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`

Explanation

`eta = 1 - (T_(2))/(T_(1)) implies (1)/(10) = 1 - (T_(2))/(T_(1))`
`(T_(2))/(T_(1)) = (9)/(10)`
Coefficient of performance of refrigerator
`eta' = (Q_(2))/(W) = (Q_(2))/(Q_(2) - Q_(2)) = (1)/((Q_(1))/(Q_(2)) - 1) = (1)/((T_(1))/(T_(2)) - 1) = (1)/((10)/(9) - 1) = 9`
`(Q_(2))/(10) = 9`
`Q_(2) = 90 J`