Question: A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)
Options:
Correct Answer: 1.25
Exam Year: 2022
Solution:
COP = Q_c / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.