A Carnot engine whose efficiency is 40% takes in heat from source maintained at a temperature of 500 K. It…
A Carnot engine whose efficiency is 40% takes in heat from source maintained at a temperature of 500 K. It is desired to have an
engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be
1200 K
750 K
600 K
800 K
Solution
Efficiency of Carnot engine, $\eta_1=40 \%=0.4$
Temperature of source $T_1=500 \mathrm{~K}$
Temperature of $\operatorname{sink} T_2=T$
Desired efficiency, $\eta_{\varepsilon}=60 \%=0.6$
Temperature of $\operatorname{sink} T_2=T$
Temperature of source $T_1=$ ?
For case 1 Efficiency, $\eta_1=1-\frac{T_2}{T_1}=1-\frac{T}{500}$
$
\begin{aligned}
& \Rightarrow & \frac{4}{10} & =\frac{500-T}{500} \\
\Rightarrow & & 200 & =500-T \\
\Rightarrow & & T & =300 \mathrm{~K}
\end{aligned}
$
For case 2 Efficiency, $\eta_2=1-\frac{T_2}{T_1}=1-\frac{T}{T_1^{\prime}}$
$
\begin{aligned}
\frac{6}{10} & =1-\frac{300}{T_1^{\prime}} \\
\Rightarrow \quad \frac{300}{T_1^{\prime}} & =1-\frac{6}{10} \Rightarrow \frac{300}{T_1^{\prime}}=\frac{4}{10} \\
\Rightarrow \quad \frac{300}{4} \times 10 & =T_1^{\prime} \Rightarrow T_1^{\prime}=750 \mathrm{~K}
\end{aligned}
$