When 20.0 mL of 1.0 M H3PO4 is added to
60.0 mL of 1.0 M NaOH at 25.0 degrees celcius in a calorimeter, the
temperature of the aqueous solution increases to 35.0 degrees
celcius.
Assuming that the specific heat of the solution is 4.18
J/(g⋅∘C), that its density is 1.00 g/mL, and that the calorimeter
itself absorbs a negligible amount of heat, calculate ΔH
in kilojoules/mol H3PO4 for the reaction.
H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)