The temperatures of equal masses of three dif | vedclass

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(C) Let the specific heats of liquids $x$,$y$,and $z$ be $s_1$,$s_2$,and $s_3$ respectively. Since the masses are equal $(m_1 = m_2 = m_3 = m)$,the pr

Vedclass · Accepted Answer

$23.84$

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(C) Let the specific heats of liquids $x$,$y$,and $z$ be $s_1$,$s_2$,and $s_3$ respectively. Since the masses are equal $(m_1 = m_2 = m_3 = m)$,the principle of calorimetry states: $m s_1 T_1 + m s_2 T_2 = (m s_1 + m s_2) T_f$.For $x$ and $y$ mixed: $s_1(10) + s_2(20) = (s_1 + s_2)(16) \implies 10s_1 + 20s_2 = 16s_1 + 16s_2 \implies 4s_2 = 6s_1 \implies s_1 = \frac{2}{3}s_2$.For $y$ and $z$ mixed: $s_2(20) + s_3(30) = (s_2 + s_3)(26) \implies 20s_2 + 30s_3 = 26s_2 + 26s_3 \implies 4s_3 = 6s_2 \implies s_3 = \frac{3}{2}s_2$.For $x$ and $z$ mixed: $s_1(10) + s_3(30) = (s_1 + s_3)T_f$.Substituting $s_1 = \frac{2}{3}s_2$ and $s_3 = \frac{3}{2}s_2$:$(\frac{2}{3}s_2)(10) + (\frac{3}{2}s_2)(30) = (\frac{2}{3}s_2 + \frac{3}{2}s_2)T_f$.$\frac{20}{3}s_2 + 45s_2 = (\frac{4+9}{6})s_2 T_f \implies \frac{20+135}{3} = \frac{13}{6}T_f$.$\frac{155}{3} = \frac{13}{6}T_f \implies T_f = \frac{155 	imes 6}{3 	imes 13} = \frac{310}{13} \approx 23.84^{\circ}C$.

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