Q. If a Carnot’s engine functions at source temperature $127℃$ and at sink temperature $87℃$, what is its efficiency
A) 10%
B) 25%
C) 40%
D) 50%
Explanation
(a) $\eta = \frac{T_{1} - T_{2}}{T_{1}} = \frac{(127 + 273) - (87 + 273)}{(127 + 273)}$
$= \frac{400 - 360}{400} = 0.1 \rightarrow 10\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.