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
foot-pounds
The slight difference in results is due to decimals.
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EXAMPLES FOR PRACTICE
1. Find the equivalent of 50,000 foot-pounds in British thermal units. Ans. 64.27, B. T. U.
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? Ans. 720,817 ft.-lb.
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
