Thermodynamics question. 10 Points for Best Answer!

I can't seem to solve this question :(

A calorimeter with a heat capacity of 128KJ/(degrees)C contains 75g of water at 22.2(degrees)C. If the temperature increases to 45.7(degrees)C, how much energy has been added?

Can you show me how you got the answer?

10 Points for best answer! :)

BY Q(total) = Q(calorimeter) + Q(water)
=>Q = [s x ∆t](calorimeter) + [ms∆t](water)
=>Q = 128000 x (45.7 - 22.2) + 0.075 x 4187 x (45.7 - 22.2)
=>Q = 3015379.59 J OR 3015.38 kJ

ΔH=CΔT
ΔT=Tfinal-Tinitial
ΔH=128KJ/(°C)*(45.7-22.2°C)
ΔH=+3,008 KJ
3 significant figures so...
ΔH=+3,010 KJ