Question

The efficiency of the Carnot engine is $1/6$. On decreasing the temperature of the sink by $65\text{K}$, the efficiency increases to $1/3$. Find the temperature of the source and the sink.

• A. $T_1=496\text{K},T_2=325\text{K}$
• B. $T_1=390\text{K},T_2=325\text{K}$
  
• C. $T_1=390\text{K},T_2=425\text{K}$
• D. $T_1=325\text{K},T_2=390\text{K}$

Answer 1

The correct option is B $T_1=390\text{K},T_2=325\text{K}$


We have,
$\eta =\frac{T_1-T_2}{T_1}$, where $T_1\text{and}{T}_2$ are the temperatures of source and sink respectively.
$\eta =\text{Effeciency}$

$\therefore \eta =\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$\Rightarrow T_2=\frac{5}{6}{T}_1.......\left(i\right)$
Now the temprature the sink is reduced by $65\text{K}$.

$\Rightarrow \text{Temprature of the sink}=(T_2-65)$

$\Rightarrow \eta =\frac{T_1-(T_2-65)}{T_1}=\frac{1}{3}$
$\Rightarrow 3T_1-3T_2+195=T_1$
$\Rightarrow 2T_1-3T_2=-195$
Now putting (i)
$\Rightarrow 2T_1-3\times \frac{5}{6}{T}_1=-195$
$\Rightarrow -\frac{1}{2}{T}_1=-195$
$\Rightarrow T_1=390\text{K}$
Putting this value in (i) to get
$\Rightarrow T_2=\frac{5}{6}\times 390=325\text{K}$