A Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^\circ \mathrm{C}$. The te

Question

A Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^\circ \mathrm{C}$. The temperature of the sink is:

Accepted Answer

The efficiency 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 in Kelvin and $T_2$ is the temperature of the sink in Kelvin.
Given, $\eta = 50\% = 0.5$
$T_1 = 327^\circ \mathrm{C} = 327 + 273 = 600 	ext{ K}$
Substituting the values:
$0.5 = 1 - \frac{T_2}{600}$
$\frac{T_2}{600} = 0.5 \implies T_2 = 300 	ext{ K}$
Converting the sink temperature back to Celsius:
$T_2 = 300 - 273 = 27^\circ \mathrm{C}$

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