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 $B$
$m{C}_A\left(16-12\right)=m{C}_B\left(19-16\right)\Rightarrow \frac{C_A}{C_B}=\frac{3}{4}...\left(1\right)$
For mixing of $C$ and $B$ :
Heat lost by liquid $B$ $=$ Heat gained by liquid $C$
$m{C}_B\left(23-19\right)=m{C}_C\left(28-23\right)$
$\Rightarrow \frac{C_B}{C_C}=\frac{5}{4}...\left(2\right)$
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,
$m{C}_A(\theta -12)=m{C}_C\left(28-\theta \right)$
$\Rightarrow \frac{C_A}{C_C}=\frac{28-\theta }{\theta -12}...\left(4\right)$
On solving equations $\left(3\right)$ and $\left(4\right)$ ,
$\frac{15}{16}=\frac{28-\theta }{\theta -12}\Rightarrow \theta =20.2^{\circ }C$