Q. An engine takes in compressed steam at $127℃$ and rejects it at $47℃$. Efficiency of the engine is
A) 60%
B) 35%
C) 20%
D) 40%
Explanation
(c) $\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{(47 + 273)}{(127 + 273)} = 1 - \frac{320}{400} = \frac{1}{5} = 20\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.