Question

The temperatures of equal masses of three liquids A, B, C are $20{}^{o}C,30{}^{o}Cand40{}^{o}C$ respectively. On mixing A and B, resulting temperature of the mixture is $26{}^{o}C$. On mixing A and C the resulting temperature of the mixture is $32{}^{o}C$. Find the final resulting temperature of the mixture B and C.

Answer 1

For mixture of A and B,
$26=\frac{m{s}_{1}20+m{s}_{2}30}{m{s}_1+m{s}_2}$
$26s_1+26s_2=20s_1+30s_2$
$\Rightarrow \frac{s_1}{s_2}=\frac{4}{6}=\frac{2}{3}$
For mixture of A and C,
$32=\frac{20s_1+40s_3}{s_1+s_3}$
$32s_1+32s_3=20s_1+40s_3$
$\Rightarrow \frac{s_1}{s_3}=\frac{8}{12}=\frac{2}{3}$
So, $\Rightarrow \frac{s_2}{s_3}=\frac{2}{3}\times \frac{3}{2}=1$
For mixture or B and C
$\theta =\frac{30s_2+40s_3}{s_2+s_3}=\frac{30+40}{2}=35{}^{o}C$