chemistry grade 12 Question?

Molar mass of CH₃OH
= (12.01 + 16.0 + 1.008×4) g/mol
= 32.04 g/mol

No. of moles of CH₃OH
= (5.000 g) / (32.04 g/mol)
= 0.1561 mol

Energy released from reaction
= Energy absorbed by water
= m c ΔT
= (4000 g) × (4.184 J/g°C) × [(29.765 - 24.000)°C]
= 96480 J
= 96.48 kJ

Energy fro the reaction
= -(96.48 kJ) / (0.1561 mol)
= -618.1 kJ/mol

first... for a temp change without a phase change, we use this equation for heat
.. Q = m * Cp * dT
and if we group like this
.. Q = (m * Cp) * dT
and look at typical units of those terms in ( )
.. (m * Cp) ===> (g * J/g°C) ----> J/°C
notice those are the same units of the "calorimeter"
meaning the 2657 J/°C is really (m * Cp) for the calorimeter.
Why do we do this you ask? because instead of measuring mass + Cp each time we
use our calorimeter, we can just measure it 1 time and stamp it on the outside of the calorimeter.

that behind us.. let's continue
**********
from an energy balance
.. heat given off by reaction = Qrxn = heat absorbed by water + heat absorbed by calorimeter
then
.. Qrxn = (m * Cp * dT)water + (m * Cp * dT)calorimeter
and using that stuff from above
.. Qrxn = (m * Cp * dT)water + ((m * Cp) * dT))calorimeter

plugging in the data
.. Qrxn = (4000g * 4.184J/g°C * (29.765ºC - 24.000ºC)) + (2657 J/ºC * (29.765ºC - 24.000ºC))
.. Qrxn = 1.118x10^5 J

and that is for 5.000g CH3OH (molar mass = 32.04 g/mol).. converting to kJ/mol
.... .... ... ... ... ... ... ... ... .. .1.118x10^5 J.. . 1 kJ... .. . 32.04 g
.. Qcombustion CH3OH = ---- ---- ---- ---- x ---- ---- x --- ---- --- = 716.4 kJ/mol
.... .... ... .... ... .... ... ... ... .... ..5.000g.. ... ...1000J.. ... .1 mol

which compares nicely to the upper heating value shown here
https://en.wikipedia.org/wiki/Heat_of_combustion