First, we need to calculate the heat absorbed by the water:
q = mCΔT
where q is the heat absorbed by the water, m is the mass of water, C is the specific heat of water, and ΔT is the change in temperature.
q = (1200 g)(4.184 J/g oC)(33.20 oC - 25.00 oC)
q = 35,220.48 J
Next, we need to calculate the total heat absorbed by the system (water + calorimeter):
Q = q + CcalΔT
where Q is the total heat absorbed by the system, Ccal is the heat capacity of the calorimeter, and ΔT is the change in temperature of the system.
Q = 35,220.48 J + (837 J/oC)(33.20 oC - 25.00 oC)
Q = 36,993.16 J
Finally, we need to calculate the heat released by the combustion of octane:
ΔH = Q/n
where ΔH is the heat released per mole of octane, Q is the total heat absorbed by the system, and n is the number of moles of octane.
n = mass/molar mass = 1.000 g/114.23 g/mol = 0.008748 mol
ΔH = 36,993.16 J/0.008748 mol
ΔH = 4,228,236 J/mol
Therefore, the heat released per mole of octane is 4,228,236 J/mol.