A Carnot engine operates between 627^{circ}C | vedclass

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Question

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

Vedclass · Accepted Answer

$44.4$

Vedclass · Answer

(C) The efficiency $(\eta)$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.Given: $T_1 = 627^{\circ}C = 627 + 273 = 900 \ K$ and $T_2 = 227^{\circ}C = 227 + 273 = 500 \ K$.Substituting the values: $\eta = 1 - \frac{500}{900} = 1 - \frac{5}{9} = \frac{4}{9}$.To express as a percentage: $\eta = \frac{4}{9} 	imes 100 \% \approx 44.44 \%$.Therefore, the correct option is $C$.

$A$ Carnot engine operates between $627^{\circ}C$ and $227^{\circ}C$. The efficiency of the Carnot engine is ....... $\%$

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