A heat engine receives `50 kcal` of heat from the source per cycle, and operates with an efficiency of `20%`. The heat rejected by engine to the sink per cycle is

Options

A. `40 kcal`
B. `25 kcal`
C. `30 kcal`
D. `50 kcal`

Answer

A

Explanation

Given `Q_1 = 50 kcal and eta = 20%`
Using `eta = W/(Q_1)`, we have `W = etaxxQ_(1)`
i.e., work obtained per cycle `W = 20% xx 50 kcal`
`=10kcal = 42 kJ`
Since `Q_(1) = Q_(2)+W` so `Q_(2) = Q-W`
i.e., heat rejected to the sink per cycle `Q_(2)`
`=50kcal-10kcal=40kcal`.