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
$\frac{40}{100}=1-\frac{T_2}{T_1}$
$\frac{T_2}{T_1}=0.6$
$\Rightarrow T_2=0.6T_1$
Now, $\frac{50}{100}=1-\frac{T_2}{T_1^{{}^{\prime }}}\Rightarrow \frac{T_2}{T_1^{{}^{\prime }}}=0.5$
$\frac{0.6T_1}{T_1^{\prime }}=0.5$
$T_1^{\prime }=\frac{0.6}{0.5}{T}_1$
$\Rightarrow T_1^{{}^{\prime }}=\frac{6}{5}{T}_1$
$T_1^{{}^{\prime }}=\frac{6}{5}T$
$[\because T_1=T]$