## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 15 - The Laws of Thermodynamics - Questions: 11

#### Answer

Heat pumps and air conditioners have exactly opposite purposes and their efficiencies are defined differently.

#### Work Step by Step

For refrigerators, the COP is defined the way it is because the point of a refrigerator is removing heat from a low-temperature reservoir. An “efficient” refrigerator removes a lot of heat from the cold reservoir per joule of input work, hence $COP = \frac{Q_{L}}{W}$. For heat pumps, the COP is defined the way it is because the point of a heat pump is delivering heat to a high-temperature reservoir. An “efficient” heat pump delivers a lot of heat to the hot reservoir per joule of input work, hence $COP = \frac{Q_{H}}{W}$.

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