Investigating Energy When Fuels Burn.

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Investigating Energy When Fuels Burn

Plan and aim

My aim is to find out how much energy is released when different alcohols are burnt.

There are many different variables including:

  • The type of alcohol.
  • Mass of alcohol in moles.
  • Mass of water in volume.
  • The height and shape of the can.
  • The starting temperature in C.
  • The temperature rise of water.

My input variable will be different types of alcohol. My output variable is the temperature rise of the water because I need this to be able to calculate the amount of energy released. I will use the following formula to calculate the amount of energy released by each alcohol:

        ENERGY = Mass of water x 4.2 (specific heat capacity) x temperature rise

I will control the mass of fuel, mass of water, height and shape of the can and the starting temperature. I am going to control the other possible variables to make it a fair experiment.

If I repeat each alcohol six times, this will allow me to calculate an average. This should help me to notice any odd results. If there are any anomalous results, they will not be included because they will make the average odd.

I will need to wear goggles and tie back long hair for safety issues. The Bunsen burner will always be kept on a safety flame so it is visible.

Prediction

I predict that as the number of the carbons in the alcohol increases, the temperature rise in the water will increase. The bond energies show that the more carbons there are in the alcohol the more energy is given out. This reaction is exothermic so it will lose energy.

Methanol

Methanol + oxygen → carbon dioxide + water

CH4O + 1.5O2  → CO2  + 2H2O

  1. C-H x 3 = 413 x 3 = 2139

C-O x 1 = 358 x 1 = 358                        = 2808.45

O-H x 1 = 464 x 1 = 464

O=O x 1.5 = 498.3 x 1.5 = 747.45

  1. C=O x 1 = 805 x 2 =  1610

O                                                = 3466

O-H x 4 = 464 x 4 = 1856

2808.45 – 3466 = -657.55

Ethanol

Ethanol + oxygen → carbon dioxide + water

C2H6O + 3O2  → 2CO2  + 3H2O

(1) C-H x 5 = 413 x 5 = 2065

C-C x 1 = 347 x 1 = 347

Join now!

C-O x 1 = 358 x 1 = 358                        = 4728.9

O-H x 1 = 464 x 1 = 464

O=O x 3 = 498.3 x 3 = 1494.9

(2) C=O x 2 = 805 x 4 = 3220

O                                                = 6004

O-H x 6 = 464 x 6 = 2784

4728.9 – 6004 = -1275.1

Propanol

Propanol + oxygen → carbon dioxide + water

C3H8O + 4.5O2  → 3CO2  + 4H2O

(1) C-H x 7 = 413 x 7 = 2891

C-C x 2 = 347 x 2 = 694

C-O x 1 = 358 x 1 = ...

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