The heat of neutralization of HCl(aq) and NaOH(aq) is -55.90 kJ/mol of water produced. If 50.00 mL of 1.16 M NaOH at 25.15 ºC is added to 25.00 mL of 1.79 M HCl at 26.34 ºC in a plastic-foam cup calorimeter, what willthe solution temperature be immediately after the neutralization reaction has occurred? You may assume that each solution has a density and specific heat that is the same as for pure water and that no heat escapes from the calorimeter.

Question

The heat of neutralization of HCl(aq) and NaOH(aq) is -55.90 kJ/mol of water produced.  If 50.00 mL of 1.16 M NaOH at 25.15 ºC is added to 25.00 mL of 1.79 M HCl at 26.34 ºC in a plastic-foam cup calorimeter, what willthe solution temperature be immediately after the neutralization reaction has occurred?  You may assume that each solution has a density and specific heat that is the same as for pure water and that no heat escapes from the calorimeter.

I have already found the number of moles: 0.058 mol OH and 0.04475 mol H, and determined that HCl is the limiting reagent.

I have also found that (0.04475 mol H)(-55.9 kJ/mol) = -2.50 kJ.

I'm confused as to where to go from here. I don't need answers, but I would like to see formulas so I can solve it myself.