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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperatu

Vedclass · Accepted Answer

$1.73:1$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.For the first case: $T_1 = 200 + 273 = 473\,K$ and $T_2 = 0 + 273 = 273\,K$.$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.For the second case: $T_1 = 0 + 273 = 273\,K$ and $T_2 = -200 + 273 = 73\,K$.$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.The ratio $\frac{\eta_2}{\eta_1} = \frac{200/273}{200/473} = \frac{473}{273} \approx 1.73$.Thus,the ratio is approximately $1.73:1$.

$A$ 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 $\left( \frac{\eta_2}{\eta_1} \right)$ of the engine in the two cases is:

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