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Question 121 of 126

Q.A refrigerator has COP of 4. How much work must be supplied to the refrigerator in order to remove 300 J of heat from its interior?

(a) 600 J
(b) 66.67 J
(c) 50 J
(d) 75 J
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2025MCQ· 1mImportance★★★★★
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Using COP = Q_cold / W, the work needed to remove 300 J of heat with a refrigerator of COP 4 is W = 300/4 = 75 J.

For a refrigerator, the Coefficient of Performance (COP) is defined as the ratio of heat removed from the cold reservoir (interior) to the work input required to do so:

COP = Q_cold / W

Given: COP = 4, Q_cold = 300 J.

Rearranging for W:

W = Q_cold / COP = 300 J / 4 = 75 J

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