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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 !!!

Respuesta :

Answer:

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

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