Q. A Carnot engine whose low-temperature reservoir is at $27℃$ has efficiency $37.5\%$. The high-temperature reservoir is at
A) $480℃$
B) $327℃$
C) $307℃$
D) $207℃$
Explanation
(d) $T_{2} = 27 + 273 = 300\ K,\eta = 37.5\%$
As $\eta = 1 - \frac{T_{2}}{T_{1}}$
$\therefore\frac{37.5}{100} = 1 - \frac{300}{T_{1}}$
Or$\ \ \ \frac{300}{T_{1}} = \frac{62.5}{100} = \frac{5}{8}$
$T_{1} = \frac{2400}{5} = 480\ K = 480 - 273 = 207℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.