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Zinaida [17]
3 years ago
8

An actual refrigerator operates with a COP that is half the Carnot COP. It removes 10 kW of heat from a cold reservoir at 250 K

and dumps the waste heat into the atmosphere at 300 K. (a) Determine the net work consumed by the refrigerator. (b) What-if Scenario: How would the answer change if the cold storage were to be maintained at 200 K without altering the rate of heat transfer
Physics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

Explanation:

Given

\left ( COP\right )_{actual}=0.5\left ( COP\right )_{ideal}

T_L=250 K

T_H=300 K

ideal COP=\frac{T_L}{T_H-T_L}=\frac{250}{300-250}=5

\left ( COP\right )_{actual}=2.5

Also COP=\frac{Desired\ effect}{Power\ supplied}

2.5=\frac{10}{W_{in}}

W_{in}=4 kW

(b)if T_L=200 K

\left ( COP\right )_{actual}=0.5\times \frac{200}{300-200}=1

thus W_{in}=10 kW

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