Question

In Carnot engine, efficiency is $40\%$ at hot reservoir temperature $T$. For efficiency $50\%$, what will be the temperature of hot reservoir?

• A. $\frac{2T}{5}$
• B. $6T$
• C. $\frac{6T}{5}$
• D. $\frac{T}{5}$

Answer 1

Answer: (C)

The efficiency, $\eta=\frac{\text { work done }}{\text { heat input }}=\frac{W}{Q}$
$\eta=1-\frac{T_{2}}{T_{1}}$
where, $T_{2}=$ temperature of sink
and $T_{1}=$ temperature of hot reservoir
$\frac{40}{100} =1-\frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}}=0.6$
$ \Rightarrow T_{2}=0.6 T_{1}$
Now, $ \frac{50}{100} =1-\frac{T_{2}}{T_{1}^{'}} \Rightarrow \frac{T_{2}}{T_{1}^{'}}=0.5$
$ \frac{0.6 T_{1}}{T_{1}'} =0.5$
$T_{1}' =\frac{0.6}{0.5} T_{1}$
$\Rightarrow T_{1}^{'}=\frac{6}{5} T_{1}$
$T_{1}^{'} =\frac{6}{5} T $
$[\because \,T_{1}=T]$