Question

If 200. mL of aqueous 0.862 M HCl is mixed with 200. mL of aqueous 0.431 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 20.48ºC. The neutralization reaction H+(aq) + OH–(aq) → H2O(l) has a ΔHº = –56.2 kJ/mol. What is the final temperature of the mixed solution? (Assume all solutions have a density of 1.00 g/cm3 and a specific heat capacity of 4.184 J/g•ºC)

Answer 1

i got that rize in temperature is 5,78 °C or that final temperature is 26,26 °C if you have solutions please check so i can tell you how its calculated

Answer 2

ok ill type in short what ive done write balanced equation of neutralization and there you see that 2 molecules of water are created.
then i calculated from concentration and volume moles of HCl, then from that and reaction you can see that moles of HCl = moles of H2O and put that in equation
Q = dH * n
where Q is heat you get, dH is enthalpy of neutralization you are given (56,2 kJ/mol)
and now that we know what heat is generated we go on to using
Q = m cp dT -> dT = Q / (m cp)
where dT is change in temperature, Q is heat generated you calculated above, m is mass which ill explain further on how to get, cp is heat capacity you got given (4.184 J/gºC)

you are given density and from there you calculate mass
density (rho) = m/V -> m = rho * V
where rho you are given (1 g/cm3) and V is volume (volume of both solutions so its 400 ml)

and when you get all that in you get that dT is 5,78 °C or that final temperature is 20,48+5,78 = 26,26 °C