Q. A Carnot engine operating between temperature $T_{1}$and $T_{2}$ has efficiency $\frac{1}{6}.$ When $T_{2}$ is lowered by 62 K, its efficiency increase to $\frac{1}{3}.$ Then $T_{1}and\ T_{2}$are, respectively
A) 372 K and 320 K
B) 330 K and 268 K
C) 310 K and 248 K
D) 372 K and 310 K
Explanation
(d) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta_{1} = 1 - \frac{T_{2}}{T_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{2}}{T_{1}} = \frac{5}{6}$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta}_{2} = 1 - \frac{T_{2} - 62}{T_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{3} = 1 - \frac{T_{2} - 62}{T_{1}}$
On solving Eqs. (i) and (ii) , we get
$T_{1} = 372\ K\ and\ T_{2} = 310\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.