The efficiency of heat engine is 30% If it gives 30 KJ heat to the heat sink, than it should have absorbed ……. KJ heat from heat source.
(A) 42.8
(B) 39
(C) 29
(D) 9
Correct option: (A) 42.8
Explanation:
η = 30%
η = 1 – (Q2 / Q1)
Q1 : heat absorbed
Q2 : heat given to sink
∴ 0.3 = 1 (Q2 / Q1)
(Q2 / Q1) = 0.7
Given : Q2 = 30 KJ
Hence (30 / 0.7) = Q1
Q1 = 42.8 KJ
i.e. heat absorbed from heat source is 42.8 KJ
