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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?

Respuesta :

Answer:15,883.63 KJ

Explanation:

Given

Power=400 W

Freezing compartment temperature is 273 K[tex](T_L)[/tex]

Outside air Temperature=306 K[tex](T_H)[/tex]

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

[tex]COP=\frac{T_L}{T_H-T_L}[/tex][tex]=\frac{Desired\ effect}{Work}[/tex]

[tex]\frac{273}{306-273}=\frac{D.E.}{1920}[/tex]

[tex]D.E.=8.27\times 1920=15883.63 KJ[/tex]

Therefore 15,883.63 KJ is removed in 80 minutes

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