Enthalpy of Neutralisation 5.3

[Jefferson]

Hi everyone,
Please help me with the following questions (in attachment).

For
(1) I averaged the two, (23.2+24.4)/2 = 23.8

(2) Not sure what to do here, so I did 33.5-23.8 = 9.7 (took value for first minute minus average initial).

(3) q = 100g * 4.18J/gK * 9.7K = 4054.6 (not sure if that value for ΔT is correct).

(4) NaOH + HCl ---> NaCl + H₂O
n(HCl) = 0.05L * 2 mol/L = 0.1
n(NaOH) = 0.05L * 1 mol/L = 0.05 <---- limiting reagent.

(5) 0.05 moles of H₂O produced
 
(6) (4054.6 * 10^-3) kJ / 0.05 mol  = -81.092 kJ (exothermic)
Ideal value is -57.2 kJ/mol. Should the value have been less than 57.2 (in magnitude) due to heat being loss in the external environment, hence the recorded heat released is lower? Where did I go wrong in previous steps?

Thanks

[r1ckworthy]

Hey Jefferson,

I think the step where you went wrong is question 2. Delta (the triangle before T) is a greek symbol that stands for 'change in'. Basically, the formula is:
Change in T= Final Temperature - Initial temperature
After that, you should calculate everything else correctly.

Not sure if this answer is too relevant, as it was posted a long time, but hope it helps anyhow.