A Carnot engine with sink's temperature at 17 | vedclass

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Question

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

Vedclass · Accepted Answer

$145$

Vedclass · Answer

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given $T_2 = 17 + 273 = 290 \, K$ and initial efficiency $\eta_1 = 0.50$.$0.50 = 1 - \frac{290}{T_1} \Rightarrow \frac{290}{T_1} = 0.50 \Rightarrow T_1 = 580 \, K$.For the new efficiency $\eta_2 = 0.60$ with the same sink temperature:$0.60 = 1 - \frac{290}{T_1'} \Rightarrow \frac{290}{T_1'} = 0.40 \Rightarrow T_1' = \frac{290}{0.40} = 725 \, K$.The change in source temperature is $\Delta T = T_1' - T_1 = 725 - 580 = 145 \, K$.

$A$ Carnot engine with sink's temperature at $17\,^{\circ}C$ has $50\%$ efficiency. By how much should its source temperature be changed to increase its efficiency to $60\%$? (in $K$)

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