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The efficiency of a carnot engine eta = 1/10. This carnot engine used as refrigerator. If work done on system is 10 J, then heat absorbed from sink will be ……. |
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Answer» 100 J `ETA=(T_1-T_2)/T_1=(Q_1-Q_2)/Q_1` COEFFICIENT of performance of refrigerator , `beta=Q_2/(Q_1-Q_2)` `beta=(1-eta)/eta` `therefore beta=(1-1/10)/(1/10)=(9/10)/(1/10)=9` but `beta=Q_2/W` `therefore Q_2=betaW` `therefore Q_2=9xx10=90` J |
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