Question

The efficiency of a heat engine is 30% If it gives 30KJ heat to the heat sink,then it should have absorbed.......KJ heat from heat source.

Answer 1

Given efficiency of heat engine = 30% = 0.3

Heat rejected Q₂ = 30 KJ

To determine Heat supplied to the engine Q₁ = ?

We know efficiency η = (Output/Input) = 1 -(Q₂/Q₁)

Q₁ = (- 30/(0.3 - 1) = 42.85 J