The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, when B and C are mixed is 23°C; what is the temperature when A and C are mixed ?

Let m be the mass of each liquid. Let the specific heats of A, B and C be s₁, s₂ and s₃ respectively.
(i) When A and B are mixed :
Heat gained by A = heat lost by B
ms₁ (16 -12) = ms₂ (19 - 16)
or ${\mathrm{s}}_2=\frac{4}{3}{\mathrm{s}}_1....\left(\mathrm{i}\right)$
(ii) When B and C are mixed :
Heat gained by B = heat lost by C
ms₂ (23-19) = ms₃ (28-23)
4s₂ = 5s₃
$\therefore {\mathrm{s}}_3=\frac{4}{5}{\mathrm{s}}_2=\frac{4}{5}\frac{4}{3}{\mathrm{s}}_1=\frac{16}{15}{\mathrm{s}}_1. ...\left(\mathrm{ii}\right)$
Let $\mathrm{\theta }$ be the resultant temperature.
Heat gained by A = heat lost by C
${\mathrm{ms}}_1(\mathrm{\theta }-12)={\mathrm{ms}}_3(28-\mathrm{\theta })$
or  ${\mathrm{s}}_1(\mathrm{\theta }-12)={\mathrm{s}}_3(28-\mathrm{\theta })=\frac{16}{15}{\mathrm{s}}_1(28-\mathrm{\theta })$
$\therefore \mathrm{\theta }=\frac{628}{31}=20{.26}^{\circ }\mathrm{C}$