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 assuming the same initial temperatures?

• A. ${30.26}^{\circ }C$
• B. ${40.26}^{\circ }C$
• C. ${20.26}^{\circ }C$
• D. ${10.26}^{\circ }C$

Answer 1

The correct option is C

${20.26}^{\circ }C$

Let $m$ be the mass of each liquid.

When $A$ and $B$ are mixed,
Heat lost by $B$ = Heat gained by $A$
$m{s}_B\Delta T_B=-m{s}_A\Delta T_A$
$s_B(19-16)=-s_A(12-16)$
$3s_B=4s_A$ . . . .(1)

When $B$ and $C$ are mixed,
Heat lost by $C$ = Heat gained by $B$
$m{s}_C\Delta T_C=-m{s}_B\Delta T_B$
$s_C(28-23)=-m{s}_B(19-23)$
$5s_C=4s_B$ . . . .(2)

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

When $A$ and $C$ are mixed,
Let $\theta$ be the final temperature attained
Heat lost by $C$ = Heat gained by $A$
$m{s}_C\Delta T_C=-m{s}_A\Delta T_A$
$s_C(28-\theta )=-s_A(12-\theta )$

Using (3), we get
$16s_A(28-\theta )$ = $15s_A(\theta -12)$
$\theta ={20.26}^{\circ }C$