Question
For an engine the sink temperature is 300 K and the source temperature is $T$. If absorbs 200 joules of heat from the source and gives 120 joules of heat to the sink. Find the temperature of the source and efficiency of the engine.
✓
Answer
$\begin{aligned}T_2 & =300 K \\T_1 & =T \text { (Assume) } \\Q_1 & =200 \text { Joule } \\Q_2 & =120 \text { Joule } \\T_1 & =? \\\eta & =?\end{aligned}$
Efficiency of engine
$\begin{aligned}\eta & =\frac{Q_1-Q_2}{Q_1}=\frac{200-120}{200}=\frac{80}{200} \\\eta & =0.4\end{aligned}$
$\text {Therefore,}\quad\eta \%=0.4 \times 100=40 \%$
We know that
$\eta=\frac{T_1-T_2}{T_1}$
$0.4=\frac{T-300}{T}$
$ \begin{array}{l} \Rightarrow \quad 0.4 T=T-300 \\ \Rightarrow \quad 300=T-0.4 T \\ \Rightarrow \quad 300=0.6 T \end{array} $
$\begin{aligned} \Rightarrow \quad T & =\frac{300}{0.6}=\frac{3000}{6}=500 K \\ T & =500 K\end{aligned}$
Efficiency of engine
$\begin{aligned}\eta & =\frac{Q_1-Q_2}{Q_1}=\frac{200-120}{200}=\frac{80}{200} \\\eta & =0.4\end{aligned}$
$\text {Therefore,}\quad\eta \%=0.4 \times 100=40 \%$
We know that
$\eta=\frac{T_1-T_2}{T_1}$
$0.4=\frac{T-300}{T}$
$ \begin{array}{l} \Rightarrow \quad 0.4 T=T-300 \\ \Rightarrow \quad 300=T-0.4 T \\ \Rightarrow \quad 300=0.6 T \end{array} $
$\begin{aligned} \Rightarrow \quad T & =\frac{300}{0.6}=\frac{3000}{6}=500 K \\ T & =500 K\end{aligned}$
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