Question

The temperature of equal masses of three different liquids A, B and C are \(12^{\circ}C\), \(12^{\circ}C\) and \(28^{\circ}C\) respectively. The temperature when A and B are mixed is \(16^{\circ}C\) and when B and C are mixed is \(23^{\circ}C\). The temperature when A and C are mixed is

• A. $18.2^{\circ}C$
• B. $22^{\circ}C$
• C. $20.2^{\circ}C$
• D. $25.2^{\circ}C$

Answer 1

The Correct Option is C

Heat gain = heat lost
$C_{A}\left(16 -12\right) = C_{B}\left(19-16\right) \Rightarrow \frac{C_{A}}{C_{B}} = \frac{3}{4}$
and $C_{B}\left(23-19\right) = C_{c}\left(28-23\right) \Rightarrow \frac{C_{B}}{C_{C}} = \frac{5}{4}$
$\Rightarrow \frac{C_{A}}{C_{C}} = \frac{15}{16}\quad\quad\ldots\left(i\right)$
If $\theta$ is the temperature when A and C are mixed then,
$C_{A}\left(\theta-12\right) = C_{C}\left(28-\theta\right)$
$\Rightarrow \frac{C_{A}}{C_{C}} = \frac{28-\theta}{\theta-12}\quad \quad \ldots \left(ii\right)$
On solving equations $\left(i\right)$ and $\left(ii\right) \theta=20.2^{\circ}C$