Q. A Carnot engine working between 450 K and 600 K has a work output of 300 J/cycle. The amount of heat energy supplied to the engine from the source in each cycle is
A) 400 J
B) 800 J
C) 1600 J
D) 1200 J
Explanation
(d) Given, $T_{1} = 600\ K,\ T_{2} = 450\ K\ and\ W = 300\ J$
Efficiency of Carnot engine
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{W}{Q} = 1 - \frac{T_{2}}{T_{1}}$
$or\ \ \ \ \frac{W}{Q} = 1 - \frac{T_{2}}{T_{1}}$
$or\ \frac{W}{Q} = 1 - \frac{450}{600}$
$or\ \ \frac{W}{Q} = \frac{1}{4}$
$or\ \ \ Q = 4W$
$or\ \ Q = 4 \times 300 \Rightarrow Q = 1200\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.