Question

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

• A. $18.2^{\circ} C$
• B. $22^{\circ} C$
• C. $20.2^{\circ} C$
• D. $25.2^{\circ} C$

Answer 1

Answer: (C)

When two liquids with different temperatures are mixed, heat is lost by the hotter liquid to the cooler liquid.
Let, the mass of liquids $A$ , $B$ and $C$ be $m$ . The specific heat capacities of liquids $A, B$ and $C$ be $C_{A}, C_{B} $ and $C_{C}$ .
For mixing of $A$ and $B$ :
Heat gained by liquid $A =$ Heat lost by liquid $B$
$mC_{A}\left( 16 - 12 \right)=mC_{B}\left( 19 - 16 \right)\Rightarrow \frac{C_{A}}{C_{B}}=\frac{3}{4}...\left(1\right)$
For mixing of $C$ and $B$ :
Heat lost by liquid $B$ $=$ Heat gained by liquid $C$
$mC_{B}\left( 23 - 19 \right)=mC_{C}\left( 28 - 23 \right)$
$\Rightarrow \frac{C_{B}}{C_{C}}=\frac{5}{4}...\left(2\right)$
On multiplying $\left(1\right)$ and $\left(2\right)$ ,
$\Rightarrow \frac{C_{A}}{C_{C}}=\frac{15}{16}...\left(3\right)$
If $\theta $ is the final temperature when $A$ and $C$ are mixed then,
$mC_{A}\left(\right. \theta - 12 \left.\right)=mC_{C}\left( 28 - \theta \right)$
$\Rightarrow \frac{C_{A}}{C_{C}}=\frac{28 - \theta }{\theta - 12}...\left(4\right)$
On solving equations $\left(3\right)$ and $\left(4\right)$ ,
$\frac{15}{16}=\frac{28 - \theta }{\theta - 12}\Rightarrow \theta =20.2^{\circ}C$