Temperature rise remains constant as heat and volume scale proportionally

Steps:

• Calculate moles in first experiment: 25 cm³ of 2.0 mol dm⁻³ NaOH and HCl each gives 0.05 mol of each reactant.
• Heat released is proportional to moles neutralized (0.05 mol), causing 12°C rise in 50 cm³ total volume.
• In second experiment, 50 cm³ of each gives 0.1 mol, so heat doubles, but total volume doubles to 100 cm³.
• Temperature rise ΔT = heat / (volume × specific heat capacity × density), so ΔT remains 12°C.

Why B is correct:

• Enthalpy of neutralization for strong acid-base is constant per mole; equal scaling of moles and volume keeps ΔT unchanged per the heat capacity formula q = m c ΔT.

Why the others are wrong:

• A: Halves ΔT incorrectly, ignoring proportional scaling.
• C: Doubles ΔT, assuming heat increase without volume effect.
• D: Quadruples ΔT, confusing moles with volume squared.

Final answer: B