The amount of carbon
soot left on the container: This affects the experiment because the carbon soot acts as an insulator and slows down the rate at which heat is transferred to the container and consequently to the water, as explained for glass in the above statement.
Volume/Mass of alcohol: As the alcohol burns this will begin to drop because carbon and oxygen are released through combustion as waste gases. If I record the change in mass of the alcohol from before it burns and after it has burned I can work out how many moles where burned and calculate how many KJ per mole are released.
Time: This should to be constant because you should not vary more than one thing in an experiment, the relative alcohol burnt and temperature increase, will produce the same result.
Distance between wick and container: This is important because if the container is to far away from the heat source virtually no energy will be transferred , however if the container is in the flame a lot of carbon will be formed on the container insulating it from gaining as much heat as it should gain hence leading to inaccurate results .
Boss & clamp: Is the closest I can get to direct heating , because the clamp is rubber insulated to protect against heat exchange and the boss doesn’t even come in direct contact with the container.
Starting temperature of water: Should be kept constant throughout the experiment because I can only change one variable at a time. However it is not very important because I am reading changes in temperature.
The Plan:
List of the Apparatus:
- Thermometer
- Container(copper)
- Water
- Alcohols
- Stand, Boss & Clamp
- Spirit Burner
- Stop Clock
- Water (300cm3)
- Measuring cylinder
- Stop Clock
- Balance
Method
- I will set the apparatus up as shown above, setting length distance to 15cm.
- After that I will weight the spirit burner with the alcohol, say I am testing methanol.
- Then I will put 250cm3 of water into the container.
- Having let the water and the container come to the same temperature, I will measure the temperature of the water.
- Then I will light the spirit burner, and allow it to burn for two minutes, using the stop clock.
- During this time the water will be stirred continuously.
- After I finished with the burner, 2 minutes having passed , I will then take a reading of the temperature of the water, and weigh the burner for a second time.
I will use the container over the glass beaker because it is made of copper, which conducts heat faster, and better than glass, therefore transferring heat to water better.
I chose the stand, boss and clamp because the clamp has rubber over its claws, which does not conduct as much heat away from the container.
I will repeat the experiment with
methanol
ethanol
- propan-1ol
- butan-1ol
- pentan-1ol
Change in Plan :
I will change the length distance to 5 cm because not a lot of the heat getting transferred to the container, but and any closer, and it would be too close to the flame, thus lots of carbon would be formed on the container, preventing some of the heat being transfer. I will only use 100cm3 of water because the school could only provide so much.
Safety:
Keep lid on burners at all times, when moving and not burning. This is to prevent evaporation and the possibility of an explosion. Keep fingers, clothes and any other body parts away from burners to prevent burns.
Prediction :
Methanol:
Total = Energy of bonds broken – energy of bonds made
Total = 5613 - 6692 = -1079J
Equation: 2CH3OH + 3O2 = 2CO2 + 4H2O – 1079J
Total energy for only one molecule: 539.5
Ethanol:
Total = Energy of bonds broken – energy of bonds made
Total 4724 - 5764 = -1040J
Equation: C2H5OH + 3O2 = 2CO2 + 3H2O - 1040J
Propan-1ol:
Total = Energy of bonds broken – energy of bonds made
Total 12591 - 16364 = -3773J
Equation: 2C3H7OH + 9O2 = 6CO2 + 8H2O -3773J
Total energy for only one molecule: 1540.5J
Butan-1ol:
Total = Energy of bonds broken – energy of bonds made
Total 8559 - 10600 = -2041J
Equation: C4H9OH + 6O2 = 4CO2 + 5H2O - 2041J
Pentan-1ol:
Total = Energy of bonds broken – energy of bonds made
Total 19569 - 26036 = -6467J
Equation: 2C5H11OH + 15O2 = 10CO2 +12H2O - 6467J
Total energy for only one molecule: 2541.5 J
The above tables show the theoretical calculation of the energy being released during the full combustion of alcohol. From the tables you can see that, X is equal to the number of carbons in the chain, and Y is the change in energy, then Y = ( -500.5 x X ) -39.
However this is only possible in perfect world where all the energy from the fuel burned would be actually transferred to the water, sadly this isn’t true and all my results will have inaccuracies, but I should be able to see a similar pattern emerging when I actually burn the alcohols.
If you double the number of carbons in the chain, then the energy released will approximately double. I think this is because, when the bond energy before the experiment doubles, the calculated energy released doubles, and when reacted, shorter carbon chains burn slower, because they have less hydrogens.
When the carbon chain is doubled, the number of carbons is doubled, which in turn doubles the hydrogens and the oxygens by two, therefore one carbon will bond with two hydrogens and two oxygens. Because we now have more bonds we will also need more energy to break them, 1918J more of energy ,to be precise, are needed for every extra carbon added , so you are almost going to get a doubling effect.
Table of Results :
To get these results I have used a very sensitive electronic scale.
Heat evolved by the can (X)= mass of can x specific heat capacity x temperature change
Heat evolved by the water (Y)= mass of water x specific heat capacity x temperature change
∆H for 1 Mole = (X+Y)÷mass of alcohol burnt x RMM÷1000
X = 31.3 x 0.44 x 27 = 371.84
Y = 100 x 4.2 x 27 = 11340
With the above formulae I can calculate the ∆H for 1 mole of each of the alcohols.
Methanol:
∆H = 11711.84 ÷ 1.2 x 0.032 = 312.3
Ethanol
∆H = 11711.84 ÷ 0.7 x 0.046 = 769.6
Propanol
∆H = 11711.84 ÷ 0.8 x 0.06 = 878.4
Butanol
∆H = 11711.84 ÷ 1.1 x 0.074 = 787.9
Pentanol
∆H = 11711.84 ÷ 0.9 x 0.088 = 1145.2
Observations :
From my results and the graph which I have plotted I can see that the experimental values which I have gathered are lower than the theoretical values which I have calculated. Nearly all results gathered in the experiment look flawed , however they all show that if you increase the number bonds in a hydrocarbon chain you increase the amount of energy that it gives out.
Analisys:
I chose the 5 smallest chains of hydrocarbons and experimented with them in a systematic order so that I can gather evidence that proves a correlation between the increase of hydrocarbon bonds in a chain and the energy given out. From my results I can see a strong positive correlation as to the fact that increasing the number of bonds in a chain of hydrocarbons increase the energy output but there is not as positive a correlation that doubling the number of bonds doubles the energy output.
Evaluation:
Overall my results were very inaccurate. Below there is a list of reasons as to why it was so inaccurate:
- Energy given off through sound and light.
- Heat conducted away by the air.
- The fact that the beaker gets hot.
- The rubber clamp transferred heat away.
- The fact that at higher temperatures, heat is lost faster to the air and out of the container, due to the bigger heat difference, making the higher temperatures more inaccurate
- By incomplete combustion
- The amount of energy you give the alcohol originally.
- The availability of alcohol for the wick to burn, if not enough then the wick would burn not the alcohol which would give an inaccurate result.
- Evaporation of water so there will be less water to heat, making the water hotter.
- The size of the wick.
- The flame size changed due to the type of alcohol, hence it was a different distance away from the beaker each time.
The equipment that I used in this experiment was very inaccurate because heat is a bad way of transferring energy without any loss of it. Any molecule will conduct heat, radiation happens and can be reduced but not completely halts. I feel that the most limiting factor of the experiment is the convection of air and to a lesser extent, of water. Also during the experiment, some of the water will have evaporated, thus the water mass/temp reading will be altered.
I feel that this experiment could have been improved by using a wider range of alcohols such as hexanol and heptanol. This would give a better graph reading and a wider range of results to support a firm conclusion.
If I had started below room temperature, so that the amount of energy gained, from room temperature might equal the energy lost at temperatures higher than room temperature. Next time reducing heat lost would be my main priority. Improving insulation techniques would be a valuable asset in obtaining the most reliable data I could.
Another error is that of incomplete combustion. Complete combustion occurs if there are lots of oxygen atoms available when the fuel burns, then you get carbon dioxide (carbons atoms bond with two oxygen atoms).
If there is a limited supply of oxygen then you get carbon monoxide (each carbon atom can only bond with one oxygen atom). This is when incomplete combustion has occurred. This is so because the carbon monoxide could react some more to make carbon dioxide. If the oxygen supply is very limited then you get some atoms of carbon released before they can bond with any oxygen atoms. This is what we call soot. Since heat is given out when bonds form, less energy is given out by incomplete combustion. So this is why it affects the outcome of the experiment. To overcome this problem, I would have to make sure a sufficient supply of oxygen was involved in the reaction.
