The temperature of equal masses of three different liquids A,B and C are ${15}^{0}C,{20}^{0}C$ and ${30}^{0}C$ respectively. When A and B are mixed their equilibrium temperature is ${18}^{0}C$ When B & C mixed, it is ${22}^{0}C$ . Equilibrium temperature when liquids A and C are mixed is $\frac{30}{11}$ times $T^{0}C$ . Then T is 

When A and B are mixed. The final temperature is 18 ⁰C.

$\therefore \text{ Heat gained by }\mathrm{A}=\text{ heat lost by }\mathrm{B}$

$\Rightarrow {\mathrm{mS}}_{\mathrm{A}}(18-15)={\mathrm{mS}}_{\mathrm{B}}(20-18)$

$\Rightarrow {\mathrm{S}}_{\mathrm{B}}=\frac{3}{2}{\mathrm{S}}_{\mathrm{A}}\dots ..\left(\mathrm{i}\right)$

When B and C are mixed. The final temperature is 22 ⁰C.

$\therefore \text{ Heat gained by }\mathrm{B}=\text{ heat lost by }\mathrm{C}$

$\Rightarrow {\mathrm{mS}}_{\mathrm{B}}(22-20)={\mathrm{mS}}_{\mathrm{C}}(30-22)$

$\Rightarrow {\mathrm{S}}_{\mathrm{C}}=\frac{{\mathrm{S}}_{\mathrm{B}}}{4}\dots ..\left(\mathrm{ii}\right)$

When A and C are mixed. The final temperature is T ⁰C.

$\Rightarrow {\mathrm{mS}}_{\mathrm{A}}(\mathrm{\theta }-15)={\mathrm{mS}}_{\mathrm{C}}(30-\mathrm{\theta })$

$\Rightarrow \mathrm{m}\frac{2}{3}{\mathrm{S}}_{\mathrm{B}}(\mathrm{\theta }-15)=\mathrm{m}\frac{{\mathrm{S}}_{\mathrm{B}}}{4}(30-\mathrm{\theta })$

$\Rightarrow \mathrm{\theta }=\frac{30\times 7}{11}{}^{\mathrm{o}}\mathrm{C}$