Question

The temperature of equal masses of three different liquids $A,B$ and $C$ are ${12}^{\circ }C,{19}^{\circ }C$ and ${28}^{\circ }C$ respectively. The temperature when $A$ and $B$ are mixed is ${16}^{\circ }C$ and when $B$ and $C$ are mixed it is ${23}^{\circ }C$. What should be the temperature when $A$ and $C$ are mixed?

Answer 1

Let $m$ be the mass of each liquid and $S_A,S_B,S_C$ be specific heats of liquids $A,B$ and $C$ respectively. When $A$ and $B$ are mixed. The final temperature is ${16}^{\circ }C$.
$\therefore$ Heat gained by $A=$ heat lost by $B$
i.e., $m{S}_A=(16-12)=m{S}_B(19-16)$
i.e., $S_B=\frac{4}{3}{S}_A.....\left(i\right)$
When $B$ and $C$ are mixed. Heat gained by $B=$ Heat lost by $C$
i.e., $m{S}_B=(23-19)=m{S}_C(28-23)$
i.e., $S_C=\frac{4}{5}{S}_B.....\left(ii\right)$
From eq. (i) and (ii)
$S_C=\frac{4}{5}\times \frac{4}{3}{S}_A=\frac{16}{15}{S}_A$
When $A$ and $C$ are mixed, let the final temperature be $\theta$
$\therefore m{S}_A(\theta -12)=m{S}_C(28-\theta )$
i.e., $\theta -12=\frac{16}{15}(28-\theta )$
By solving, we get,
$\theta =\frac{628}{31}={20.26}^{\circ }C$.