Question
The efficiency of a Carnot engine for the following temperature differences same is :
(i) 100 K and 500 K
(ii) TK and 900 K
Find the temperature $T$ of the sink.
(i) 100 K and 500 K
(ii) TK and 900 K
Find the temperature $T$ of the sink.
✓
Answer
Efficiency of engine at first temperature
$\begin{array}{l}\eta_1=\frac{T_1-T_2}{T_1} \\\eta_1=\frac{500-100}{500}=\frac{400}{500}=\frac{4}{5} \\\eta_1=\frac{4}{5}\end{array}$
Let the efficiency of engine at second temperature be $\eta_2$
But
$\begin{array}{l}\eta_2=\frac{900-T}{900} \\\text {But}\quad\eta_1=\eta_2\end{array}$
$\begin{aligned}\therefore \frac{4}{5} & =\frac{900- T }{900} \\ 3600 & =4500-5 T \\\therefore 5 T & =4500-3600=900 \\ T & =\frac{900}{5}=180 K\end{aligned}$
$\begin{array}{l}\eta_1=\frac{T_1-T_2}{T_1} \\\eta_1=\frac{500-100}{500}=\frac{400}{500}=\frac{4}{5} \\\eta_1=\frac{4}{5}\end{array}$
Let the efficiency of engine at second temperature be $\eta_2$
But
$\begin{array}{l}\eta_2=\frac{900-T}{900} \\\text {But}\quad\eta_1=\eta_2\end{array}$
$\begin{aligned}\therefore \frac{4}{5} & =\frac{900- T }{900} \\ 3600 & =4500-5 T \\\therefore 5 T & =4500-3600=900 \\ T & =\frac{900}{5}=180 K\end{aligned}$
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