The heat capacity (calorimeter constant) of the calorimeter is 34.85 kJ/ ∘C, what is the heat of combustion per gram of the material?

Question

A 2.333 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 25.17 ∘C to 30.62 ∘C.

The heat capacity (calorimeter constant) of the calorimeter is 34.85 kJ/ ∘C, what is the heat of combustion per gram of the material?

heat of combustion: ? kJ/g

Answer 1

When the material is combusted, the heat generated is absorbed by the calorimeter. The way to calculate the amount of heat absorbed by the calorimeter is to use the equation

q = C∆T

q = heat absorbed = ?

C = calorimeter constant = 34.85 kJ/º

∆T = change in temperature = 30.62º - 25.17º = 5.45º

Solving for q we have...

q = (34.85 kJ/º)(5.45º) = 189.9 kJ

Since the question asks for the heat of combustion PER GRAM of material, we now simply divide this value by the grams of material (2.333 g) to get...

189.9 kJ / 2.333 g = 81.41 kJ/g (4 sig.figs.)

Answer 2

The starting point for solving this problem is that the amount of heat gained by the calorimeter is equal to the amount of heat produced in the combustion reaction, therefore, you can figure out how much heat the calorimeter gained (in kJ) and since this is equal to the amount of heat that the reaction produced, you can then divide this kJ value by the amount of material you started with in grams to get the heat of combustion in kJ/g.

So how do you calculate the heat gained by the calorimeter? Well, you know the heat capacity of the calorimeter, so you can calculate the amount of heat gained by the calorimeter by multiplying the heat capacity and the difference between the initial and final temperatures.