Page:International Library of Technology, Volume 93.djvu/68

 The specific heat of air at constant pressure is .23751; hence, the total number of heat units required is
 * .07496 × 100 × .23751 = 1.78 heat units;
 * 1.78 × 778 = 1,384.8 foot-pounds

Since the outer work required 402.19 foot-pounds, the inner work will require
 * 1,384.8 - 402.19 = 982.61 foot-pounds

This shows that, in the case of air and gases, the outer work is a little less than half the inner work. Since the force of cohesion has no perceptible effect in the case of gases, the inner work tends only to raise the temperature, or, in other words, to increase the vibratory movement of the molecules. Consequently, if the piston in Fig. 12 were fastened down, so that the volume of the gas would remain the same, there would be no outer work, and the total work required to raise the temperature 100° would be 982.61 foot-pounds; or, to raise the temperature one degree, 9.8261 foot-pounds. The inner work may also be calculated by using the specific heat at constant volume, as indicated in Art. 60. Thus, the inner work is The slight difference in results is due to decimals.

1. Find the equivalent of 50,000 foot-pounds in British thermal units.

2. The performance of a certain amount of work requires 926.5 British thermal units; what is the amount of work done, in foot-pounds?

3. A pull of 125 pounds is required to move a loaded car on a level track at a uniform velocity: (a) How many foot-pounds of