Enthalpy change of neutralisation help

The question says 'work out the enthalpy change for the neutralisation of 1 mole of KOH by H2SO4' (i.e you're working out the enthalpy change per mole of the reaction, so your answer will only be in KJ - the 'per mole' is assumed as it has been asked in the question).

Energy change = 70 x 4.18 x (36-19.5) = 4827.9 J or 4.8279 kJ (this was correct)

But we need to work out how many moles of KOH we reacted:

MolesKOH = (35/1000) x 2.4 = 0.084

Hence, our energy change per mole of KOH reacted is (-4.8279/0.084) = -57.475 kJ/mol (don't forget the neagtive because the reaction is exothermic). So your answer is -57.5 kJ (3 s.f)

Ohhhh okay that makes sense thank you
The energy change of 4.2879 is endothermic because is that the energy that you're putting into the water??

This question looks at neutralisation reactions.
(a) A student carries out an experiment to determine the enthalpy change for a neutralisation
reaction.
The student measures out 35.0 cm3 of 2.40 mol dm–3 KOH and 35.0 cm3 of 1.20 mol dm–3
H2SO4. The temperature of each solution is 19.5 ºC.
The student mixes the solutions. The KOH is all neutralised and the maximum temperature
reached is 36.0 °C.

(ii) Calculate the enthalpy change for the neutralisation of 1 mol KOH by H2SO4.
Assume that the density of the mixture is 1.00 g cm–3 and that the specific heat
capacity for the solution is the same as for water.

The question says to give your answer in KJ not KJ mol ^-1 so why did they also work out the moles of water and then divide their energy in KJ by the moles??

For the energy I got : +4.8279 KJ