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

In Carnot engine efficiency is $40 \%$ at hot reservoir temperature $T$. For efficiency $50 \%$ what will be temperature of hot reservoir?
  1. $\frac{T}{5}$
  2. $\frac{2 \mathrm{~T}}{5}$
  3. $6 \mathrm{~T}$
  4. $\frac{6 \mathrm{~T}}{5}$

Solution

$\eta=1-\frac{T_{2}}{T_{1}}$
$\mathrm{T}_{1}=\mathrm{T}$ (Temperature of hot reservoir) For $\eta=40 \%$
$\frac{40}{100}=1-\frac{T_{2}}{T} \Rightarrow \frac{T_{2}}{T}=\frac{3}{5} \Rightarrow T_{2}=\frac{3}{5} T$
For $\eta=50 \%$,
$\frac{50}{100}=1-\frac{\frac{3}{5} T}{T_{1}} \Rightarrow T_{1}=\frac{6 T}{5}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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