Question

A 1.533 g

1.533 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.37 

∘

C

24.37 ∘C to 28.46 

∘

C.

28.46 ∘C.

The heat capacity (calorimeter constant) of the calorimeter is 33.97 kJ/ 

∘

C,

33.97 kJ/ ∘C, what is the heat of combustion per gram of the material?

Answer 1

$q=c\Delta t$

q - heat absorbed

c = heat capacity of calorimeter = 33.97 kJ/°C

$\Delta t=t2-t1=28.46-24.37=4.09$

$q=c\Delta t=33.97\times4.09=138.9373$

when absorbed, the sign is positive, when isolated, it is negative

the heat of combustion per gram of the material:

$\Delta H=\frac{q}{m}=\frac{-138.9373}{1.533}=-90.630985$ kJ/g