A refrigerator extracts 65.0 J of heat from a heat reservoir at 0.00°C. If the coefficient of performance of the refrigerator is 2.50, then how much heat is discharged out of the refrigerator?

Question

Quizplus · Accepted Answer

The coefficient of performance of the refrigerator is defined as the ratio of heat extracted from the cold reservoir to the work input:

COP = Qc/W

where Qc is the amount of heat extracted from the cold reservoir and W is the work input.

We can rearrange this equation to solve for Qc:

Qc = COP * W

In this case, we know that COP = 2.50 and that the work input is equal to the heat extracted from the cold reservoir (since the system is in steady state), which is 65.0 J. Therefore:

Qc = 2.50 * 65.0 J = 162.5 J

This is the amount of heat that would need to be input into the refrigerator to extract 65.0 J of heat from the cold reservoir. Since we want to know how much heat is discharged out of the refrigerator, we need to subtract the amount of heat input from the amount of heat extracted:

Qh = Qc - W

Qh = 162.5 J - 65.0 J = 97.5 J

Therefore, the answer is B) 91.0 J.

A Refrigerator Extracts 65