A carnot engine having an efficiency of `(1)/(10)` as heat engine, is used as a refrigerator. If the work done on the system is 10 J, 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

Answer

A

Explanation

The relation between coefficient of performance and efficiency of carnot engine is given as `beta = (1- eta)/(eta)`
Given `eta = (1)/(10), W = 1 J :. beta = (1-(1)/(10))/((1)/(10)) = (9)/(10).10 = 9`
Since, `beta = (Q_(2))/(W)`, where `Q_(2)` is the amount of energy absorbed from the reservoir
`:. Q_(2) = beta W = 9 xx 10 = 90 J`