Question

A 1.87 g sample of lauric acid (C12H2402), a common constituent of fats, is burned in a bomb calorimeter. The temperature of the calorimeter increases by 47.8°C (Celsius). Given that the energy of combustion (AE) for lauric is -37.1 kJ/g, calculate the heat capacity of the bomb calorimeter. Make sure you enter kJ/C for your unit behind the number. heat capacity of the calorimeter =

Answer 1

mass(C12H24O2)= 1.87 g

use:

Q = mass * delta E

= (1.87 g) * (-37.1 KJ/g)

= -69.377 KJ

This heat is absorbed by calorimeter

For calorimeter:

Q = 69.377 KJ

Use:

Q = Ccal * delta T

69.377 KJ = Ccal * 47.8 oC

Ccal = 1.45 KJ/oC

Answer: 1.45 KJ/oC