Carnot devices: A Carnot refrigerator has a coefficient of performance of 2.5. The refrigerator consumes 50 W of power. How much heat is removed from the interior of the refrigerator in 1 hour?
A) 7.5 kJ
B) 450 kJ
C) 180 kJ
D) 720 kJ
E) 72 kJ

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat removed from the interior of the refrigerator to the work done by the refrigerator. In this case, the COP is 2.5 and the refrigerator consumes 50 W of power. Therefore, the heat removed from the interior of the refrigerator in 1 hour is:Heat removed = COP * Power * Time= 2.5 * 50 W * 1 hour= 125 W-hr= 125 J/s * 3600 s= 450 kJ

Carnot Devices: a Carnot Refrigerator Has a Coefficient of Performance