Refrigerator Coefficient Of Performance

Question

A refrigerator extracts 360 J of heat from its cold interior while the compressor consumes 90 J of electrical work per cycle, so what is its coefficient of performance?

RankGuru · Accepted Answer

As outlined in NCERT Class 11, Chapter 12 (Thermodynamics), a refrigerator is essentially a heat engine run in reverse: it uses external work to pump heat from a cold region to a warmer one. Its performance is measured not by efficiency but by the coefficient of performance, $	ext{COP} = \dfrac{Q_C}{W}$, where $Q_C$ is the heat extracted from the cold space and $W$ is the work input supplied to the compressor. Substituting $Q_C = 360$ J and $W = 90$ J gives $	ext{COP} = \dfrac{360}{90} = 4$, which means each joule of work removes four joules of heat from the cold interior. The option 0.25 is wrong because it inverts the ratio, computing $W/Q_C$ instead. The option 5 is wrong because it uses $Q_H/W = (360+90)/90 = 5$, which is the heat-delivered or heat-pump quantity, not the refrigerator's cold-side COP. The option 0.2 is wrong because it inverts that heat-pump value. A plausibility check confirms the answer: unlike engine efficiency, which is always below 1, a refrigerator's COP can exceed 1 because the device merely transports heat rather than producing work; a value of 4 is realistic for a domestic refrigerator, and the joule units cancel to leave a dimensionless ratio as expected.

Refrigerator Coefficient Of Performance

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