Question

A reversible heat engine converts $\frac{1}{6}th$ of heat which it absorbs from source into work. When the temperature of sink is reduced by ${62}^{o}C$, its efficiency is doubled. The temperature of the source is

• A. $372K$
• B. $272K$
• C. $172K$
• D. $72K$

Answer 1

The correct option is A $372K$

Efficiency of the engine is given by : $\eta$ = $1-\frac{T_C}{T_H}$ = $\frac{W_{Done}}{Q_{Supplied}}$
given that it converts 1/6 th of heat into work .
$\therefore$ $\eta$ = $\frac{1}{6}$ = $1-\frac{T_C}{T_H}$
$\frac{T_C}{T_H}$ = $\frac{5}{6}$
$6\times$$T_C$ = $5\times$$T_H$ ..... 1

When the sink temperature is reduced by $62$ , effficience is doubled.
$\therefore$ $\frac{1}{3}$ = $1-\frac{T_C-62}{T_H}$

$\frac{T_C-62}{T_H}$ = $\frac{2}{3}$
$3T_C$ - 186 = $2T_H$ ..... 2
solving equations 1 and 2,
[eqn 1] - 2$\times$[eqn 2]

$T_H=372K$