- JEE Mains 2021
- Hard
- MCQ
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 ...... \(^{\circ} {C}\)
Answer & Solution
(C) \(23.84\)
Step-by-step Solution
Detailed explanation
\({m}_{1}={m}\quad\quad \quad {m}_{2}={m}\quad \quad \quad \quad {m}_{3}={m}\)
\({T}_{1}=10^{\circ} {C}\quad \quad {T}_{2}=20^{\circ} {C}\quad \quad {T}_{3}=30^{\circ} {C}\)
\({s}_{1}\quad \quad \quad \quad \quad \quad {s}_{2}\quad \quad \quad \quad \quad \quad \quad {s}_{3}\)
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
\({m}_{1} {s}_{1} {T}_{1}+{m}_{3} {S}_{3} {T}_{3}=\left({m}_{1} {s}_{1}+{m}_{3} {S}_{3}\right) {Tf}\)
\(\frac{2}{3} {s}_{2} \times 10+\frac{2}{3} {s}_{2} \times 20=\left(\frac{2}{3} {s}_{2}+\frac{3}{2} {s}_{2}\right) {T}_{{f}}\)
\({T}_{{f}}=23.84^{\circ} {C}\)
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