JEE PYQ: Thermodynamics Question 30
Question 30 - 2020 (08 Jan Shift 2)
Three containers $C_1$, $C_2$ and $C_3$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)
| $C_1$ | $C_2$ | $C_3$ | $T$ |
|---|---|---|---|
| $1l$ | $2l$ | – | 60$°$C |
| – | $1l$ | $2l$ | 30$°$C |
| $2l$ | – | $1l$ | 60$°$C |
| $1l$ | $1l$ | $1l$ | $\theta$ |
The value of $\theta$ (in $°$C to the nearest integer) is __________.
Show Answer
Answer: 50
Solution
Let $\theta_1, \theta_2, \theta_3$ be the temperatures of container $C_1, C_2$ and $C_3$ respectively.
Using principle of calorimetry in container $C_1$:
$(\theta_1 - 60) = 2 \times ms(60 - \theta)$
$\theta_1 - 60 = 120 - 2\theta$
$\Rightarrow \theta_1 = 180 - 2\theta$ …(i)
For container $C_2$: $ms(\theta_2 - 30) = 2ms(30 - \theta)$
$\theta_2 = 90 - 2\theta_3$ …(ii)
For container $C_3$: $2ms(\theta_1 - 60) = ms(60 - \theta)$
$\Rightarrow 2\theta_1 + \theta_3 = 180$ …(iii)
Also, $\theta_1 + \theta_2 + \theta_3 = 30$ …(iv)
Adding (i), (ii) and (iii):
$3\theta_1 + 3\theta_2 + 3\theta_3 = 450$
$\Rightarrow \theta_1 + \theta_2 + \theta_3 = 150 \Rightarrow 3\theta = 150 \Rightarrow \theta = 50°$C
BETA
