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

Explanation

`eta = 1 - (T_(2))/(T_(1)) implies (1)/(10) = 1 (T_(2))/(T_(1)) implies (T_(2))/(T_(1)) = (9)/(10)`
`(Q_(2))/(Q_(1)) = (T_(2))/(T_(1)) = (9)/(10) implies Q_(1) = (10)/(9) Q_(2)`
`W = Q_(1) - Q_(2)`
`10 = (10)/(9) Q_(1) - Q_(2) = (1)/(9) Q_(2)`
`Q_(2) = 90 J`