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 (environment) is ...................
(A) 40 KJ
(B) 20 J
(C) 30 KJ
(D) 20 KJ

Answer 1

Answer:

(D)20kJ

Explanation:

Efficiency=(Output/Input)×100%

Input=50,000J

Output=x(Unknown)

Efficiency=40%

Put the values in the formula:

40%=(x/50,000)×100%

40/100=x/50,000

0.4=x/50,000

0.4×50,000=x

x=20,000J or 20kJ