23. In changing a solid body to a liquid, either by melting or by dissolving it, or in changing a liquid to vapor or gas, a large amount of heat may be applied without changing the temperature. Thus, a pound of ice at 32° will absorb 142.65 heat units in changing to water having the same temperature. Again, the pound of water thus produced may be heated to the boiling point, 212°, by the addition of 180.531 heat units. But, in order to convert the water at 212° into steam at the same temperature, 966.069 heat units must be added. Thus, a pound of steam at 212° contains 966.069 + 180.531 + 142.65 = 1,289.25 heat units more than a pound of ice at 32°, although the difference in temperature is only 180°.

The heat which is thus absorbed in the process of melting or vaporization, is called latent heat.

The temperature at which a body changes from a solid to a liquid state is called the temperature of fusion; and the number of B. T. U. required to effect this change in a body weighing 1 pound is called the latent heat of fusion. The temperature at which a body changes from a liquid state to a vapor (gas) is called the temperature of vaporization; and the heat required to effect this change in one pound of the liquid is called the latent heat of vaporisation.

When a vapor changes back to a liquid, it is said to condense, and when a liquid changes back to a solid, it is said to freeze; in either case, an amount of heat equal to the latent heat of vaporization or of fusion, as the case may be, must be abstracted from (given up by) the body before the change can be effected.

24. The following table shows the amounts of latent heat required for the fusion or vaporization of 1 pound of various substances, they having first been raised to the temperature at which the change takes place, and the pressure being 1 atmosphere, or 14.7 pounds per square inch:

## Table 6

 Substance. Temperature of Fusion. Temperature of Vaporization. LatentHeat ofFusion.B. T. U. Latent Heat ofVaporization.B. T. U. Water...... 32° 212° 142.65 966.069 Mercury..... -37.8° 662° 5.09 157 Sulphur..... 228.3° 824° 13.26 Tin........ 446° 25.65 Lead.... 626° 9.67 Zinc....... 680° 1,900° 50.63 493 Alcohol..... Unknown 173° 372 14° 313° 124 Linseed oil.... 600°

The temperature of vaporization in the above table is the boiling' point of the liquid under the ordinary atmospheric pressure of 14.7 pounds per square inch.

The variation of the boiling point by changes in pressure differs greatly in various liquids. The temperature of fusion, or the melting point, is similarly affected by changes in pressure, but the amount of the variation is unimportant for all the purposes of heating and ventilation.

## Example

If 5 pounds of ice, having a temperature of 10° below zero, be mixed with hot water having a temperature of 200°, what weight of hot water will be required to melt the ice and bring the temperature of the mixture up to 60°?

## Solution

The temperature of the ice is 10 + 32 = 42° below the freezing point. The amount of heat required to raise its temperature to the freezing point will be 42 X .504 (specific heat) X 5 = 105.84 heat units. To liquefy the ice at 32° will require 5 X 142.65 (latent heat of fusion) = 713.25 heat units. Then, to change the ice into water at 32°, 105.84 + 713.25 = 819.09 heat units must be applied to it. To bring its temperature up to 60°, there must be (60-32) X 5 = 140 heat units added, which makes the total quantity of heat required, 819.09+ 140 = 959.09 heat units.

The hot water is to be cooled from 200° down to 60º; therefore, each pound will give up 200-60 = 140 heat units; 959.09 ÷140 = 6.85 pounds of hot water will be required to furnish the 959.09 heat units. Ans.