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Question 104 of 126

Q.A refrigerator has COP of 3. How much work must be supplied to a refrigerator in order to remove 200 J of heat from its interior ?

(a) 33.33 J
(b) 44.44 J
(c) 66.67 J
(d) 50 J
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019MCQ· 1mImportance★★★★★
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Using COP = Qc/W with Qc = 200 J and COP = 3, the work required is W = 200/3 = 66.67 J.

For a refrigerator, the coefficient of performance (COP) is defined as the ratio of the heat Qc extracted from the cold reservoir (the refrigerator's interior) to the work W that must be supplied to run the refrigerator:

COP = Qc / W

Given Qc = 200 J and COP = 3:

W = Qc / COP = 200 / 3 = 66.666... J is approximately 66.67 J

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