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**Hint:**The equation shows the relationship between the heat energy and the temperature which are different for different materials and shows that the specific heat is a value and describes how they relate to the heat energy. By using the specific heat capacity formula, the temperature is determined.

**Useful formula**

Specific heat capacity formula,

$Q = mc\Delta T$

Where, $Q$ is the heat energy, $m$ is the mass of the substance, $c$ is the specific heat, $\Delta T$ is the temperature difference

**Complete step by step solution**

the data that are given in the problem is;

temperature of the liquid $A = {12^ \circ }\,C$

temperature of the liquid $B = {19^ \circ }\,C$

temperature of the liquid $C = {28^ \circ }\,C$

All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.

1. The liquids $A$ and $B$ are mixed together. Then, the final temperature is ${16^ \circ }\,C$.

When the liquids $A$ and $B$ are mixed together,

Heat gained by the liquid $A$ is equal to the heat loss by the liquid $B$

By using specific heat capacity formula,

${Q_A} = {Q_B}$

Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_B}$ is the heat energy of liquid $B$.

${Q_A} = {Q_B}$

${m_A}{c_A}\left( {16 - 12} \right) = {m_B}{c_B}\left( {19 - 16} \right)\,.............\left( 1 \right)$

Where, ${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$,${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$.

All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.

Then, the equation (1) be changed as,

$m{c_A}\left( {16 - 12} \right) = m{c_B}\left( {19 - 16} \right)$

By cancelling the same term $m$ on both sides,

${c_A}\left( {16 - 12} \right) = {c_B}\left( {19 - 16} \right)$

On further simplifying,

$4{c_A} = 3{c_B}$

Therefore,

${c_B} = \dfrac{4}{3}{c_A}\,.............\left( 2 \right)$

2. The liquids $B$ and $C$ are mixed together. Then, the final temperature is ${23^ \circ }\,C$.

When the liquids $B$ and $C$ are mixed together,

Heat gained by the liquid $B$ is equal to the heat loss by the liquid $C$

By using specific heat capacity formula,

${Q_B} = {Q_C}$

Where, ${Q_B}$ is the heat energy of liquid $B$, ${Q_C}$ is the heat energy of liquid $C$.

${Q_B} = {Q_C}$

${m_B}{c_B}\left( {23 - 19} \right) = {m_C}{c_C}\left( {28 - 23} \right)\,.............\left( 3 \right)$

Where, ${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.

All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.

Then, the equation (3) be changed as,

$m{c_B}\left( {23 - 19} \right) = m{c_C}\left( {28 - 23} \right)$

By cancelling the same term $m$ on both sides,

${c_B}\left( {23 - 19} \right) = {c_C}\left( {28 - 23} \right)$

On further simplifying,

$4{c_B} = 5{c_C}$

Therefore,

${c_C} = \dfrac{4}{5}{c_B}\,................\left( 4 \right)$

Substituting the equation (3) in equation (4), then,

${c_C} = \dfrac{4}{5} \times \dfrac{4}{3}{c_A}$

On multiplying,

${c_C} = \dfrac{{16}}{{15}}{c_A}\,...............\left( 5 \right)$

3. The liquids $A$ and $C$ are mixed together. Then, the final temperature is $T$.

Heat gained by the liquid $A$ is equal to the heat loss by the liquid $C$

By using specific heat capacity formula,

${Q_A} = {Q_C}$

Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_C}$ is the heat energy of liquid $C$

${Q_A} = {Q_C}$

${m_A}{c_A}\left( {T - 12} \right) = {m_C}{c_C}\left( {28 - T} \right)\,............\left( 6 \right)$

Where,${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.

All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.

Then, the equation (6) be changed as,

$m{c_A}\left( {T - 12} \right) = m{c_C}\left( {28 - T} \right)$

By cancelling the same term $m$ on both sides,

${c_A}\left( {T - 12} \right) = {c_C}\left( {28 - T} \right)\,.................\left( 7 \right)$

On substituting the equation (5) in equation (7), then,

${c_A}\left( {T - 12} \right) = \dfrac{{16}}{{15}}{c_A} \times \left( {28 - T} \right)\,$

By cancelling the same terms on each side,

\[\left( {T - 12} \right) = \dfrac{{16}}{{15}}\left( {28 - T} \right)\,\]

On further,

\[15\left( {T - 12} \right) = 16\left( {28 - T} \right)\,\]

By multiplying,

$15T - 180 = 448 - 16T$

Takin the $T$ on one side and other terms in other side,

$15T + 16T = 448 + 180$

On further calculation,

$31T = 628$

Then,

$T = \dfrac{{628}}{{31}}$

On dividing,

$

T = {20.25^ \circ }\,C \\

T \simeq {20.3^ \circ }\,C \\

$

Thus, the temperature when $A$ and $C$ are mixed is ${20.3^ \circ }\,C$

**Hence, the option (C) is correct.**

**Note:**When the liquids are mixed together so the heat energy equation for the two liquids are equated. And all the three liquids are having the same mass then the mass value gets cancelled in the equations. The liquid $A$ and $C$ are mixed together then the final temperature is assumed as $T$.

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