Question

A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$. When the temperature of cold reservoir raised by $x$, its efficiency decreases to $\frac{1}{6}$. The value of $x$, if the temperature of hot reservoir is $99^{\circ} C$, will be :

• A. $62\, K$
• B. $16.5\, K$
• C. $33\, K$
• D. $66 \,K$

Answer 1

Answer: (A)

$ T _{ H }=99^{\circ} C =99+273 $
$ =372 K$
$ 1-\frac{ T _{ C }}{ T _{ H }}=\frac{1}{3}$
$ \frac{ T _{ C }}{ T _{ H }}=\frac{2}{3} (1) \Rightarrow T _{ C }=\frac{2}{3} \times 372 $
$ =2 \times 124=248 K $
$ 1-\frac{ T _{ C }+ X }{ T _{ H }}=\frac{1}{6}$
$ \frac{5}{6}=\frac{ T _{ C }+ X }{ T _{ H }} $
$ \frac{5}{6}=\frac{248+ X }{372}$
$248+X=5 \times 62 $
$ X =310-248=62 K$