The temperature of equal masses of three different liquids $\mathrm{A}, \mathrm{B}$ and C are $12^{\

Question

The temperature of equal masses of three different liquids $\mathrm{A}, \mathrm{B}$ and C are $12^{\circ} \mathrm{C}, 19^{\circ} \mathrm{C}$ and $28^{\circ} \mathrm{C}$ respectively. The temperature when $A$ and $B$ are mixed is $16^{\circ} \mathrm{C}$ and when $B$ and $C$ are mixed $23^{\circ} \mathrm{C}$. The temperature when A and C are mixed is\

Professor PC Thomas Classes · Answer

Heat gain $=$ heat lost\ $C_{A}(16-12)=C_{B}(19-16) \Rightarrow \frac{C_{A}}{C_{B}}=\frac{3}{4}$\ and $\mathrm{C}_{\mathrm{B}}(23-19)=\mathrm{C}_{\mathrm{C}}(28-23) \Rightarrow \frac{\mathrm{C}_{\mathrm{B}}}{\mathrm{C}_{\mathrm{C}}}=\frac{5}{4}$\ $\Rightarrow \frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{C}_{\mathrm{C}}}=\frac{15}{16}$\ If $	heta$ is the temperature when A and C are mixed then,\ $\mathrm{C}_{\mathrm{A}}(	heta-12)=\mathrm{C}_{\mathrm{C}}(28-	heta)$\ $	herefore \frac{28-	heta}{	heta-12}=\frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{C}_{\mathrm{C}}}=\frac{15}{16}$. Solving, $	heta=20.2^{\circ} \mathrm{C}$\