Question

The efficiency of a Carnot engine is $\frac{1}{6}$. In reducing the sink temperature by 65 K, the efficiency becomes $\frac{1}{3}$. The initial and final temperature between which the engine works, are

• A. ${117}^{o}C,{52}^{o}C$
• B. ${217}^{o}C,{52}^{o}C$
• C. ${317}^{o}C,{52}^{o}C$
• D. ${17}^{o}C,{52}^{o}C$

Answer 1

The correct option is A ${117}^{o}C,{52}^{o}C$

Efficiency of a carnot engine : $\eta$ = $1-\frac{T_C}{T_H}$
given $\eta$ = $\frac{1}{6}$

$1-\frac{T_C}{T_H}$ = $\frac{1}{6}$
$1-\frac{1}{6}$ = $\frac{T_C}{T_H}$
$\frac{5}{6}$ = $\frac{T_C}{T_H}$
5$\times T_H$ = 6$\times T_C$ ...... 1

In reducing the sink temperature by 65K, the efficiency
becomes $\frac{1}{3}$
$1-\frac{T_C-65}{T_H}$ = $\frac{1}{3}$
$1-\frac{1}{3}$ = $\frac{T_C-65}{T_H}$
$\frac{2}{3}$ = $\frac{T_C-65}{T_H}$
2$\times T_H$ = 3$\times T_C$ - 195 ...... 2

From equations 1 and 2 ,
[eqn 1] - 2x[eqn 2]
$T_H=2\times 195=390K={117}^{\circ }$C

Substituting $T_H$ back in equation 1 ,

$T_C$ = $\frac{5}{6}$$T_H$ = $\frac{5}{6}$$390$ = 325 K = ${52}^{\circ }$C

Answer = ${117}^{\circ }$C, ${52}^{\circ }$C