Question

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

• A. ${31}^{\circ }\text{C}$
• B. ${20.26}^{\circ }\text{C}$
• C. ${19.5}^{\circ }\text{C}$
• D. ${28}^{\circ }\text{C}$

Answer 1

The correct option is B ${20.26}^{\circ }\text{C}$

Let $m$ be the mass of each liquid. Let the specific heats of $A,B$ and $C$ be $s_1,s_2$ and $s_3$ respectively.
(i) When $A$ and $B$ are mixed:
Heat gained by $A=$ Heat lost be $B$
$\Rightarrow m{s}_1(16-12)=m{s}_2(19-16)$
or $s_2=\frac{4}{3}{s}_1...\left(i\right)$

(ii) When $B$ and $C$ are mixed:
Heat gained by $B=$ Heat lost be $C$
$\Rightarrow m{s}_2(23-19)=m{s}_3(28-23)$
$\Rightarrow 4s_2=5s_3$
$\therefore s_3=\frac{4}{5}{s}_2=\frac{4}{5}\times \frac{4}{3}{s}_1=\frac{16}{15}{s}_1...\left(ii\right)$

(iii) When $A$ and $C$ are mixed:
Let $\theta$ be the resultant temperature.
Heat gained by $A=$ Heat lost by $C$
$\Rightarrow m{s}_1(\theta -12)=m{s}_3(28-\theta )$
or $s_1(\theta -12)=s_3(28-\theta )=\frac{16}{15}{s}_1(28-\theta )$
$\therefore \theta =\frac{628}{31}={20.26}^{\circ }\text{C}$