Question

A refrigerator, whose coefficient of performance is 7,
extracts heat from the low temperature compartment at the
rate of 250 J/cycle. Work done per cycle required to operate
the refrigerator is nearly
A) 50 J
B) 36 J
C) 27 J
D) 63 J​

Answer 1

Given : A refrigerator whose coefficient of performance is 7. extract heat from the temperature compartment at the rate of 250 J/cycle.

To find : work done per cycle required to operate the refrigerator.

solution : work done per cycle, C . O . P = Q₂/W

⇒7 = (250 J/cycle)/W

⇒W = 250/7 = 35.7 ≈ 36 J/Cycle

Therefore work done per cycle required to operate the refrigerator is nearly 36 J . so option (B) 36 J, is correct choice.

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Answer 2

Answer:36J

Explanation: