Calorimetry, finding final temperature

I was looking for some help on understanding this practice question from my text book, I know how to figure out enthalpy, but this is my first temperature change question and I am stumped. Any guidance would be greatly appreciated.



> 
> Nitric acid is neutralized with potassium hydroxide in the following reaction:
> 
> 
> $$\ce{NHO3(aq) + KOH3(aq) -> KNO3(aq) + H2O(l)}$$
> 
> 
> $$\Delta\_\mathrm rH= -53.4\ \mathrm{kJ/mol}\ \ce{HNO3}$$.
> 
> 
> $55.0\ \mathrm{mL}$ of a $1.30\ \mathrm{mol/L}$ solutions of both reactants, at $21.40\ \mathrm{^\circ C}$, are mixed in a calorimeter. What is the final temperature of the mixture? Assume that the density of both solutions is $1.00\ \mathrm{g/mL}$. Also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. No heat is lost to the calorimeter itself.
> 
> 
> 


I know the answer is 29.7 degrees Celsius, I'm just not sure how to get there. I tried juggling around my equation "delta t r H= m c delta t" but since I don't know the final temperature i don't think the delta t works in this case. I've tried omitting the delta t but I get no where near 29.7 degrees Celsius.


So I took my formula Q=mc delta t and rearranged it, I now have delta t= Q / mc. Delta t= -53.4 kJ/mol / (55.0g)(4.19 J/g C) I changed the 55mL to g as the question says to assume the density is 1.00 g/mL. I also used the heat capacity as water given in my course data booklet). And I get -0.2317. I tried doubling the mass as there is two solutions but that doesn't work either




n(HNO3)= 0.055L \* 1.30 mol\L = 0.0715 mol  

Delta H = 0.0715 mol \* -53.4 kJ/mol =3.818kJ  
 
Delta H = mc delta t  

3818J = (110g)(4.19J/g\*C)(delta t)  
 
Delta t = 8.3\*C  
 
Ti + Tf = 21.4\*C + 8.3\*C = 29.7\*C 


 -1