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Q.
A carnot engine whose efficiency is $40 \%$, receives heat at $500\, K$. If the efficiency is tobe $50 \%$, the source temperature for the same exhaust temperature is
TS EAMCET 2020
Solution:
We know that, efficiency of a Carnot engine is given by
$\eta=\left(1-\frac{T_{2}}{T_{1}}\right)$
Given, $\eta_{1}=0.4$ and $T_{1}=500\, K$
(temperature of source)
So, $0.4=1-\frac{T_{2}}{500}$
$\Rightarrow \frac{T_{2}}{500}=1-0.4=0.6 $
$\Rightarrow T_{2}=0.6 \times 500=300\, K $
(temperature of sink)
Now, in the second gas,
$\eta=0.5$ and $T_{2}=300 \,K $
$\therefore 0.5=\left(1-\frac{300}{T_{1}}\right) $
$\Rightarrow \frac{300}{T_{1}}=1-0.5=0.5$
$ \Rightarrow T_{1}=\frac{300}{0.5}=600\, K$