Q. A Carnot engine has same efficiency between (i) 100 K and 500 K, (ii) $T$ K and 900 K. The value of $T$ is
A) 180 K
B) 90 K
C) 270 K
D) 360 K
Explanation
(a) $\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{100}{500} = 1 - \frac{T}{900}$
$\therefore\ \ \frac{T}{900} = \frac{1}{5}\ \ or\ \ \ T = 180\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.