If a Carnot engine functions at a source temp | vedclass

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(D) 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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$10$

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(D) 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 = 127 + 273 = 400 	ext{ K}$$T_2 = 87 + 273 = 360 	ext{ K}$Now,substitute these values into the efficiency formula:$\eta = \frac{T_1 - T_2}{T_1} = \frac{400 - 360}{400}$$\eta = \frac{40}{400} = 0.1$To express this as a percentage,multiply by $100$:$\eta = 0.1 	imes 100 = 10\%$

If a Carnot engine functions at a source temperature of $127^{\circ}C$ and a sink temperature of $87^{\circ}C$,what is its efficiency in $\%$?

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