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 litres) are taken from each container and mixed (assume no loss of heat during the process)

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

The value of $\theta$ (in ${}^{\circ }C$ to the nearest integer) is _______.

We know that: $Q=ms\Delta \theta$
Given:

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

Let $C_1$ is at ${\theta }_1,C_2$ is at ${\theta }_2$ and $C_3$ is at ${\theta }_3$
Then according to table and applying law of calorimetry

For $1^{st}$ case

$ms({\theta }_1-60)=2ms(60-{\theta }_2)$

${\theta }_1-60=120-2{\theta }_2$

${\theta }_1=180-2{\theta }_2\dots \left(i\right)$
For $2^{nd}$ case

$ms({\theta }_2-30)=2ms(30-{\theta }_3)$

${\theta }_2=90-2{\theta }_3\dots \left(ii\right)$

For $3^{rd}$ case

$2ms({\theta }_1-60)=ms(60-{\theta }_3)$

$2{\theta }_1-120=60-{\theta }_3$

$2{\theta }_1+{\theta }_3=180\dots \left(iii\right)$

Adding equation $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$3({\theta }_1+{\theta }_2+{\theta }_3)=9\theta$

$\theta ={50}^{\circ }C$