Need helped finding expected temperature change?

18.1°C is the experimental temperature change.
You are asked to calculate the expected temperature change (i.e. the calculated temperature change) based on the data given in the question.

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HCl is a strong acid, while NaOH is a strong alkali.
Initial no. of moles of H⁺ = Initial no. of moles of HCl = (4.0 mol/L) × (19.9/1000 L) = 0.0796 mol
Initial no. of moles of OH⁻ = Initial no. of moles of NaOH = (4.0 mol/L) × (39.4/1000L) = 0.1576 mol

The heat of neutralization (or enthalpy change of neutralization) of a strong acid and a strong alkali is -57.62 kJ/mol, i.e.
H⁺(aq) + OH⁻(aq) → H₂O(l) …… ΔH = -57.62 kJ/mol
Mole ratio H⁺ : OH⁻ : H₂O = 1 : 1 : 1
As (Initial no. of moles of H⁺) < (Initial no. of moles of OH⁻), H⁺ is the limiting reactant.
No. of moles of H₂O formed = Initial no. of moles of H⁺ = 0.0796 mol

Heat released from reaction = (57.62 × 1000 J/mol) × (0.0796 mol)
Heat absorbed by the solution = m c ΔT = {[(39.4 + 19.9) mL] × (1 g/mL)} × (4.18 J/g°C) × ΔT

Heat absorbed by the solution = Heat released from reaction
(39.4 + 19.9) × 1 × 4.18 × ΔT = 57.62 × 1000 × 0.0796
59.3 × 4.18 × ΔT = 57.62 × 1000 × 0.0796
ΔT = 57.62 × 1000 × 0.0796 / (59.3 × 4.18) °C
Expected temperature change, ΔT = 18.5°C