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aleksandr82 [10.1K]
3 years ago
5

20 J of work is done per cycle on a refrigerator with a coefficient of performance of 4.0.

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

A)Qa=80 J

B)Qr= 100 J

Explanation:

Given that

W= 20 J

COP = 4

Heat rejected from cold reservoir = Qa

Heat exhausted to hot reservoir = Qr

The COP of refrigerator is given as

COP=\dfrac{Qa}{W}

4=\dfrac{Qa}{20}

Qa= 4 x 20 J

Qa=80 J

By using first law of refrigerator

Qr= Qa + W

Qr= 80 + 20 J

Qr= 100 J

A)Qa=80 J

B)Qr= 100 J

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