Question

A 3.15 g sample of lauric acid (c12h24o2), a common constituent of fats, is burned in a bomb calorimeter. the temperature of the calorimeter increases by 80.6oc (celsius). given that the energy of combustion (∆e) for lauric is -37.1 kj/g, calculate the heat capacity of the bomb calorimeter. heat capacity of the calorimeter =

Answer 1

The heat capacity of the bomb calorimeter is calculated to be approximately 1.45 kJ/°C using the energy of combustion for lauric acid and the measured temperature increase.

Step-by-step explanation:

To calculate the heat capacity of the bomb calorimeter using the data provided for the combustion of lauric acid, we apply the concept that the energy released by the substance (q) is absorbed by the calorimeter, which increases its temperature.

First, we calculate the total energy released (q) by the sample when burned:

q = (ΔE) × mass = –(-37.1 kJ/g) × 3.15 g = –116.865 kJ

Since the calorimeter absorbed all the energy released, we can write:

q = Ccalorimeter × ΔT

Solving for Ccalorimeter, we get:

Ccalorimeter = q / ΔT

Ccalorimeter = –116.865 kJ / 80.6°C

The negative sign on q can be disregarded for this calculation since the temperature increase means the calorimeter absorbed the heat.

Therefore, the heat capacity of the calorimeter is:

Ccalorimeter ≈ –116.865 kJ / 80.6°C ≈ 1.45 kJ/°C