Question #281791

A Carnot refrigerator rejects 2500 kJ of heat at 80°C while using 1100 kJ of work


1
Expert's answer
2021-12-21T12:15:36-0500

Find the heat absorbed:


Qa=W+Qr=1100+2500=3600 kJ.Q_a=W+Q_r=1100+2500=3600\text{ kJ}.

The efficiency of the refrigerator is


η=WQa=11003600=0.31.\eta=\frac{W}{Q_a}=\frac{1100}{3600}=0.31.


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