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