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)

The value of $\theta$ (in ${ }^{0}C$  to the nearest integer) is -----------

$\begin{array}{l}First Mixing : \left(1\right)s \left({\theta }_1-60\right)+\left(2\right)s\left({\theta }_2-60\right)+\left(0\right)s\left({\theta }_3-60\right)=0\text{}\\ \Rightarrow {\theta }_1+2{\theta }_2=180 \dots \dots \dots \dots . \left(1\right)\\ \\ Second Mixing:\left(0\right)s \left({\theta }_1-30\right)+\left(1\right)s\left({\theta }_2-30\right)+\left(2\right)s\left({\theta }_3-30\right)=0\text{}\\ \Rightarrow {\theta }_2+2{\theta }_3=90 \dots \dots \dots \dots . \left(2\right)\\ \\ Third mixing: \left(2\right)s \left({\theta }_1-60\right)+\left(0\right)s\left({\theta }_2-60\right)+\left(1\right)s\left({\theta }_3-60\right)=0\text{}\\ \Rightarrow 2{\theta }_1+{\theta }_3=180 \dots \dots \dots \dots . \left(3\right)\\ \\ Solving equations \left(1\right) \left(2\right) and \left(3\right) we get\\ \text{}\begin{array}{l}{\theta }_1=80, {\theta }_2=50, {\theta }_3=20\\ \\ Fourth mixing: \left(1\right)s \left({\theta }_1-\theta \right)+\left(1\right)s\left({\theta }_2-\theta \right)+\left(1\right)s\left({\theta }_3-\theta \right)=0\\ \Rightarrow \left(80-\theta \right)+\left(50-\theta \right)+\left(20-\theta \right)=0\\ \Rightarrow \theta =\frac{150}{3}=50\end{array}\end{array}$