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Physics Question on Conductance

The temperature of equal masses of three different liquids A, B and C are 12∘C12^{\circ}C, 12∘C12^{\circ}C and 28∘C28^{\circ}C respectively. The temperature when A and B are mixed is 16∘C16^{\circ}C and when B and C are mixed is 23∘C23^{\circ}C. The temperature when A and C are mixed is

A

18.2∘C18.2^{\circ}C

B

22∘C22^{\circ}C

C

20.2∘C20.2^{\circ}C

D

25.2∘C25.2^{\circ}C

Answer

20.2∘C20.2^{\circ}C

Explanation

Solution

Heat gain = heat lost
CA(16−12)=CB(19−16)⇒CACB=34C_{A}\left(16 -12\right) = C_{B}\left(19-16\right) \Rightarrow \frac{C_{A}}{C_{B}} = \frac{3}{4}
and CB(23−19)=Cc(28−23)⇒CBCC=54C_{B}\left(23-19\right) = C_{c}\left(28-23\right) \Rightarrow \frac{C_{B}}{C_{C}} = \frac{5}{4}
⇒CACC=1516…(i)\Rightarrow \frac{C_{A}}{C_{C}} = \frac{15}{16}\quad\quad\ldots\left(i\right)
If θ\theta is the temperature when A and C are mixed then,
CA(θ−12)=CC(28−θ)C_{A}\left(\theta-12\right) = C_{C}\left(28-\theta\right)
⇒CACC=28−θθ−12…(ii)\Rightarrow \frac{C_{A}}{C_{C}} = \frac{28-\theta}{\theta-12}\quad \quad \ldots \left(ii\right)
On solving equations (i)\left(i\right) and (ii)θ=20.2∘C\left(ii\right) \theta=20.2^{\circ}C