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  • A refrigerator operates on a Carnot cycle between two reservoirs at temperatures <img alt="\mathrm{T_{\text {hot }}=300 \mathrm{~K} \, \, and\, \, T_{\text {cold }}=200 \mathrm{~K}}" src="https://e

A refrigerator operates on a Carnot cycle between two reservoirs at temperatures \mathrm{T_{\text {hot }}=300 \mathrm{~K} \, \, and\, \, T_{\text {cold }}=200 \mathrm{~K}}. Calculate the coefficient of performance (COP) of the refrigerator.
 

Option: 1

16.10 \mathrm{k}


Option: 2

450.56 \mathrm{k}


Option: 3

20.98 \mathrm{k}


Option: 4

5.20 \mathrm{k}


Answers (1)

best_answer

The coefficient of performance (COP) of a refrigerator is given by the formula:
\mathrm{COP}=\frac{Q_{\text {cold }}}{W}
where \mathrm{Q_{\text {cold }}} is the heat removed from the cold reservoir and W is the work done by the refrigerator.
In a Carnot cycle, the efficiency \mathrm{(\eta)} is given by:

\mathrm{\eta=1-\frac{T_{\text {cold }}}{T_{\text {hot }}}}
Since the COP is the inverse of the efficiency, we have:
\mathrm{\mathrm{COP}=\frac{1}{\eta}=\frac{T_{\text {hot }}}{T_{\text {hot }}-T_{\text {cold }}}}

Substitute the given values:

\mathrm{\mathrm{COP}=\frac{300 \mathrm{~K}}{300 \mathrm{~K}-200 \mathrm{~K}}=\frac{300 \mathrm{~K}}{100 \mathrm{~K}}=3}

Therefore, the coefficient of performance of the refrigerator is COP =3.

Posted by

Devendra Khairwa

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