Could you please look over my work and see if I did this correctly? Thanks
When 3.200 g of liquid ethyl alcohol (C2H5OH) is burned in a bomb calorimeter containing 3.500 kg of water, the temp rises from 22.84 degrees C to 28.36 degrees C. The calorimeter constant is 25550 J/ degree C.
a) calculate the delta E for this combustion process. Call it delta E comb. Convert it to kj/mole C2H5OH.
This is what i did:
delta E comb = 5.52 degrees C (delta T) x 2550 j/degree C
delta E comb = 14076 J
= 14.076 kJ
3.200 g (1 mole ethyl/46.08 g)= 0.069 mol
14.076kJ/0.069 mol= 204 mol
but since it's combustion its negative, so -204 kJ/mol.
b) balance the reaction for the combustion process, forming liquid H20 and CO2 (g).
C2H5OH (l) + 3O2 (g)---> 2CO2 (g) + 3H2O (l)