Please help me figure out where I went wrong.

Question

HNO3 + KOH -> KNO3 + H2O enthalpy change = -53.4 kJ/mol

55.0 mL of 1.30 mol/L solutions of both reactants, at 21.4°C, are mixed in a calorimeter. What is the final temperature of the mixture? Assume the density of both solutions is 1.00 g/mL and that the specific heat capacity of both solutions is the same as water's. No heat is lost to the calorimeter itself.

n=cxv
=1.30 mol/L x 0.055 L
= 0.0715 mol

Enthalpy change = -Q/n
-53.4 kJ/mol = -Q/0.0715 mol
-53.4 kJ x 0.0715 mol = -Q
-3.82 = -Q
3.82 = Q

Q=mc(delta)T
3.82 = (55.0g + 55.0g)(4.184 J/g°c)(Tf-21.4°c)
Tf-21.4 = 460.24 - 3.82
Tf = 456.42 + 21.4°C
Tf = 477.82°C

The answer should be 29.7°C

Answer 1

I just realised I made the stupidest mistake ever. I guess this is what happens when you try to do chemistry while sick.

The last bit of math doesn't make sense; it should be

3.82 = (460.24)(Tf-21.4°C)
3.82 = Tf460.24 - 9849.136
-Tf460.24 = -9849.136 - 3.82
Tf = -9852.956/-460.24
Tf = 21.4°C

The answer is still wrong, but I wanted to fix that embarrassing display of math.

Answer 2

I checked it over for what feels like the tenth time and I see that I should've converted the 3.82 kJ to 3820 J; I ended up with the correct answer using that.

But I feel like my method was a bit over-complicated. Is there a simpler method that I'm just overlooking?

Answer 3

I worked the problem and came up with the right answer but before I posted iZ looked at your solution and caught the errors. Yes, that 3.82 kJ should be 3818 J. I think the error is in the way you did the math It is almost ALWAYS easier to substitute the numbers and not do the algebra.
3818 = 110g x 4.184 x (Tf- 21.4)
At this point just keep going.
3818 = 460.2Tf - 9849.14
3818 + 9849.14 = 460.2Tf
Tf = 29.7