At a constant volume

Question

At constant volume, the heat of combustion of a particular compound, compound A, is - 3262k / m * ol When 1.029 g of compound A (molar mass = 115.09g / m * ol ) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C.

What is the heat capacity (calorimeter constant) of the calorimeter?

C =

KJFC

Suppose a 3.419 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 25.77 °C to 30.58 °C

What is the heat of combustion per gram of compound B

heat of combustion:

kJ / g

Answer 1

moles A combusted = 1.029 g x 1 mol / 115.09 g = 8.941x10^-3 moles

heat of combustion = 8.941x10^-3 moles x -3262 kJ/mol = -29.16 kJ

Calorimeter constant (C_cal) = 29.16 kJ / 3.661º = 7.965 kJ/º = C_cal

∆T = 30.58º - 25.77º = 4.81º

q = 4.81º x 7.965 kJ/º = 38.31 kJ

Heat of combustion per gram = 38.31 kJ / 3.419 g = 11.21 kJ/g