Question

Carnot engine has efficiency $\frac{1}{5}$. Efficiency becomes $1/3$ when the temperature of sink is decreased by $50K$. Temperature of the sink (in $K$ ) is

Answer 1

Answer: 300

The efficiency of the Carnot engine
$\eta =\frac{\text{ Work output }}{\text{ Heat input }}-\frac{W}{Q_H}$
or $\eta =\frac{Q_H-Q_L}{Q_H}=1-\frac{Q_L}{Q_H}$
Also we can show that
$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$
$\therefore \eta =1-\frac{T_L}{T_H}$
where $T_L$ is temperature of $\sin k$ and $T_H$ is the temperature of hot reservoir.
According to question
$\frac{1}{3}=1-\frac{T_L}{T_H}$ ... (i)
$\frac{1}{3}=1-\frac{T_L-50}{T_H}$ ... (ii)
From Eq. (i)
$\Rightarrow T_H=\frac{5}{4}{T}_L$
Substituting value of $T_H$ in Eq. (ii), we get
$\frac{1}{3}=1-\frac{5}{6}$
or $\frac{4\left(T_L-50\right)}{5T_1}=\frac{2}{3}$
or $T_L-50=\frac{2}{3}\times \frac{5}{4}{T}_L$
or $T_L-\frac{5}{6}{T}_L=50$
$\therefore T_L=50\times 6=300K$