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Step-by-Step Solution
Step 1: Understand the Problem
We have three different liquids X, Y, and Z, each having the same mass but different specific heats (let them be $s_1$, $s_2$, and $s_3$, respectively). Their initial temperatures are given:
Liquid X: $10^\circ \mathrm{C}$
Liquid Y: $20^\circ \mathrm{C}$
Liquid Z: $30^\circ \mathrm{C}$
When mixed in pairs (each pair having equal masses of the liquids), the mixture temperatures are:
Mixing X + Y: $16^\circ \mathrm{C}$
Mixing Y + Z: $26^\circ \mathrm{C}$
We want to find the final temperature when X and Z are mixed.
Step 2: Establish Equations for Mixture Temperatures
Equation from mixing X and Y
The energy balance for mixing equal masses of X (with specific heat $s_1$ and temperature $10^\circ\mathrm{C}$) and Y (with specific heat $s_2$ and temperature $20^\circ\mathrm{C}$) to get final temperature $16^\circ\mathrm{C}$ is:
$ m \, s_1 \times 10 + m \, s_2 \times 20 \;=\; \bigl(m\,s_1 + m\,s_2\bigr) \times 16
$
Since $m$ is the same and can be factored out from all terms, we get:
$ s_1 \times 10 + s_2 \times 20 = (s_1 + s_2) \times 16.
$
Rearranging gives a relationship between $s_1$ and $s_2$:
$ s_1 = \tfrac{2}{3} s_2. \quad (1)$
Equation from mixing Y and Z
Similarly, for mixing Y (with specific heat $s_2$, temperature $20^\circ\mathrm{C}$) and Z (with specific heat $s_3$, temperature $30^\circ\mathrm{C}$) to get final temperature $26^\circ\mathrm{C}$, the energy balance is:
$ m \, s_2 \times 20 + m \, s_3 \times 30 \;=\; \bigl(m\,s_2 + m\,s_3\bigr) \times 26.
$
Again, factoring out $m$ and rearranging leads to:
$ s_2 \times 20 + s_3 \times 30 = (s_2 + s_3)\times 26,
$
which simplifies to:
$ s_3 = \tfrac{3}{2} s_2. \quad (2)$
Step 3: Use the Ratios to Find the Temperature When X and Z Are Mixed
We now mix X (specific heat $s_1$, temperature $10^\circ\mathrm{C}$) and Z (specific heat $s_3$, temperature $30^\circ\mathrm{C}$). The energy balance gives:
$ m \, s_1 \times 10 + m \, s_3 \times 30 \;=\; \bigl(m\,s_1 + m\,s_3\bigr)\,T_f,
$
where $T_f$ is the desired final temperature. Substituting $s_1 = \tfrac{2}{3} s_2$ from (1) and $s_3 = \tfrac{3}{2} s_2$ from (2), the equation becomes:
$ \left(\tfrac{2}{3} s_2\right) \times 10 + \left(\tfrac{3}{2} s_2\right) \times 30
\;=\; \Bigl(\tfrac{2}{3} s_2 + \tfrac{3}{2} s_2\Bigr)\,T_f.
$
Performing the multiplication on the left side:
$ \tfrac{20}{3} s_2 + 45 s_2 = \tfrac{20}{3} s_2 + \tfrac{135}{3} s_2 = \tfrac{155}{3} s_2.
$
On the right side, we calculate:
$ \tfrac{2}{3} s_2 + \tfrac{3}{2} s_2 = \tfrac{4}{6} s_2 + \tfrac{9}{6} s_2 = \tfrac{13}{6} s_2.
$
Thus:
$ \tfrac{155}{3} s_2 = \left( \tfrac{13}{6} s_2 \right) T_f.
$
Canceling $s_2$ and solving for $T_f$:
$
T_f = \dfrac{\tfrac{155}{3} s_2}{\tfrac{13}{6} s_2}
= \dfrac{155}{3} \times \dfrac{6}{13}
= \dfrac{155 \times 6}{3 \times 13}
= \dfrac{930}{39}
= 23.84^\circ \mathrm{C}.
$
Step 4: Final Answer
The temperature of the mixture when liquids X and Z are mixed is $23.84^\circ \mathrm{C}.$
