# Energy Content of Fuels Investigation Lab Report

1504 Words Oct 1st, 2010 7 Pages
Fuels Investigation Aim: To find out which fuel gives out the most energy. Planning
We will be using 6 different fuels to heat up 100ml of water, and find out the changes of the temperature. We will measure the temperatures of the water before and after the experiment. We will burn heat the water for exactly 2 minutes, and check the changes in temperature. The change in temperature will allow us to work out the energy given off the fuel by using this formula: Mass of water x 4.2 (water’s specific heat capacity) x temperature change = energy transferred from the fuel to the water When the fuels are burnt, energy is given off. I will be calculating the energy given off using the formula above. The specific heat capacity is the energy
Types of bonding | Energy in bond (kilojoules per mole) | C – C | 348 KJ | C – H | 412 KJ | C – O | 360 KJ | C = O | 805 KJ | O – H | 463 KJ | O = O | 498 KJ | The balance equations and theoretical energy given off: - Methanol
2CH3OH + 3O2  2CO2 + 4H2O 6 x C-H (412) + 2 x C-O (360) + 3 x O-H (463) + 3 x O=O (498)  4 x C=O (805) + 8 x H-O (463)
= 5612 – 6924
= -1312KJ/Mole Ethanol
C2H5OH + 3O2  2CO2 + 3H2O 5 x C-H (412) + C-C (348) + C-O (360) + O-H (463) + 3 x O=O (498)  4 x C=O (805) + 6 X O=H (463)
= 4725 – 5998
= -1273KJ/mole
Propanol
2C3H7OH + 9O2  6CO2 + 8H2O 14 x C-H (412) + 4 x C-C (348) + 2 x C-O (360) + 2 x O-H (463) + 9 x O=O (498)  12 x C=O (805) + 16 x O=H (463)
= 13288 – 17068
= -1890KJ/mole Butanol
C4H9OH + 6O2  4CO2 + 5H2O 9 x C-H (412) + 3 x C-C (348) + C-O (360) + O-H (463) + 6 x O=O (498)  8 x C=O (805) + 10 x O=H (463)
= 8563 – 11070
= -2507KJ/mole Pentanol
2C5H11OH + 15O2  10CO2 + 12H2O 22 x C-H (412) + 8 x C-C (348) + 2 x C-O (360) + 2 x O-H (463) + 15 x O=O (498)  20 x C=O (805) + 24 x O=H (463)
= 20964 – 27212
= -3124KJ/mole Hexanol
C6H13OH + 9O2  6CO2 + 7H2O 13 x C-H (412) + 5 x C-C (348) + C-O (360) + O-H (463) + 9 x O=O (498)  12 x C=O (805) + 14 x O=H (463)
= 12401 – 16142
= -3741KJ/mole At the Left side of the equation, its show the energy (KJ/mole) taken in by the reaction,