Three different liquids A , B , and C have temperatures of 12^ ^ ^...

Let the specific heat capacities of liquids $A$, $B$, and $C$ be $s_A$, $s_B$, and $s_C$ respectively. Let the mass of each liquid be $m$.

When liquids $A$ and $B$ are mixed, the equilibrium temperature is 16^{}^{{}^{\circ}} \text{C}. By the principle of calorimetry, heat gained by $A$ equals heat lost by $B$:

$m s_A (16 - 12) = m s_B (19 - 16)$

$4 s_A = 3 s_B \Rightarrow s_A = \dfrac{3}{4} s_B$

When liquids $B$ and $C$ are mixed, the equilibrium temperature is 23^{}^{{}^{\circ}} \text{C}. Heat gained by $B$ equals heat lost by $C$:

$m s_B (23 - 19) = m s_C (28 - 23)$

$4 s_B = 5 s_C \Rightarrow s_C = \dfrac{4}{5} s_B$

Let $T$ be the equilibrium temperature when liquids $A$ and $C$ are mixed. Heat gained by $A$ equals heat lost by $C$:

$m s_A (T - 12) = m s_C (28 - T)$

Substituting the values of $s_A$ and $s_C$ in terms of $s_B$:

$\dfrac{3}{4} s_B (T - 12) = \dfrac{4}{5} s_B (28 - T)$

$\dfrac{3}{4} (T - 12) = \dfrac{4}{5} (28 - T)$

Multiplying both sides by $20$:

$15(T - 12) = 16(28 - T)$

$15T - 180 = 448 - 16T$

$31T = 628$

T = \dfrac{628}{31} \approx 20.258^{}^{{}^{\circ}} \text{C}

Rounding to one decimal place, the temperature is 20.3^{}^{{}^{\circ}} \text{C}.