A refrigerator uses 40 J of work to extract heat from a heat reservoir at 0.00°C. If the coefficient of performance of the refrigerator is 2.1, then how much heat is extracted from the heat reservoir at 0.00°C?
A) 48 J
B) 65 J
C) 84 J
D) 90 J

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) of the refrigerator is given by COP = heat removed from the refrigerated space / work done on the refrigerator. In this case, COP = 2.1, so the heat removed from the refrigerated space is 2.1 times the work done on the refrigerator.

Since the refrigerator is extracting heat from a reservoir at 0.00°C, we know that the heat removed is equal to the change in entropy of the reservoir divided by its temperature. Using the specific heat capacity of water, we can calculate the change in entropy for a given amount of heat.

ΔS = Q / T = (40 J) / (273 K) = 0.147 J/K

The heat removed by the refrigerator is then:

Q = COP x work done = 2.1 x 40 J = 84 J

Therefore, the best choice is C (84 J) as this is the amount of heat that is extracted from the heat reservoir at 0.00°C.

A Refrigerator Uses 40 J of Work to Extract Heat