A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 40 kJ/min. Determine the rate of heat transfer to the kitchen air in kilojoules per minute to three significant digits. Pay attention to the units asked for in the answer!

Question

Physics

Hayley Barajas

19 January, 16:50

A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 40 kJ/min. Determine the rate of heat transfer to the kitchen air in kilojoules per minute to three significant digits. Pay attention to the units asked for in the answer!

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Answers (1)

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Mackenzie · Answer 1

73.3 kJ / min

Explanation:

COP or coefficient of performance of a refrigerator is defined as ratio of heat extracted from the refrigerator to electrical imput to the refrigerator

If Q₁ be the heat extracted out and Q₂ be the heat given out to the surrounding

Imput energy = Q₂ - Q₁

so COP = Q₁ / Q₂ - Q₁

Given

COP = 1.2

Q₁ = 40kJ

Substituting the values

1.2 = 40 / (Q₂ - 40)

1.2 (Q₂ - 40) = 40

1.2 Q₂ = 2.2 X 40

Q₂ = 73.3 kJ / min