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).
$$\begin{array}{|c|c|c|c|}\hline C_1 & C_2 & C_3 & T \\ \hline 1\ l & 2\ l & - & 60^\circ\text{C} \\ \hline - & 1\ l & 2\ l & 30^\circ\text{C} \\ \hline 2\ l & - & 1\ l & 60^\circ\text{C} \\ \hline 1\ l & 1\ l & 1\ l & \theta \\ \hline \end{array}$$
The value of $\theta$ (in $^\circ$C to the nearest integer) is ______.
Numerical value type. Enter your answer.
Answer: 50
For water alone, the mixture temperature is the volume-weighted mean. Let the containers be at $T_1, T_2, T_3$:
$$T_1 + 2T_2 = 180,\qquad T_2 + 2T_3 = 90,\qquad 2T_1 + T_3 = 180$$
Adding the three equations: $3(T_1 + T_2 + T_3) = 450$, so $T_1 + T_2 + T_3 = 150$.
Equal volumes of all three give
$$\theta = \frac{T_1 + T_2 + T_3}{3} = 50^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics