Energy Content of Fuels Investigation Lab Report - 1504 Words

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†¦show more content†¦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 ï‚ ® 2CO2 + 4H2O 6 x C-H (412) + 2 x C-O (360) + 3 x O-H (463) + 3 x O=O (498) ï‚ ® 4 x C=O (805) + 8 x H-O (463) = 5612 – 6924 = -1312KJ/Mole Ethanol C2H5OH + 3O2 ï‚ ® 2CO2 + 3H2O 5 x C-H (412) + C-C (348) + C-O (360) + O-H (463) + 3 x O=O (498) ï‚ ® 4 x C=O (805) + 6 X O=H (463) = 4725 – 5998 = -1273KJ/mole Propanol 2C3H7OH + 9O2 ï‚ ® 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) ï‚ ® 12 x C=O (805) + 16 x O=H (463) = 13288 – 17068 = -1890KJ/mole Butanol C4H9OH + 6O2 ï‚ ® 4CO2 + 5H2O 9 x C-H (412) + 3 x C-C (348) + C-O (360) + O-H (463) + 6 x O=O (498) ï‚ ® 8 x C=O (805) + 10 x O=H (463) = 8563 – 11070 = -2507KJ/mole Pentanol 2C5H11OH + 15O2 ï‚ ® 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) ï‚ ® 20 x C=O (805) + 24 x O=H (463) = 20964 – 27212 = -3124KJ/mole Hexanol C6H13OH + 9O2 ï‚ ® 6CO2 + 7H2O 13 x C-H (412) + 5 x C-C (348) + C-O (360) + O-H (463) + 9 x O=O (498) ï‚ ® 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,Show MoreRelatedTwo Shaft Gas Turbine Lab Report1546 Words   |  7 PagesTwo-Shaft Gas Turbine 1st Law Demonstration Lab Report CCTD101B THE UNIVERSITY OF TRINIDAD TOBAGO FOR: MR REAN MAHARAJ March 25, 2012 Authored by: Odia Pollard (55628) 1 Contents AIM ................................................................................................................................................................ 2 METHOD..............................................................................................................................................Read MoreBricks6391 Words   |  26 Pagesutilisation of alternative materials in the manufacture of mineral wool insulation Compiled by Dr Evaggelia Petavratzi John Barton, School of Civil Engineering University of Leeds (DEFRA Project Code WRT_177) September 2007 -2- Contents 1 Scope ............................................................................................................................................. 4 2 The brick manufacturing sector in the UK..................................................Read MoreChemistry: Glucose and Sports Drinks2098 Words   |  9 PagesDetermination of Sugars in Sports Drinks: A Spectrophotometric Analysis By: Crystle Culling Student Number: 2686923 Class: Tuesday 12pm (odd weeks) Page | 1 1015MSC Lab Report Crystle Culling 2686923 Abstract The concentration of sugars in two well-known sports drinks, Powerade and Gatorade, were determined by monitoring an enzyme-catalysed reaction sequence involving the appearance of NADPH. 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