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

Explanation

Here, `eta=1/10`
Work done on the system, `W=10J`
As `eta=W/(Q_(1))`
`:. Q_(1)=W/(eta)= 10/(1//10)= 100J`
`:.` Heat rejected by the engine at lower temp.
`Q_(2)=Q_(1)-W=100-10=90J`
When the same engine is used as a refrigerator between the same two temperature, the heat absorbed from the reservoir at lower temperature is same as the heat rejected by the engine at lower temperature `= 90J`.