Q. A Carnot engine takes $3 \times 10^{6}$ cal of heat from a reservoir at $627℃$ and gives it to a sink at $27℃$. The work done by the engine is
A) $4.2 \times 10^{6}\ J$
B) $8.4 \times 10^{6}\ J$
C) $16.8 \times 10^{6}\ J$
D) Zero
Explanation
(b) Given, $T_{1} = 627 + 273 = 900K$
$\ \ \ \ \ \ \ \ \ \ \ \ Q_{1} = 3 \times 10^{6}\ cal$
$T_{2} = 27 + 273 = 300\ K$
$\therefore\ \ \ \ \ \ \ \frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$
$\Rightarrow \ \ \ \ \ \ Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{300}{900} \times 3 \times 10^{6}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 10^{6}\ cal\ $
Work done $= Q_{1} - Q_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3 \times 10^{6} - 1 \times 10^{6} = 2 \times 10^{6}cal\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times 4.2 \times 10^{6}\ J = 8.4 \times 10^{6}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.