Question

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

Answer 1

Let $m$ be the mass of each liquid. Let the specific heat of A,B,C be $S_1,S_2,S_3$ respectively.

When B and A are mixed

Heat gain by A = Heat lost by B

$\Rightarrow m{S}_1(16-12)=m{S}_2(19-16)\Rightarrow S_2=\frac{4}{3}{S}_1$

When B and C are mixed

Heat gain by B = Heat lost by C

$\Rightarrow m{S}_2(23-19)=m{S}_3(28-23)\Rightarrow 4S_2=5S_3\Rightarrow S_3=\frac{4}{5}{S}_2\Rightarrow S_3=\frac{4}{5}\times \frac{4}{3}{S}_1=\frac{16}{15}{S}_1$

When A and C are mixed

Let $\theta$ be resultant temperature

Heat gain by A = Heat lost by C

$\Rightarrow m{S}_1(\theta -12)=m{S}_3(28-\theta )\Rightarrow S_1(\theta -12)=\frac{16}{15}{S}_1(28-\theta )\Rightarrow 15\theta -180=448-16\theta \Rightarrow 31\theta =628\Rightarrow \theta =20.258℃$