Question

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

• A. ${18.2}^{o}C$
• B. ${22}^{o}C$
• C. ${20.2}^{o}C$
• D. ${24.2}^{o}C$

Answer 1

The correct option is C ${20.2}^{o}C$

When A and B is mixed

$m{C}_A\Delta T=m{C}_B\Delta T\frac{C_A}{C_B}=\frac{19-16}{16-12}=\frac{3}{4}$

When B and C is mixed

$m{C}_B\Delta T=m{C}_C\Delta T{C}_B(23-19)=C_C(28-23)\frac{C_B}{C_C}=\frac{5}{4}$

When A and C is mixed.

Let final temp be x

$m{C}_A\Delta T_1=m{C}_C{\Delta }_2\frac{C_A}{C_C}=\frac{28-x}{x-12}\frac{C_A}{C_B}\times \frac{C_B}{C_C}=\frac{28-x}{x-12}\frac{3}{4}\times \frac{5}{4}=\frac{28-x}{x-12}$

On solving

$x=20.2°C$