Question

A reversible heat engine converts one-fourth of heat input into work. When the temperature of the sink is reduced by ${150}^0$, its efficiency is doubled. The temperature of the source is:

• A. ${450}^0$
• B. $327$ K
• C. $500$ K
• D. ${600}^0$

Answer 1

The correct option is A ${450}^0$

Efficiency of heat engine, ${\eta }_1=\frac{T_2-T_1}{T_2}=\frac{1}{6}$

Where, $T_2$is source temperature and $T_1$is sink temperature.

When the temperature of sink is reduced by ${150}^0$, the efficiency is doubled. So, new efficiency is

$2{\eta }_1=\frac{T_2-(T_1-150)}{T_2}$

$2{\eta }_1=\frac{T_2-T_1}{T_2}+\frac{150}{T_2}$

$2{\eta }_1={\eta }_1+\frac{150}{T_2}$

$2\times \frac{1}{6}=\frac{150}{T_2}$

$\frac{1}{3}=\frac{150}{T_2}$

$\frac{1}{3}=\frac{150}{T_2}$

$T_2=450{}^0$