Question

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

Answer 1

Correct Answer - C
(c) The efficiency `(eta)` of a Carnot engine and the coefficient
of performance `(beta)`
of a refrigerator are related as
`beta=(1-eta)/eta Here, eta=1/10 :. Beta=(1-1/10)/(1/10)=9`.
Also, Coefficient of performance `(beta)` is given by `beta=(Q_2)/W,`
where `Q_2` is the energy absorbed from the reservoir.
or, `9=(Q_2)/10 :. Q_2=90J`