A refrigerator extracts 65 J of heat from a heat reservoir at 0.00$^{\circ}$C.If the coefficient of performance of the refrigerator is 2.5,then how much work is needed to run the refrigerator?
A) 26 J
B) 48 J
C) 75 J
D) 90 J

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) of a refrigerator is given by COP = Qc/W, where Qc is the heat extracted from the cold reservoir and W is the work done. Rearranging gives W = Qc/COP. Substituting the given values, W = 65 J / 2.5 = 26 J.

A Refrigerator Extracts 65 J of Heat from a Heat ∘^{\circ}∘

A Refrigerator Extracts 65 J of Heat from a Heat $^{\circ}$