## Conceptual Physics (12th Edition)

Published by Addison-Wesley

# Chapter 18 - Think and Explain: 60

#### Answer

To increase the efficiency, it is better to decrease the temperature of the cold reservoir by a given amount, than to increase the temperature of the hot reservoir by the same amount.

#### Work Step by Step

The simplest way to see this is by picking some numbers. As discussed on page 345, Carnot's equation states that the ideal efficiency of a heat engine is $\frac{T_{hot} – T_{cold}}{T_{hot}}$. For exam­ple, let the hot temperature be 500K and the cold temperature be 300K. $$ideal \; efficiency = \frac{500K - 300K}{500K} = 0.4$$ Now let the hot temperature stay at the original 500K, but the cold temperature be decreased by 100K, to 200K. The efficiency increases from its original value. $$ideal \; efficiency = \frac{500K - 200K}{500K} = 0.6$$ On the other hand, now let the hot temperature be increased by 100K to be 600K, while the cold temperature stays at 300K. The efficiency does increase from its original value, but by a smaller amount. $$ideal \; efficiency = \frac{600K - 300K}{600K} = 0.5$$ As stated, to increase the efficiency, it is better to decrease the temperature of the cold reservoir by a given temperature increment than to increase the temperature of the hot reservoir by the same amount.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.