The temperature of equal masses of three different liquids

Question:

The temperature of equal masses of three different liquids $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$ respectively. The temperature of mixture when $x$ is mixed with $y$ is $16^{\circ} \mathrm{C}$ and that when $\mathrm{y}$ is mixed with $\mathrm{z}$ is $26^{\circ} \mathrm{C}$. The temperature of mixture when $\mathrm{x}$ and $\mathrm{z}$ are mixed will be :

 

  1. $28.32^{\circ} \mathrm{C}$

  2. $25.62^{\circ} \mathrm{C}$

  3. $23.84^{\circ} \mathrm{C}$

  4. $20.28^{\circ} \mathrm{C}$


Correct Option: , 3

Solution:

when $\mathrm{x} \& \mathrm{y}$ are mixed, $\mathrm{T}_{\mathrm{f}_{1}}=16^{\circ} \mathrm{C}$

$\mathrm{m}_{1} \mathrm{~s}_{1} \mathrm{~T}+\mathrm{m}_{2} \mathrm{~s}_{2} \mathrm{~T}_{2}=\left(\mathrm{m}_{1} \mathrm{~s}_{1}+\mathrm{m}_{2} \mathrm{~s}_{2}\right) \mathrm{Tf}_{1}$

$\mathrm{s}_{1} \times 10+\mathrm{s}_{2} \times 20=\left(\mathrm{s}_{1}+\mathrm{s}_{2}\right) \times 16$

$\mathrm{s}_{1}=\frac{2}{3} \mathrm{~s}_{2} \quad \ldots .$ (i)

when $\mathrm{y} \& \mathrm{z}$ are mixex, $\mathrm{T}_{\mathrm{f}_{2}}=26^{\circ} \mathrm{C}$

$\mathrm{m}_{2} \mathrm{~s}_{2} \mathrm{~T}+\mathrm{m}_{3} \mathrm{~s}_{3} \mathrm{~T}_{3}=\left(\mathrm{m}_{3} \mathrm{~s}_{3}+\mathrm{m}_{3} \mathrm{~s}_{3}\right) \mathrm{Tf} \mathrm{f}_{2}$

$\mathrm{s}_{2} \times 20+\mathrm{s}_{3} \times 30=\left(\mathrm{s}_{2}+\mathrm{s}_{3}\right) \times 26$

$\mathrm{s}_{3}=\frac{3}{2} \mathrm{~s}_{2}$ ............(II)

when $\mathrm{x} \& \mathrm{z}$ are mixex

$\mathrm{m}_{1} \mathrm{~s}_{1} \mathrm{~T}_{1}+\mathrm{m}_{3} \mathrm{~s}_{3} \mathrm{~T}_{3}=\left(\mathrm{m}_{1} \mathrm{~s}_{1}+\mathrm{m}_{3} \mathrm{~s}_{3}\right) \mathrm{Tf}$

$\frac{2}{3} s_{2} \times 10+\frac{2}{3} s_{2} \times 20=\left(\frac{2}{3} s_{2}+\frac{3}{2} s_{2}\right) T_{f}$

$\mathrm{T}_{\mathrm{f}}=23.84^{\circ} \mathrm{C}$

$\operatorname{Ans}(3)$

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