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Question: Answered & Verified by Expert
Three liquids with masses $m_1, m_2, m_3$ are thoroughly mixed. If their specific heats are $c_1, c_2, c_3$ and their temperatures $T_1, T_2, T_3$ respectively, then the temperature of the mixture is
PhysicsThermal Properties of MatterJEE Main
Options:
  • A $\frac{c_1 T_1+c_2 T_2+c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$
  • B $\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$
  • C $\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 T_1+m_2 T_2+m_3 T_3}$
  • D $\frac{m_1 T_1+m_2 T_2+m_3 T_3}{c_1 T_1+c_2 T_2+c_3 T_3}$
Solution:
2274 Upvotes Verified Answer
The correct answer is: $\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}$
Let the final temperature be $T^{\circ} \mathrm{C}$.
Total heat supplied by the three liquids in coming down to $0^{\circ} \mathrm{C}=m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3$ ...(i)
Total heat used by three liquids in raising temperature from $0^{\circ} \mathrm{C}$ to $T^{\circ} \mathrm{C}$
$=m_1 c_1 T+m_2 c_2 T+m_3 c_3 T$ ...(ii)
By equating (i) and (ii) we get
$\begin{aligned} & \left(m_1 c_1+m_2 c_2+m_3 c_3\right) T \\ = & m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3 \\ \Rightarrow & T=\frac{m_1 c_1 T_1+m_2 c_2 T_2+m_3 c_3 T_3}{m_1 c_1+m_2 c_2+m_3 c_3}\end{aligned}$

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