When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of the calorimeter increases by 10.0 c?

I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture. 

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
Q = +6970 kJ

Man i dont know this answer

A, it is falling action my have a good day