
At some time you will have sat down to what you hoped would be a hot meal only to find that. parts.of it have gone cold. You will also have met with the scalding pudding that you have to sit and look at while it cools off before you can attempt to eat any of it, and the jam sponge where the jam is so much hotter than the sponge.

Using a thermocouple or small probe digital thermometer, investigate the cooling of different foods. Also find out just how hot food has to be before it becomes difficult to eat, and how cold it must be before it is unpleasant.

Be careful not to burn yourself

How does the surface area and texture of the food affect the cooling?
If possible, investigate the rate of rise of temperature during the thawing of some frozen food such as a loaf of bread taken from a freezer.

HEALTH AND SAFETY
If the food is to be eaten at any point in the experiment make sure that all the apparatus with which it may come into contact is cleaned carefully before starting the experiment.

Example problem
An electrical immersion heater with a power of 50 W is used to heat a glass beaker of mass 75 g containing 200 g of water. If the initial temperature of the beaker and the water is 15oC calculate the temperature after 3 minutes of heating. You can assume that no heat energy is lost to the surroundings.

Electrical energy input = VIt = Power x time = 50x3x60 = 9000 J
Heat energy gained by the water and the glass = 0.075x600x(T – 15) + 0.20x4200x(T – 15)
= 45T – 675 + 840T – 12600 = 885T-13275
Therefore:     9000 + 13275 = 22275 = 885T and so T = 25.2oC