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20 J of work is done per cycle on a refrigerator with a coefficient of performance of 4.0.
Part A: How much heat is extracted from the cold reservoir per cycle?
Part B: How much heat is exhausted to the hot reservoir per cycle?

Respuesta :

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

[tex]COP=\dfrac{Qa}{W}[/tex]

[tex]4=\dfrac{Qa}{20}[/tex]

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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