Question

Three containers $C_1,C_2$ and $C_3$ have water at different temperatures. The table below shows the final temperature $T$ when different amount of water (given in litres) are taken from each container and mixed (assume no loss of heat during the process)

$C_1$	$C_2$	$C_3$	$T$
$1\text{L}$	$2\text{L}$	-	${60}^{\circ }\text{C}$
-	$1\text{L}$	$2\text{L}$	${30}^{\circ }\text{C}$
$2\text{L}$	-	$1\text{L}$	${60}^{\circ }\text{C}$
$1\text{L}$	$1\text{L}$	$1\text{L}$	$\theta$

The value of $\theta$ (in °C to the nearest integer) is

Answer 1

Let ${\theta }_1,{\theta }_2,{\theta }_3$ be the temperatures of the container $C_1,C_2$ and $C_3$ respectively.
Using principle of calorimetry in case 1, we have
$ms({\theta }_1-60)=2ms(60-{\theta }_2)$
$\Rightarrow {\theta }_1-60=120-2{\theta }_2$
$\Rightarrow {\theta }_1=180-2{\theta }_2...\left(i\right)$

For case 2,
$ms({\theta }_2-30)=2ms(30-{\theta }_3)$
$\Rightarrow {\theta }_2=90-2{\theta }_3...\left(ii\right)$

For case 3,
$2ms({\theta }_1-60)=ms(60-{\theta }_3)$
$\Rightarrow 2{\theta }_1-120=60-{\theta }_3$
$\Rightarrow 2{\theta }_1+{\theta }_3=180...\left(iii\right)$
For case 4,
${\theta }_1+{\theta }_2+{\theta }_3=3\theta ....\left(iv\right)$
Adding $\left(i\right),\left(ii\right)$ and $\left(iii\right)$
$3({\theta }_1+{\theta }_2+{\theta }_3)=450$
$\Rightarrow {\theta }_1+{\theta }_2+{\theta }_3=150$
On comparision with $\left(iv\right)$
$\Rightarrow 3\theta =150\Rightarrow \theta ={50}^{\circ }C$