Question

Three liquids with masses $m_1$, $m_2$, $m_3$ are thoroughly mixed. If there specific heats are $s_1$, $s_2$, $s_3$ and their temperatures ${\theta }_1$, ${\theta }_2$, ${\theta }_3$, respectively, then the temperature of the mixture is

• A. $\frac{s_1{\theta }_1+s_2{\theta }_2+s_3{\theta }_3}{m_1{s}_1+m_2{s}_2+m_3{s}_3}$
• B. $\frac{m_1{s}_1{\theta }_1+m_2{s}_2{\theta }_2+m_3{s}_3{\theta }_3}{m_1{s}_1+m_2{s}_2+m_3{s}_3}$
• C. $\frac{m_1{s}_1{\theta }_1+m_2{s}_2{\theta }_2+m_3{s}_3{\theta }_3}{m_1{\theta }_1+m_2{\theta }_2+m_3{\theta }_3}$
• D. $\frac{m_1{\theta }_1+m_2{\theta }_2+m_3{\theta }_3}{s_1{\theta }_1+s_2{\theta }_2+s_3{\theta }_3}$

Answer 1

The correct option is B $\frac{m_1{s}_1{\theta }_1+m_2{s}_2{\theta }_2+m_3{s}_3{\theta }_3}{m_1{s}_1+m_2{s}_2+m_3{s}_3}$

For the composite system, energy conservation yields no net energy flow in or out of the system.
Let final temperature be T
Then, heat absorbed by A+heat absorbed by B+heat absorbed by C=0

$\Rightarrow m_1{c}_1(T-T_1)+m_2{c}_2(T-T_2)+m_3{c}_3(T-T_3)=0$

$\Rightarrow T=\frac{m_1{c}_1{T}_1+m_2{c}_2{T}_2+m_3{c}_3{T}_3}{m_1{c}_1+m_2{c}_2+m_3{c}_3}$