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Question: Answered & Verified by Expert
Two liquids $A$ and $B$ are at $32^{\circ} \mathrm{C}$ and $24^{\circ} \mathrm{C}$. When mixed in equal masses the temperature of the mixture is found to be $28^{\circ} \mathrm{C}$. Their specific heats are in the ratio of
PhysicsThermal Properties of MatterJEE Main
Options:
  • A $3: 2$
  • B $2: 3$
  • C $1: 1$
  • D $4: 3$
Solution:
2743 Upvotes Verified Answer
The correct answer is: $1: 1$
Temperature of mixture $\theta_{\text {mix }}=\frac{\theta_A c_A+\theta_B c_B}{c_A+c_B}$
$\begin{aligned} \Rightarrow & 28=\frac{32 \times c_A+24 \times c_B}{c_A+c_B} \\ \Rightarrow & 28 c_A+28 c_B=32 c_A+24 c_B \Rightarrow \frac{c_A}{c_B}=\frac{1}{1}\end{aligned}$

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