In a Carnot engine,when {T_2} = {0^o}C and {T | vedclass

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.First,con

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$0.577$

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.First,convert the temperatures to Kelvin: $T(K) = T(^oC) + 273.15$. For calculation,we use $273$.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 $T_2 = -200 + 273 = 73 \ K$.$\eta_2 = 1 - \frac{73}{273} = \frac{273 - 73}{273} = \frac{200}{273}$.Now,calculate the ratio: $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx 0.577$.

In a Carnot engine,when ${T_2} = {0^o}C$ and ${T_1} = {200^o}C$,its efficiency is ${\eta _1}$. When ${T_1} = {0^o}C$ and ${T_2} = -{200^o}C$,its efficiency is ${\eta _2}$. What is the ratio ${\eta _1}/{\eta _2}$?

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