A Carnot engine is made to work between 200^{ | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature

Vedclass · Accepted Answer

$1 : 1.73$

Vedclass · Answer

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature in Kelvin.For the first case: $T_H = 200 + 273 = 473 \ K$ and $T_L = 0 + 273 = 273 \ K$.$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.For the second case: $T_H = 0 + 273 = 273 \ K$ and $T_L = -200 + 273 = 73 \ K$.$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.The ratio of efficiencies is $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx \frac{1}{1.73}$.Thus,the ratio is $1 : 1.73$.

$A$ Carnot engine is made to work between $200^{\circ}C$ and $0^{\circ}C$ first and then between $0^{\circ}C$ and $-200^{\circ}C$. The ratio of efficiencies of the engine in the two cases is

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