Latent Heat of Water

DEFINITION. - The number of units of heat required to convert one grgmme of ice at 0° C. into water, without altering its temperature, is called the latent heat of water.

A weighed quantity of water at a known temperature is contained in the calorimeter. Some pieces of ice are then dropped in arid the fall of temperature noted. When the ice is all melted the water is weighed again, and the increase gives the mass of ice put in. From these data, knowing the water equivalent of the calorimeter, we can calculate the latent heat of the water.

The ice must be in rather small pieces, so as to allow it to melt quickly. It must also be as dry as possible. We may attain this by breaking the ice into fragments and putting it piece by piece into the calorimeter, brushing off from each piece as it is put in all traces of moisture with a brush or piece of flannel.

The ice may be lifted by means of a pair of crucible tongs with their points wrapped in flannel. These should have been left in the ice for some little time previously, to acquire the temperature of 0° C.

Another method is to put the ice into a small basket of fine copper gauze and leave it to drain for a few moments, while the ice is stirred about with a glass rod, previously cooled down to 0° C. by being placed in ice. The basket is put into the calorimeter with the ice. The water equivalent of the basket must be allowed for, being determined from its mass and specific heat.

Care must be taken not to put so much ice into the water that it cannot all be melted.

The formula from which the latent heat is found is obtained as follows: Let M be the mass of water initially, τ its temperature; let m be the mass of ice put in, which is given by the increase in mass of the calorimeter and contents during the experiment; let θ be the temperature when all the ice is melted, m₁ the water equivalent of the calorimeter, and L the latent heat

Then the heat given out by the water, calorimeter, etc., in cooling from τ to θ is

This has melted a mass m of ice at 0° C, and raised the temperature of the water formed from 0° to θ°. The heat required for this is

The temperature of the water used should be raised above that of the room before introducing the ice, and noted just before the ice is immersed. It is well to take a quantity of ice such that the temperature of the water at the end of the experiment may be as much below that of the room as it was above it initially. We may calculate this approximately, taking the latent heat of ice as 80.

Thus suppose we have 45 grammes of water at 20°, and that the temperature of the room is 15°. Then the water is to be cooled down to 10°, or through 10°.

Thus the heat absorbed from water will be 450 units.

Let us suppose we have x grammes of ice. This is melted, and the heat absorbed thereby is 80x. It is also raised in temperature from 0° to 10°, and the heat absorbed is 10x.

Thus we should require about 5 grammes of ice. (If in practice we did not know the latent heat of the substance experimented upon at all, we should for this purpose determine it approximately, then use our approximate result to determine the right quantity of the substance to employ in the more accurate experiment.)

Experiment. - Determine the latent heat of ice.