Back to 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators

15.5 Exercise 14

Avg difficulty 2.5

A 4-ton air conditioner removes 5.06×107 J5 . \text{06} \times \text{10}^{7} \text{ J} (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating ( EER \text{EER}) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per 3.60×106 J3 . \text{60} \times \text{10}^{6} \text{ J} (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating ( EER \text{EER}) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input.

OpenStax College Physics 2e, 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators, exercise 14

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Adapted from College Physics © 2022 Rice University. Licensed under CC BY 4.0. Sources: College Physics 2e, College Physics for AP® Courses 2e. Imported at revision f98d7a7.