A refrigerator has a coefficient of performance equal to 4.2. How much work must be done on the operating gas in the refrigerator in order to remove 250 J of heat from the interior compartment?
A) 1050 J
B) 480 J
C) 250 J
D) 120 J
E) 60 J

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) for a refrigerator is given by COP = Qc/W, where Qc is the heat removed from the interior and W is the work done. Rearranging gives W = Qc/COP. Substituting the given values, W = 250 J / 4.2 = 59.52 J, which rounds to 60 J.

A Refrigerator Has a Coefficient of Performance Equal to 4