A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is

Options

A. 100 J
B. 99 J
C. 90 J
D. 10 J

Answer

C

Explanation

For a carnot engine as a refrigerator, coefficient of performance `beta=(1-eta)/eta=(1-1//10)/(1//10)=(9//10)/(1//10)=9.` In a carnot refrigerator heat `(Q_(L))` is absorbed from the refrigerator, work W is done on it and heat `Q_(H)` is transferred to the source and `Q_(L)=Wxxbeta=10xx9=90J`.