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 10 J, the amount of energy absorbed from the reservoir at lower temperature is

Options

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

Answer

C

Explanation

For the heat engine, absorbed energy from heat reservoir of higher temperature = `Q_(1)` and work obtained,
Now, `eta = W/(Q_1) = 10/(Q_1) , "given", eta = 1/10` then
`10/(Q_1) = 1/10 or Q_(1) = 100 J`
So, energy released at lower temperature,
`Q_(2) = Q_(1) - W = 100 - 10 = 90 J`
Reverse phenomenon happens in case of the refrigerator.
So, when the work done on the system is 10 J, then the amount of energy absorbed from the reservoir at lower temperature = 90 J.