Question

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

• A. 30.26°C
• B. 40.26°C
• C. 20.26°C
• D. 10.26°C

Answer 1

The correct option is C

20.26°C

Let $m$ = mass of each liquid, when $A$ and $B$ are mixed,

Heat lost by $B$ = Heat gained by $A$

$\Rightarrow$ $m{s}_B$ (19 - 16) = $m{s}_A$ (16 - 12) $\Rightarrow$ $3s_B$ = $4S_A$ . . . .(1)

When $B$ and $C$ are mixed, Heat lost by $C$ = Heat gained by $B$

$\Rightarrow$ $m{s}_C$ (28 - 23) = $m{s}_B$ (23 - 19) $\Rightarrow$ $5s_C$ = $4S_B$ . . . .(2)

From (1) and (2), we get $16s_A$ = $12s_B$ = $15s_C$. . . .(3)

When $A$ and $C$ are mixed, Let $\theta$ = final temperature

Heat lost by $C$ = heat gained by $A$

$\Rightarrow$ $m{s}_C$ $(28-\theta )$ = $m{s}_B$ $(\theta -12)$

Using (3), we get

$\Rightarrow$ $15s_C(28-\theta )$ = $15s_A(\theta -12)$ $\Rightarrow$ $16s_A(28-\theta )$ = $15s_A(\theta -12)$

On solving for $\theta$, we get $\theta$ = $\frac{16\times 28+12\times 15}{16+15}$ $\Rightarrow$ $\theta$ = ${20.26}^{\circ }C$