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The tempertaure of equal masses of three different liquids $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $12^{\circ} \mathrm{C}, 19^{\circ} \mathrm{C}$ and $28^{\circ} \mathrm{C}$ respectively. The temperature when $\mathrm{A}$ and B are mixed is $16^{\circ} \mathrm{C}$ and when $\mathrm{B}$ and $\mathrm{C}$ are mixed is $23^{\circ} \mathrm{C} .$ The temperature when $\mathrm{A}$ and $\mathrm{C}$ are mixed is
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Verified Answer
The correct answer is:
$20.2^{\circ} \mathrm{C}$
Heat gain $=$ heat lost
$$
\begin{array}{l}
C_{A}(16-12)=C_{B}(19-16) \Rightarrow \frac{C_{A}}{C_{B}}=\frac{3}{4} \\
\text { and } C_{B}(23-19)=C_{c}(28-23) \Rightarrow \frac{C_{B}}{C_{C}}=\frac{5}{4} \\
\Rightarrow \frac{C_{A}}{C_{C}}=\frac{15}{16}
\end{array}
$$
If $\theta$ is the temperature when $\mathrm{A}$ and $\mathrm{C}$ are mixed then,
$$
\begin{array}{l}
C_{A}(\theta-12)=C_{C}(28-\theta) \\
\Rightarrow \frac{C_{A}}{C_{C}}=\frac{28-\theta}{\theta-12}
\end{array}
$$
On solving equations (i) and (ii) $\theta=20.2^{\circ} \mathrm{C}$
$$
\begin{array}{l}
C_{A}(16-12)=C_{B}(19-16) \Rightarrow \frac{C_{A}}{C_{B}}=\frac{3}{4} \\
\text { and } C_{B}(23-19)=C_{c}(28-23) \Rightarrow \frac{C_{B}}{C_{C}}=\frac{5}{4} \\
\Rightarrow \frac{C_{A}}{C_{C}}=\frac{15}{16}
\end{array}
$$
If $\theta$ is the temperature when $\mathrm{A}$ and $\mathrm{C}$ are mixed then,
$$
\begin{array}{l}
C_{A}(\theta-12)=C_{C}(28-\theta) \\
\Rightarrow \frac{C_{A}}{C_{C}}=\frac{28-\theta}{\theta-12}
\end{array}
$$
On solving equations (i) and (ii) $\theta=20.2^{\circ} \mathrm{C}$
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