How do I solve this problem

Question

When 1.956 g of sucrose ( molar mass 342.3 g/mole) is burned in a bomb calorimeter, the temperature of the calorimeter increases From 22.41 Celsius to 26.63 Celsius. If the heat capacity of the calorie meter is 4.900 KJ/ Celsius, what is the heat of combustion of sucrose?

Answer 1

q = mCcal∆T

q = heat = ?

m = mass = 1.956 g

C_cal = heat capacity of the calorimeter = 4.900 kJ/º = 4900 J/º

∆T = change in temperature = 4.22º

q = (1.956 g)(4900 J/º)(4.22º)

q = 40,446 J

To express the heat of combustion on a per mole basis, we find the moles of sucrose present in 1.956 g.

1.956 g sucrose x 1 mol/342.3 g = 0.005714 moles

∆Hcombustion = 40,446 J/0.005714 moles x 1 kJ/1000 J = 7078 kJ/mole