Question

The efficiency of a carnot engine is $\frac{1}{6}$. On decreasing the temperature of the sink by $65K$, the efficiency increases to $\frac{1}{3}$. The temperature of the sink (in K) is

Answer 1

We know that the efficiency of a carnot engine is
$\eta =\frac{T_2-T_1}{T_2}$; where $T_1$ and $T_2$ are the temperatures of sink and source respectively.
$\therefore \eta =\frac{T_2-T_1}{T_2}=\frac{1}{6}$
$\Rightarrow 5T_2-6T_1=0\dots \dots \left(1\right)$
Now the temperature of the sink is reduced by 65 K
$\therefore$ temp. of the sink $=(T_1-65)$
$\therefore \eta =\frac{T_2-(T_1-65)}{T_2}=\frac{1}{3}$ $\Rightarrow 2T_2-3T_1=-195\dots \dots \left(2\right)$
On solving eqns. (1) and (2), we get,
$T_1=325\text{K}$
$T_2=390\text{K}$
So, the temperature of the sink $\left(T_1\right)$ is 325 K.