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

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$1/2$

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(A) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.First,convert the temperatures from Celsius to Kelvin:$T_1 = 727 + 273 = 1000 \ K$$T_2 = 227 + 273 = 500 \ K$Now,substitute these values into the efficiency formula:$\eta = 1 - \frac{500}{1000} = 1 - 0.5 = 0.5$Therefore,the maximum possible efficiency is $1/2$.

The source temperature of a heat engine is $727^{\circ}C$ and the sink temperature is $227^{\circ}C$. What is the maximum possible efficiency of this engine?

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