Question

The temperatures of equal masses of three liquids A, B, C are ${20}^o$C, ${30}^o$C and ${40}^o$C respectively. On mixing A and B, resulting temperature of the mixture is ${26}^o$C. On mixing A and C the resulting temperature of mixture is ${32}^o$C. Find the final resulting temperature of the mixture B and C.

Answer 1

Let $m$ be the mass of each liquid. Let specific heat of A,B,C are $S_a,S_b,S_c$ respectively.

When A and B are mixed

Heat gain by A = Heat loss by B

$\Rightarrow m{S}_a(26-20)=m{S}_b(30-26)\Rightarrow S_a=\frac{4}{6}{S}_b=\frac{2}{3}{S}_b$

When A and C are mixed

Heat gain by A = Heat loss by C

$\Rightarrow m{S}_a(32-20)=m{S}_c(40-32)\Rightarrow S_a=\frac{8}{12}{S}_c\Rightarrow \frac{2}{3}{S}_b=\frac{8}{12}{S}_c\Rightarrow \frac{2}{3}{S}_b=\frac{2}{3}{S}_c\Rightarrow S_b=S_c$

When B and C are mixed

Heat gain by B = Heat loss by C

$\Rightarrow m{S}_b(\theta -30)=m{S}_c(40-\theta )\left[Letfinaltemperaturebe\theta \right]\Rightarrow \theta -30=40-\theta \Rightarrow 2\theta =70\Rightarrow \theta =35℃$