JEE PYQ: Thermodynamics Question 28
Question 28 - 2020 (07 Jan Shift 2)
Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_1$ and $T_2$. The temperature of the hot reservoir of the first engine is $T_1$ and the temperature of the cold reservoir of the second engine is $T_2$. $T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_1$ and $T_2$, if both the engines perform equal amount of work?
(1) $T = \frac{2T_1 T_2}{T_1 + T_2}$
(2) $T = \frac{T_1 + T_2}{2}$
(3) $T = \sqrt{T_1 T_2}$
(4) $T = 0$
Show Answer
Answer: (2)
Solution
Let $Q_H =$ Heat taken by first engine, $Q_L =$ Heat rejected by first engine, $Q_2 =$ Heat rejected by second engine.
Work done by 1st engine = work done by 2nd engine
$W = Q_H - Q_L = Q_L - Q_2 \Rightarrow 2Q_L = Q_H + Q_2$
$2 = \frac{Q_H}{Q_L} + \frac{Q_2}{Q_L}$
Let T be the temperature of cold reservoir of first engine. Then in Carnot engine:
$\frac{Q_H}{Q_L} = \frac{T_1}{T}$ and $\frac{Q_L}{Q_2} = \frac{T}{T_2}$
$\Rightarrow 2 = \frac{T_1}{T} + \frac{T_2}{T}$ using (i)
$\Rightarrow 2T = T_1 + T_2 \Rightarrow T = \frac{T_1 + T_2}{2}$
BETA
