Q. Three designs are proposed for an engine operating between 500 K and 300 K. For 1 kcal of heat input, design $A$ claims to produce 3000 J of work, design $B$ claims to produce 2000 J of work and design $C$ claims to produce 1680 J of work. The design which is possible
A) $A$ only
B) $B$ only
C) $C$ only
D) All of these
Explanation
(c) Maximum value of efficiency
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{500} = \frac{2}{5}$
As $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{W}{Q_{1}}$
$W = \eta Q_{1} = \frac{2}{5} \times 1000\ cal$
$= 400 \times 4.2\ J = 1680\ J$
As no engine can produce more than 1680 J, disigns A and B are not possible.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.