JEE PYQ: Thermodynamics Question 10
Question 10 - 2021 (25 Feb Shift 2)
A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be __________.
Show Answer
Answer: 208
Solution
$\eta = \frac{W}{Q_s} = \frac{1}{4}$
$\frac{1}{4} = 1 - \frac{T_2}{T_1}$
$\frac{T_2}{T_1} = \frac{3}{4}$
When the temperature of the sink is reduced by 52K then its efficiency is doubled: $\frac{1}{2} = 1 - \frac{(T_2 - 52)}{T_1}$
$\frac{T_2 - 52}{T_1} = \frac{1}{2}$
$\frac{3}{4} - \frac{52}{T_1} = \frac{1}{2}$
$\frac{52}{T_1} = \frac{1}{4}$
$T_1 = 208$ K
BETA
