q = Ccal*dT = 4.84 kJ*(27.1-23.7) = ? kJ
heat combustion is -?kJ/0.009298 = -?kJ/mol
A 0.009298-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.7oC to 27.1oC. If the heat capacity of the calorimeter is 4.84 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole?
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