In order to calibrate a bomb calorimeter , 3000 g of water is added to the calorimeter. next,

1.08 x 10-2mol of ethanol (C2H6O) is combusted in the presence of excess oxygen according to
the equation,
C2H6O (l) + 3 O2 (g) > 3 H2O (l) + 2 CO2 (g)
DE = - 1365 kJ/mol .
Assuming that the temperature of the water increases by 0.966 oC calculate the heat capacity of the
empty calorimeter (the calorimeter without water). (Remember that the specific heat of water is 4.18
J/g K)

1 answer

How much heat is added to the calorimeter and water with the combustion of the ethanol. That's 0.0108 mol x 1365 kJ/mol = ? and convert to J.
Then J heat = (mass H2O x specific heat H2O x delta T) + Ccal*delta T
Solve for Ccal.