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The temperature of equal masses of three different liquids $A, B$ and $C$ are $12°C, 19°C$ and $28°C$ respectively. The temperature when $A$ and $B$ are mixed is $16°C$ and when $B$ and $C$ are mixed is $23°C$. The temperature when $A$ and $C$ are mixed is........ $^oC$
$18.2$
$22$
$20.2$
$25.2$
Solution
(c) Heat gain $=$ heat lost
$C_A(16 -12) = C_B (19 -16) $
==> $\frac{{{C_A}}}{{{C_B}}}$ = $\frac{3}{4}$
and $C_B(23 -19) = C_C (28 -23) $
==>$\frac{{{C_B}}}{{{C_C}}}$ = $\frac{5}{4}$
==> $\frac{{{C_A}}}{{{C_C}}} = \frac{{15}}{{16}}$ …$(i)$
If $\theta$ is the temperature when $A$ and $C$ are mixed then,
${C_A}(\theta – 12)$ = ${C_C}(28 – \theta )$
==> $\frac{{{C_A}}}{{{C_C}}}$$ = \frac{{28 – \theta }}{{\theta – 12}}$ …$(ii)$
On solving equation $(i)$ and $(ii)$
$\theta$ $= 20.2°C.$
