Conclusion:
In 2 minutes the ethanol increased its temperature by 1°c from 5°c to 6°c.
Evaluation:
We realized from my preliminary experiment that 40cm3 of water were too much to heat up in a reasonable time. From my experiment we decided that 3cms was a suitable distance of the water from the flame. In my main experiment we will use a water volume of 30cm3.
Main Experiment
Method:
We are going to find out which alcohol is the best fuel for raising the temperature of water. We will do this by investigating how much fuel it takes to raise the temperature of a boiling tube containing 30cm3 of water by 20ºC. We will set the fuel in a jar with cotton wool and a wick so that it will create a controlled flame when it is set alight. The boiling tube of water will be set at a distance of 3cm from the flame.
We will use the following alcohols:
Methanol: CH3OH
Structure
Ethanol: C2H5OH
Structure
Propan-1-ol: C3H7OH
Structure
Propan-2-ol: C3H7OH
Structure
Butan-1-ol: C4H9OH
Structure
Equipment:
Thermometer
Boiling Tube
Clamp and stand
Burner
Fuels
30cm3 of water
Weighing Scales
We will record the mass of the every fuel and the burner with the lid on at the start of every experiment, at the end to find out how much fuel has been used up during the investigation.
Fair Testing
The variables that we are going to keep the same are:
Boiling tube.
Amount of cold water. (30cm3)
Distance that the boiling tube is held from the flame. (3cm)
Equal amounts of each alcohol in the burner.
Thermometer.
Each alcohol will have its own burner.
We are going to cool down the boiling tube before each use so that we do not start each investigation with a different temperature as this could change the speed that the water will heat up.
We will use the same thermometer but we will cool it down after each experiment.
This is to reset the thermometer to the same temperature before each new experiment.
We are going to keep the water volume the same at 30cm3 so that it will not take a long time to heat up. If the volume of the water is too large then the experiment will go to slow and more fuel will be used.
The energy will have to be spread out more amongst the water particles and the temperature will rise slower. The opposite is true if there is less water.
The distance of the boiling tube from the flame has to be kept the same to make sure that an equal amount of energy reaches the water by convection, conduction and radiation.
We are going to keep the lid on the top of the burner at all times so that no fuel is lost due to evaporation.
We are going to change:
The type of fuel.
Prediction:
We think that methanol will be the least efficient for heating water, alcohol out of the five I mentioned above. To prove my prediction we will calculate the amount of energy released
ΔH is when the alcohol reacts with oxygen.
Methanol
Reactants: energy is needed to break the bonds.
We will use the following bond energies.
C - H 412kJ/mol
C - O 360kJ/mol
O - H 463kJ/mol
C = O 743kJ/mol
O = O 496kJ/mol
C - C 347kJ/mol
(C-H x 3) + C-O + O-H + (O=O x 3/2)
1236 + 360 + 463 + 744 = 2803kJ/mol
Products: energy is given out to form bonds.
(4 x O-H) + (C=O x 2)
(4 x 463) + (2 x 743)
1852 + 1486 =3338kJ/mol
ΔH = Reactants – Products.
2803 – 3338 = -535kJ/mol
Ethanol
Reactants: energy needed to break the bonds.
(C-H x 5) + C-C + C-O + O-H + (O=O x 3)
(5 x 412) + 347 + 360 + 463 + (3 x 496) =
2060 + 347 + 360 + 463 + 1488=4718kJ/mol
Products: energy given out to form bonds.
(6 x O-H) + (C=O x 4)
(6 x 463) + (4 x 743)
2972 + 2778 =5750kJ/mol
ΔH = Reactants – Products.
4718 – 5750 = -1032kJ/mol
Propan 1 ol / Propan 2 ol
Reactants: energy needed to break the bonds.
(C-H x 7) + (C-C x 2) + C-O + O-H + (O=O x 9/2)
(7 x 412) + (347 x 2) + 360 + 463 + (9/2 x 496) =
2884 + 694 + 360 + 463 + 2232= 6633kJ/mol
Products: energy given out to form bonds.
(8 x O-H) + (C=O x 6)
(8 x 463) + (6 x 743)
3704 + 4458 =8162kJ/mol
ΔH = Reactants – Products.
6633 – 8162 = -1529kJ/mol
Butanol:
Reactants: energy needed to break the bonds.
(C-H x 9) + (C-C x 3) + C-O + O-H + (O=O x 6)
(9 x 412) + (347 x 3) + 360 + 463 + (6 x 496) =
3708 + 1041 + 360 + 463 + 2976 =8548kJ/mol
Products: energy given out to form bonds.
(10 x O-H) + (C=O x 8)
(10 x 463) + (8 x 743)
4630 + 5944 =10574/mol
ΔH = Reactants – Products.
8548 – 10574 = -2026kJ/mol
My calculations showed my prediction that methanol would be the worst fuel for heating water.
We can see that the smaller the alcohol the less energy it gives out. methanol has the lowest energy needed to break the bonds but it only releases a slightly bigger amount of energy when the new bonds are formed.
The bigger the alcohol, the bigger the differences in reactant and product energies and the more exothermic the reaction is.
Results:
To raise the temperature by 20oC requires this much energy:
Energy = mcθ
Where: m = mass of water.
c = specific heat capacity of water.
θ = Temperature change.
Energy = 0.03 x 4200 x 20 = 2520J = 2.52kJ
To calculate the amount of energy per gram of fuel:
Energy = Energy supplied to water ÷ mass of fuel burnt
To calculate the energy per mole of fuel ΔH:
ΔH = energy per gram x molar mass of fuel.
Methanol. Molar mass = 32g/mol
Energy per gram = 2.52 ÷ 0.47 = 5.36kJ/g
Energy per mole = 5.36 x 32 =171.57kJ/mol
ΔH = 171.57kJ/mol
Ethanol. Molar mass = 46g/mol
Energy per gram = 2.52 ÷ 0.41 = 6.15kJ/g
Energy per mole = 6.15 x 46 =282.73kJ/mol
ΔH = 282.73kJ/mol
Propan-1-ol Molar mass = 60g/mol
Energy per gram = 2.52 ÷ 0.38 = 6.63kJ/g
Energy per mole = 6.63 x 60 =397.89kJ/mol
ΔH = 397.89kJ/mol
Propan-2-ol Molar mass = 60g/mol
Energy per gram = 2.52 ÷ 0.34 = 7.41kJ/g
Energy per mole = 7.41 x 60 =444.71kJ/mol
ΔH = 444.71kJ/mol
Butanol. Molar mass = 74g/mol
Energy per gram = 2.52 ÷ 0.17 = 14.82kJ/g
Energy per mole = 14.82 x 74 = 1067.29kJ/mol
ΔH = 1067.29kJ/mol
Conclusion:
My calculations show that butanol was the best fuel because it has the highest ΔH at 1067.29kJ/mol. My graph shows that the bigger the alcohol molecule, the highest the ΔH. This proves my prediction right that methanol would be the least efficient fuel for heating water. My graph shows that the smaller the alcohol, the less energy it gives out. Methanol needed the least energy to break the bonds but it only released a slightly bigger amount of energy when new bonds were formed.
The larger the alcohol, the larger the differences in reactant and product energies and the more exothermic the reaction is.
Evaluation:
On my graph, the first four alcohols on my actual ΔH line gave a good relationship between ΔH and the amount of carbons. The butanol had a much higher ΔH than expected. All my actual ΔH was lower than my predicted ΔH. This was because in my experiment energy was lost to the surroundings and not all of it was transferred to the water.
I could try to improve the experiment to reduce energy loss by lowering the boiling tube so that it is closer to the flame or prevent the wick itself from being burned. Think that the butanol was higher than the others were because the experiment had to be done in a fume cabinet because of safety reasons. The fume cupboard takes out toxic gases and brings in more air. This could improve the rate of reaction, as there is more oxygen to be combusted and release more energy.
The only way I could improve this test is by doing all of the alcohols experiments in the fume cupboard.
To extend this experiment further I could look at other organic series to see if the same relationship between molecule size and ΔH exists: e.g. Alkanes (Methane, Ethane and Propane).
