Three containers C 1 , C 2 and C 3 have water at different temperatures. The table below shows the final…

Let ${\theta }_1$, ${\theta }_2$ and ${\theta }_3$ be the temperatures of water in three containers.

$1{\theta }_1+2{\theta }_2+0{\theta }_3=\left(1+2\right)60$
${\theta }_1+2{\theta }_2=180\mathrm{}$ …. (1)
$0\times {\theta }_1+1\times {\theta }_2+2\times {\theta }_3=\left(1+2\right)30$
${\theta }_2+2{\theta }_3=90\mathrm{}$ …… (2)
$2\times {\theta }_1+0\times {\theta }_2+1\times {\theta }_3=\left(2+1\right)60$
$2{\theta }_1+{\theta }_3=180\mathrm{}$ …. (3)
and ${\theta }_1+{\theta }_2+{\theta }_3=\left(1+1+1\right)\theta$
from (1)+(2)+(3)
$3{\theta }_1+3{\theta }_2+3{\theta }_3=450⟹{\theta }_1+{\theta }_2+{\theta }_3=150$
from (4) equation $150=3\theta ⟹\theta =50 °\mathrm{C}$