Solution
Let us analyze the energy balance during the dissolution process. The total heat content (enthalpy) of the system before mixing plus any heat of reaction must equal the total heat content after mixing.
Let $H_{mix}$ be the constant heat of dissolution (energy released/absorbed due to chemical potential change) which is independent of temperature. By conservation of energy:
$$ \text{Heat of Ingredients} + \text{Heat of Reaction} = \text{Heat of Solution} $$
Case 1: Dissolving at temperature $T_1$
$$ (C_1 T_1 + C_2 T_1) + H_{mix} = C_3 (T_1 + \Delta T_1) $$ $$ (C_1 + C_2) T_1 + H_{mix} = C_3 T_1 + C_3 \Delta T_1 \quad \dots(1) $$
Case 2: Dissolving at temperature $T_2$
$$ (C_1 T_2 + C_2 T_2) + H_{mix} = C_3 (T_2 + \Delta T_2) $$ $$ (C_1 + C_2) T_2 + H_{mix} = C_3 T_2 + C_3 \Delta T_2 \quad \dots(2) $$
Subtract equation (1) from equation (2) to eliminate $H_{mix}$:
$$ (C_1 + C_2)(T_2 – T_1) = C_3(T_2 – T_1) + C_3(\Delta T_2 – \Delta T_1) $$
Dividing the entire equation by $C_3$:
$$ \frac{C_1 + C_2}{C_3}(T_2 – T_1) = (T_2 – T_1) + (\Delta T_2 – \Delta T_1) $$
Rearranging to solve for the unknown $\Delta T_2$:
$$ \Delta T_2 = \Delta T_1 + \frac{C_1 + C_2}{C_3}(T_2 – T_1) – (T_2 – T_1) $$ $$ \Delta T_2 = \Delta T_1 + (T_2 – T_1) \left[ \frac{C_1 + C_2}{C_3} – 1 \right] $$ $$ \Delta T_2 = \Delta T_1 + (T_2 – T_1) \left[ \frac{C_1 + C_2 – C_3}{C_3} \right] $$