will be found that the lead ball is the hottest. Since all the balls are in a position to absorb equal amounts of heat, it is evident that the lead ball requires less heat to raise its temperature to a given point than the other balls. Different substances require different amounts of heat to raise equal weights of those substances 1°, water requiring more than most other substances. Hence, it is taken as a standard of comparison. Thus, the ratio between the quantity of heat required to warm a body 1° and the quantity of heat required to warm an equal weight of water 1° is called the specific heat of the body. This ratio is always expressed as a decimal. Hence, if the specific heat of a certain substance is said to be .1375, it is understood that the amount of heat required to raise a given weight of that substance 1° is only .1375 times the amount of heat required to raise the same weight of water 1°.
59. In Table II are given the specific heats of a number of substances.
60. Calculations involving the specific heats of various substances may be worked out by means of the following rules:
Rule I. — To find the number of British thermal units required to raise, or to be abstracted to lower, the temperature of a body a given number of degrees, multiply the weight of the body, in pounds, by the specific heat, and by the number of degrees Fahrenheit.
Or, (1)
in which
= number of British thermal units;
= specific heat;
= weight, in pounds;
= higher temperature, in degrees F.;
= lower temperature, in degrees F.
Example 1. — How many British thermal units are required to raise 20 pounds of lead from 50° F. to 400° F., the specific heat of lead being .0314?
Solution. — Substituting values in formula 1,
= .0314 × 20 × (400 - 50) = 219.8 B. T. U. Ans.
