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 in ${}^{\circ }C$ when A and C are mixed? Answer upto 2 decimal places.

Answer 1

Given ${\theta }_A={12}^{\circ }C,{\theta }_B={19}^{\circ }C$ and ${\theta }_C={28}^{\circ }$.
Let $s_A,s_B$ and $s_C$ are the specific heats of respective liquids.
When liquid A and B are mixed, the temperature of mixture becomes ${16}^{\circ }C$, then
$m{s}_A(16-12)=m{s}_B(19-16)$ or $s_B=\frac{4}{3}{s}_A....\left(1\right)$
When liquid B and C are mixed, the temperature of mixture becomes ${23}^{\circ }C$, then
$m{s}_B(23-19)=m{s}_C(28-23)$ or $s_B=\frac{5}{4}{s}_C....\left(2\right)$
From (1) and (2), we get $s_A=\frac{15}{16}{s}_C$
Now when A and C are mixed, let equilibrium temperature of mixture is $\theta$, then
$m{s}_A(\theta -12)=m{s}_C(28-\theta )$
or $\frac{15}{16}{s}_C(\theta -12)=s_C(28-\theta )$
or $31\theta =628={20.26}^{\circ }C$

Why this question? Key concept: Heat gained by the cold body = heat lost by the hot body. Importance in JEE: Every year JEE asks some very simple application based questions. This is one such question. So, watch out for such questions.