Q. A Carnot engine is made to work between $200℃$ and $0℃$ first and then between $0℃$ to $- 200℃$. The ratio of efficiencies of the engine in the two cases is
A) 1:2
B) 1:1
C) 1.73:1
D) 1:1.73
Explanation
(d) $T_{1} = 200℃ = 200 + 273 = 473\ K$
$T_{2} = 0℃ = 0 + 273 = 273\ K$
$\eta_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{273}{473} = \frac{200}{473}$
Again, $T_{1}^{'} = 0℃ = (0 + 273)K = 273\ K$
$T_{2}^{'} = - 200℃ = ( - 200 + 273)K = 73\ K$
$\eta_{2} = 1 - \frac{T_{2}}{T_{1}^{'}} = 1 - \frac{73}{273} = \frac{200}{273}$
$\frac{\eta_{1}}{\eta_{2}} = \frac{200}{473} \times \frac{273}{200} = \frac{273}{473} = \frac{1}{1.732}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.