Final temperature from neutralization heat and solution heat capacity Steps:

• Calculate moles HNO3 = (20/1000) × 0.40 = 0.008 mol; moles NaOH = (30/1000) × 0.20 = 0.006 mol; limiting is NaOH at 0.006 mol reacting.
• Heat released Q = 0.006 × 57 × 1000 = 342 J (exothermic, so positive for warming).
• Total heat capacity = 50 cm³ × 4.2 J cm⁻³ °C⁻¹ = 210 J °C⁻¹.
• ΔT = Q / heat capacity = 342 / 210 = 1.6 °C; final T = 25.0 + 1.6 = 26.6 °C (matches A after precise computation).

Why A is correct:

• A uses Q = n × |ΔH| for limiting n and Q = C ΔT with C = V × specific heat capacity per volume, per conservation of energy.

Why the others are wrong:

• B ignores limiting reactant, using average moles (0.007 mol) for excess heat.
• C uses excess HNO3 moles (0.008 mol) as if fully neutralized.
• D adds moles of both reactants, overestimating heat by double-counting.

Final answer: A