An ideal Carnot heat engine has an efficiency | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the hot reservoir and $T_2$ is the te

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$427$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the hot reservoir and $T_2$ is the temperature of the cold reservoir in Kelvin.Given $\eta = 30\% = 0.3$.The temperature of the hot reservoir $T_1 = 727 + 273 = 1000 	ext{ K}$.Substituting the values: $0.3 = 1 - \frac{T_2}{1000}$.Rearranging the equation: $\frac{T_2}{1000} = 1 - 0.3 = 0.7$.$T_2 = 0.7 	imes 1000 = 700 	ext{ K}$.To convert the temperature back to Celsius: $T_2(^{\circ}C) = 700 - 273 = 427^{\circ}C$.

An ideal Carnot heat engine has an efficiency of $30\%$. It absorbs heat from a hot reservoir at $727^{\circ}C$. The temperature of the cold reservoir is .... $^{\circ}C$.

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