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. 99 J
B. 90 J
C. 1 J
D. 100 J

Answer

B

Explanation

`eta=(1)/(10)=1-(T_2)/(T_1)implies(T_2)/(T_1)=(9)/(10)`
`eta=(1)/((T_1)/(T_2)-1)=(1)/((10)/(9)-1)=9`
`eta'=(Q_2)/(W)`
`9=(Q_2)/(10)`
`Q_2=90J`