To Find Out Which Fuel Gives Out the Most Energy.
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 needed (Joules) to heat 1 gram of water by one degrees Celsius.
Preliminary Experiment
To help me decide the vital elements of the experiment, I made a trial run using methanol to check if my chosen quantities and set up would work.
Results:
Height of can from burner: 5cm
Mass of burner at start = 204.47g. Mass of burner finish = 203.6g. Mass of fuel burned = 0.87g
Start temp = 19oC Finish temp = 23oC, Temp rise = 4oC
Following the pilot experiment I made some minor adjustments:
Height of can above wick down to 4cm to improve contact with flame and stir the water, because the rise in temperature is too little.
Apparatus
* Glass Beaker
* 100ml Water
* Spirit Burner
* Clamp Stand
* 6 Fuels
* Thermometer
* Top-pan Balance
* Ruler
Method
. Set the experiment up as above
2. Fill the container with 100ml of tap water.
3. Measure the temperature of the water.
4. Weigh the burner before burning.
5. Heat the water for 2 minutes.
6. Record the temperature of the water after 2 minutes.
7. Weigh the burner again and record the change in weight.
8. Repeat the procedures above for the other fuels.
Safety
Goggles will be used during the experiment, because it involves glassware and poisonous alcohol. Also, if any alcohol is spilt, all fire would be put off and the alcohol would wiped immediately.
Fair Test
For each test, I am making sure that the beaker/can is the ...
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4. Weigh the burner before burning.
5. Heat the water for 2 minutes.
6. Record the temperature of the water after 2 minutes.
7. Weigh the burner again and record the change in weight.
8. Repeat the procedures above for the other fuels.
Safety
Goggles will be used during the experiment, because it involves glassware and poisonous alcohol. Also, if any alcohol is spilt, all fire would be put off and the alcohol would wiped immediately.
Fair Test
For each test, I am making sure that the beaker/can is the same distance (4cm) to the tip of the fuel burner. Also, I am going to make sure that the wick of the burner is always the same length(1.5cm), this means that the flames are the same size, so the degree of heating is going to be equal. I am going to change the water after every test, so the water always starts at similar temperatures, preventing from falsifying the rising in temperature. I am going to use the same amount of water for each test. Also, I am going to prevent the opening of the burner, because some of the fuels might evaporate, and would affect the weight of the burner. I am going to weigh the burner just before the test, so I could get the accurate weight in case of any evaporation. I am going to use an accurate scale that goes at least to one decimal point, when the burner is not in use, it must be covered to prevent evaporation. Also, the container of water must be the same every time, so the heated surface area is the same. I would repeat each test 3 times to get an accurate result in case if there were any anomalous results.
Prediction
I predict that the molecule with the most bondings, i.e. Hexanol is going to give off the most energy, because it needs more energy to bond the molecules together, so this means that there are more energy between the bonds. So when the fuel is burnt, the ones with more bondings would give off more energy, so temperature rise would be greater, because more energy is given off by the bondings. When alcohols combust with oxygen/air, they produce heat, So my prediction is that Methanol is going to give off the least energy, and as the bonding gets bigger, the more energy is given off, so Hexanol would give off the most energy. Here are the chemical formulas and the balanced equations:
Methanol - CH3OH + 3O2
Ethanol - C2H5OH
Propanol - C3H7OH
Butanol - C4H9OH
Pentanol - C5H11OH
Hexanol - C6H13OH
The energy given off per mole can be worked out theoretically with the energy per mole for each bonding.
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
4 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
3 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, and at the right hand side is the energy (KJ/mole) release, so if the right side is taken away from the left, the difference shows the energy given off. From the theoretical energy release, I can see that they support my prediction, the energy release generally increase as the numbers of atoms increase in the molecules.
Results
Here are the tables of results. In the table, anomalous results are show by a *. The ' 4th ' row is used only if a repeat of the fuel was needed. In the last column of each table I have given the energy released per gram in the fuel.
This is worked out by using the following equation:
00 (ml of water) x 4.2 (Specific Heat Capacity) x Rise in temperature
Mass of fuel burnt
Methanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.85
3
6424
2nd
0.84
2
6000
3rd
0.89
3
5009
4th
-
-
-
Average
0.86
2.7
5811
Ethanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.35
0
2000
2nd
0.35
8
9600
3rd
0.77 *
2
6546
4th
0.62
6
0839
Average
0.44
1.3
0813
Propanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.68
26
6059
2nd
0.88
26
2409
3rd
0.64
20
3125
4th
-
-
-
Average
0.73
24
3864
Butanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.52
24
9385
2nd
0.49
6
3714
3rd
0.47
4
2511
4th
-
-
-
Average
0.49
27
5203
Pentanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.31
9
2194
2nd
0.37
9
0216
3rd
0.34
9
1118
4th
-
-
-
Average
0.34
9
1176
Hexanol
Mass of fuel burnt (g)
Rise in temperature of water (°C)
Energy released per gram of fuel (j)
st
0.31
9
2194
2nd
0.43
3
2698
3rd
0.36
2
4000
4th
-
-
-
Average
0.37
1.3
2964
Energy Per Mole
I am going to use the average energy per gram to work out the energy per mole.
Methanol - Formula mass: 32
5811J/gram x 32 = 185952J/mole = 185.95KJ/mole
Ethanol - Formula mass: 46
0812J/gram x 46 = 497352J/mole = 497.35KJ/mole
Propanol - Formula mass: 60
3864J/gram x 60 = 831840J/mole = 831.84KJ/mole
Butanol - Formula mass: 74
5203J/gram x 74 = 1125022J/mole = 1125KJ/mole
Pentanol - Formula mass: 88
1176J/gram x 88 = 983488J/mole = 983.5KJ/mole
Hexanol - Formula mass: 102
2964J/gram x 102 = 1322328J/mole = 1322.3KJ/mole
