Question

If a heat engine absorbs 50 KJ heat from a heat source and has efficiency of 40%, then the heat released by it in heat sink is 

(A) 40 KJ 

(B) 30 KJ 

(C) 20 J 

(D) 20 KJ

Answer 1

Correct option: (B) 30 KJ

Explanation:

Given: heat absorbed = Q = 50 KJ

η = 40% = 0.4

η = 1 – (Q₂ / Q₁)

0.4 = 1 – (Q₂ / Q₁)

(Q₂ / Q₁) = 0.6

Given : heat given by heat source = Q₁ = 50KJ

∴ Q₂ = 0.6 Q₁ 

Q₂ = 0.6 × 50

Q₂ = 30 KJ 

i.e. heat released by sink = 30 KJ