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3 years ago

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The motor in a refrigerator has a power output of 400 W. If the freezing compartment is at 273 K and the outside air is at 306 K

, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 80.0 minutes?

1 answer:

Aleksandr [31]3 years ago

7 0

Answer:15,883.63 KJ

Explanation:

Given

Power=400 W

Freezing compartment temperature is 273 K $(T_L)$

Outside air Temperature=306 K $(T_H)$

Time =80 minutes

Energy Required to deliver 400 w power in 80 minutes

E=1920 KJ

and we know COP of refrigerator is given by

$COP=\frac{T_L}{T_H-T_L}$ $=\frac{Desired\ effect}{Work}$

$\frac{273}{306-273}=\frac{D.E.}{1920}$

$D.E.=8.27\times 1920=15883.63 KJ$

Therefore 15,883.63 KJ is removed in 80 minutes

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