In a Carnot engine,when the temperatures are | vedclass

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)

Vedclass · Accepted Answer

$0.58$

Vedclass · Answer

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)$.Case $1$: $T_{source} = 200^{\circ}C = 473 \ K$ and $T_{sink} = 0^{\circ}C = 273 \ K$.$\eta_1 = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$.Case $2$: $T_{source} = 0^{\circ}C = 273 \ K$ and $T_{sink} = -200^{\circ}C = 73 \ K$.$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.Now,calculating the ratio:$\frac{\eta_1}{\eta_2} = \frac{200}{473} 	imes \frac{273}{200} = \frac{273}{473} \approx 0.577 \approx 0.58$.

In a Carnot engine,when the temperatures are $T_2 = 0^{\circ} C$ and $T_1 = 200^{\circ} C$,its efficiency is $\eta_1$. When the temperatures are $T_1 = 0^{\circ} C$ and $T_2 = -200^{\circ} C$,its efficiency is $\eta_2$. Then the value of $\frac{\eta_1}{\eta_2}$ is:

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