What Is the Molar Enthalpy of Neutralization of NaOH?

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The molar enthalpy of neutralization for NaOH when reacting with H2SO4 is calculated using the formula n(molar enthalpy) = vc(change in temperature), where c is the specific heat capacity. The initial temperatures of NaOH and H2SO4 are 20.6°C and 21.3°C, respectively, with a final temperature of 28°C. The calculation yields a molar enthalpy of 29 kJ/mol, but there are concerns about the accuracy of averaging the initial temperatures and the need for the exact volume of sulfuric acid used. A full heat balance approach is recommended, assuming the specific heat of the solution is similar to that of pure water for better accuracy. Verification of the calculation and methodology is sought for correctness.
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Homework Statement



WHat is the molar enthalpy of neutralization of NaOH (aq) when 50.0 mL of aqueous 1.00 mol/L of NaOH (aq) reacts with an excess of 1.00 mol/L H2SO4 (aq)?

Initial temperature of H2SO4 - 21.3 degrees celcius
Initial Temperature of NaOH - 20.6 degrees celcius

Final temperature of solution - 28 degrees celcius


Homework Equations



n(molar enthalpy) = vc(change in temperature)

Where c = 4.19 J/mL x degrees celcius

The Attempt at a Solution



molar enthalpy = ( vct) / (n)

= ( 50 mL x 4.19 (28 - (21.3 + 20.6 / 2)) / (1.00 mol/L x 50.0 mL)

= 29 KJ / mol

I'm not entirely sure if this answer and its units are correct, or if averaging the two initial temperatures was correct. Any verification of my answer would be great.
 
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Averaging is not a good idea, and I don't think you can solve the question not knowing exact volume of sulfuric acid used. You should do full heat balance and assume solution to have specific heat of pure water - while this is not exactly true it would be probably close enough to the reality.
 
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