Q. A Carnot’s engine operates with source at $127℃$ and sink at $27℃.$ If the source supplies 40 kJ of heat energy, the work done by the engine is
A) 30 kJ
B) 10 kJ
C) 4 kJ
D) 1 kJ
Explanation
(b) Efficiency, $\eta = 1 - \frac{T_{2}}{T_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{(273 + 27)}{(273 + 127)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{300}{400} = \frac{1}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{Work\ done}{Heat\ supplied}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{4} = \frac{W}{40}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ W = 10\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.