The efficiency of a Carnot cycle is $\frac{1}{6}$. By lowering the temperature of sink by 65 K , it…
The efficiency of a Carnot cycle is $\frac{1}{6}$. By lowering the temperature of sink by 65 K , it increases to $\frac{1}{3}$. The initial and final temperature of the sink are
$400 \mathrm{~K}, 310 \mathrm{~K}$
$525 \mathrm{~K}, 65 \mathrm{~K}$
$309 \mathrm{~K}, 235 \mathrm{~K}$
$325 \mathrm{~K}, 260 \mathrm{~K}$
Solution
The efficiency of carnot engine is
$\eta=1-\frac{T_2}{T_1} \Rightarrow \frac{1}{6}=1-\frac{T_2}{T_1} \Rightarrow T_2=\frac{5}{6} T_1$ ....(i)
Again,
$\frac{1}{3}=1-\frac{T_2-65}{T_1} \Rightarrow \frac{T_2-65}{T_1}=\frac{2}{3}$ ....(ii)
From eq(i) and (ii), we get
$\frac{5 T_1}{6}-65=\frac{2 T_1}{3} \Rightarrow T_1=390 \mathrm{~K}$
$\therefore \quad$ Initial temperature of sink, $\mathrm{T}_2=\frac{5}{6} \mathrm{~T}_1=\frac{5}{6} \times 390=325 \mathrm{~K}$
Final temperature of $\sin \mathrm{k}, \mathrm{T}_2-65=(325-65)=260 \mathrm{K}$.