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)

$\frac{6T}{5}$

The efficiency, $\eta =\frac{workdone}{heatinput}=\frac{W}{Q}$

$\eta =1-\frac{T_2}{T_1}$

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, $\frac{50}{100}=1-\frac{T_2}{T_1^{\prime }}\Rightarrow \frac{T_2}{T_1^{\prime }}=0.5$

$\frac{0.6T_1^{\prime }}{T_1^{\prime }}=0.5$

$T_1^{\prime }=\frac{0.6}{0.5}{T}_1\Rightarrow T_1^{\prime }=\frac{6}{5}{T}_1$

$T_1=\frac{6}{5}T$ $[\because T_1=T^{\prime }]$