Question

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

• A. ${18.2}^C$
• B. ${22}^C$
• C. ${20.3}^C$
• D. ${24.2}^C$

Answer 1

The correct option is C ${20.3}^C$

let $m_1=m_2=m_3=m$
Let $s_1,s_2,s_3$ be the respective specific heats of the three liquids.
When A and B are mixed, temperature of mixed = ${16}^{o}C$
A heat gained by A = heat lost by B
$\therefore m{s}_1(16-12)=m{s}_2(19-16)$
$4s_1=3s_2$.......(i)
When B and C are mixed , temperature of mixture$={23}^{o}C$.
As heat gained by B = heat lost by C,
$m{s}_2(23-19)=m{s}_3(28-23)$
$\therefore 4s_2=5s_3$......(ii)
From (i) and (ii) , $=\frac{3}{4}{s}_2=\frac{15}{16}{s}_3$
When A and C are mixed, suppose temperature of mixture$=t$
Heat gained by A = Heat lost by C
$m{s}_1(t-12)=m{s}_3(28-t)$
$\frac{15}{16}{s}_2(t-12)=s_3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\frac{628}{3}={20.3}^{o}C$