The temperature of equal masses of three diff | vedclass

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Question

${X} \quad \quad\quad \quad \quad\quad {Y} \quad\quad \quad \quad \quad \quad {Z}$

${m}_{1}={m}\quad\quad \quad {m}_{2}={m}\quad \quad \quad \quad

Vedclass · Accepted Answer

$23.84$

Vedclass · Answer

${X} \quad \quad\quad \quad \quad\quad {Y} \quad\quad \quad \quad \quad \quad {Z}$

${m}_{1}={m}\quad\quad \quad {m}_{2}={m}\quad \quad \quad \quad {m}_{3}={m}$

${T}_{1}=10^{\circ} {C}\quad \quad {T}_{2}=20^{\circ} {C}\quad \quad {T}_{3}=30^{\circ} {C}$

${s}_{1}\quad \quad \quad \quad \quad \quad {s}_{2}\quad \quad \quad \quad \quad \quad \quad {s}_{3}$

when ${x} \;and\; {y}$ are mixed, ${T}_{{f}_{{l}}}=16^{\circ} {C}$

${m}_{1} {S}_{1} {T}+{m}_{2} {s}_{2} {T}_{2}=\left({m}_{1} {S}_{1}+{m}_{2} {S}_{2}ight) {Tf}_{1}$

${s}_{1} 	imes 10+{s}_{2} 	imes 20=\left({s}_{1}+{s}_{2}ight) 	imes 16$

${s}_{1}=\frac{2}{3} {s}_{2}$

When $y \;and\; z$ are mixex, $T_{f_{2}}=26^{\circ} {C}$

${m}_{2} {s}_{2} {T}+{m}_{3} {s}_{3} {T}_{3}=\left({m}_{3} {s}_{3}+{m}_{3} {s}_{3}ight) {T} {f}_{2}$

${s}_{2} 	imes 20+{s}_{3} 	imes 30=\left({s}_{2}+{s}_{3}ight) 	imes 26$

${s}_{3}=\frac{3}{2} {s}_{2} \quad \ldots . . 	ext { (ii) }$

when $x \;and\; z$ are mixex

${m}_{1} {s}_{1} {T}_{1}+{m}_{3} {S}_{3} {T}_{3}=\left({m}_{1} {s}_{1}+{m}_{3} {S}_{3}ight) {Tf}$

$\frac{2}{3} {s}_{2} 	imes 10+\frac{2}{3} {s}_{2} 	imes 20=\left(\frac{2}{3} {s}_{2}+\frac{3}{2} {s}_{2}ight) {T}_{{f}}$

${T}_{{f}}=23.84^{\circ} {C}$

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