The efficiency of a heat engine that works be | vedclass

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(C) Step $1$: Find the temperature where Celsius and Fahrenheit scales coincide.Let the temperature be $T_1$. The relation is $\frac{C}{5} = \frac{F-3

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$60$

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(C) Step $1$: Find the temperature where Celsius and Fahrenheit scales coincide.Let the temperature be $T_1$. The relation is $\frac{C}{5} = \frac{F-32}{9}$. Setting $C = F = T_1$:$\frac{T_1}{5} = \frac{T_1-32}{9} \Rightarrow 9T_1 = 5T_1 - 160 \Rightarrow 4T_1 = -160 \Rightarrow T_1 = -40^{\circ}C$.In Kelvin,$T_1 = -40 + 273.15 = 233.15 \ K$.Step $2$: Find the temperature where Kelvin and Fahrenheit scales coincide.Let the temperature be $T_2$. The relation is $\frac{K-273.15}{5} = \frac{F-32}{9}$. Setting $K = F = T_2$:$\frac{T_2-273.15}{5} = \frac{T_2-32}{9} \Rightarrow 9T_2 - 2458.35 = 5T_2 - 160 \Rightarrow 4T_2 = 2298.35 \Rightarrow T_2 \approx 574.59 \ K$.Step $3$: Calculate the efficiency $\eta$ of the Carnot engine.$\eta = 1 - \frac{T_{cold}}{T_{hot}} = 1 - \frac{233.15}{574.59} \approx 1 - 0.405 = 0.595 \approx 60 \%$.

The efficiency of a heat engine that works between the temperatures where Celsius-Fahrenheit scales coincide and Kelvin-Fahrenheit scales coincide is (approximately) (in $\%$)

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