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The temperature of equal masses of three different liquids ${x}, {y}$ and ${z}$ are $10^{\circ} {C}, 20^{\circ} {C}$ and $30^{\circ} {C}$ respectively. The temperature of mixture when ${x}$ is mixed with ${y}$ is $16^{\circ} {C}$ and that when ${y}$ is mixed with $z$ is $26^{\circ} {C}$. The temperature of mixture when $x$ and $z$ are mixed will be ...... $^{\circ} {C}$
$28.32$
$25.62$
$23.84$
$20.28$
Solution
${X} \quad \quad\quad \quad \quad\quad {Y} \quad\quad \quad \quad \quad \quad {Z}$
${m}_{1}={m}\quad\quad \quad {m}_{2}={m}\quad \quad \quad \quad {m}_{3}={m}$
${T}_{1}=10^{\circ} {C}\quad \quad {T}_{2}=20^{\circ} {C}\quad \quad {T}_{3}=30^{\circ} {C}$
${s}_{1}\quad \quad \quad \quad \quad \quad {s}_{2}\quad \quad \quad \quad \quad \quad \quad {s}_{3}$
when ${x} \;and\; {y}$ are mixed, ${T}_{{f}_{{l}}}=16^{\circ} {C}$
${m}_{1} {S}_{1} {T}+{m}_{2} {s}_{2} {T}_{2}=\left({m}_{1} {S}_{1}+{m}_{2} {S}_{2}\right) {Tf}_{1}$
${s}_{1} \times 10+{s}_{2} \times 20=\left({s}_{1}+{s}_{2}\right) \times 16$
${s}_{1}=\frac{2}{3} {s}_{2}$
When $y \;and\; z$ are mixex, $T_{f_{2}}=26^{\circ} {C}$
${m}_{2} {s}_{2} {T}+{m}_{3} {s}_{3} {T}_{3}=\left({m}_{3} {s}_{3}+{m}_{3} {s}_{3}\right) {T} {f}_{2}$
${s}_{2} \times 20+{s}_{3} \times 30=\left({s}_{2}+{s}_{3}\right) \times 26$
${s}_{3}=\frac{3}{2} {s}_{2} \quad \ldots . . \text { (ii) }$
when $x \;and\; z$ are mixex
${m}_{1} {s}_{1} {T}_{1}+{m}_{3} {S}_{3} {T}_{3}=\left({m}_{1} {s}_{1}+{m}_{3} {S}_{3}\right) {Tf}$
$\frac{2}{3} {s}_{2} \times 10+\frac{2}{3} {s}_{2} \times 20=\left(\frac{2}{3} {s}_{2}+\frac{3}{2} {s}_{2}\right) {T}_{{f}}$
${T}_{{f}}=23.84^{\circ} {C}$