· The height of the beaker from the wick
· Same set of scales
· Weigh the spirit burner with the lid on.
The variable that must be changed is… wwdd ddw esdddds aydd ddba ndd kcdd dduk.
· The type of alcohol used
The formulae of the alcohols that I will be using are…
· Methanol CH OH
· Ethanol C H OH
· Propanol C H OH
· Butanol C H OH
· Pentanol C H OH
Prediction
I predict that the more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C H OH. In this formula you have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds you require a certain amount of energy which I will show in a table.. wwdd ddw esdddds aydd ddba ndd kcdd dduk;
TYPE OF BOND ENERGY REQUIRED TO BREAK THE BOND(j)
C-H 410
C-O 360
O-H 510
O=O 496
C=O 740
C-C 350
To separate C-H bond you need to apply 410 joules of energy. There are five such bonds in ethanol so you multiply 410 by five to get 2050 joules. You do these calculations for all the other types of bonds that make up ethanol, add them all together and you get 3270 joules. All of the other alcohols can be broken up in this way. Below is a table showing the energy required to break up the bonds in each alcohol. AtJX Visit essaybank ag co ag uk ag for more ag Do not ag redistribute AtJX
Type of alcohol Energy required to break the bonds in the alcohol (j)
Methanol 2100
Ethanol 3270
Propanol 3740
Butanol 4560
Pentanol 5380
As you can see a longer molecule takes more energy to break it´s bonds, in this case Pentanol. Compared to a smaller molecule, methanol which requires less energy to do so. I can come to predict that the longer the molecular structure in the alcohol the more energy it will take to remove the bonds. So when I come to predicting results I can safely say that pentanol will evolve more energy than methanol simply because it has more bonds to break. wwac acw esacacs ayac acba nac kcac acuk:
Type of alcohol Before exp 1 After exp 1 Before exp 2 After exp 2 Before exp 3 After exp 3 Average temperature difference(ºC) Average increase in weight (g)
Methanol Temp(ºC) 23 73 23 73 23 73 50
Weight(g) 153.12 152.12 162.84 159.56 156.35 154.09 -2.18
Ethanol Temp(ºC) 26 76 23 73 21 71 50
Weight(g) 169.78 166.76 169.21 163.56 176.00 172.91 -3.92
Propanol Temp(ºC) 23 73 24 74 23 73 50 wwgd gdw esgdgds aygd gdba ngd kcgd gduk.
Weight(g) 191.56 189.21 168.11 157.00 189.01 187.32 -5.05
Butanol Temp(ºC) 23 73 21 71 24 74 50
Weight(g) 159.76 156.71 98.12 84.32 187.56 181.58 -7.61
Pentanol Temp(ºC) 24 74 22 72 23 73 50
Weight(g) 164.85 157.23 167.94 155.38 161.83 153.19 -9.61
Analysis
As you can see in the obtaining section I have recorded all the information from the results that I need to complete the calculations…
· Energy evolved
· Energy per gram
· Energy per mole
First of all I will calculate the amount of energy evolved. I will achieve this by using this simple formula…
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Energy evolved = Mass x Rise in temperature x SHC
Energy evolved = 100g x 50ºC x 4.2
As you can see all of the alcohols will have the same amount energy evolved because all the numbers that are filled in the formula remain constant for each alcohol and the same numbers are applied for each individual alcohol. Below is a table showing the amount of energy evolved in each case…
Type of alcohol Energy evolved in (j) Energy evolved in (kj) wwgf gfw esgfgfs aygf gfba ngf kcgf gfuk:
Methanol 21 000 21
Ethanol 21 000 21
Propanol 21 000 21
Butanol 21 000 21
Pentanol 21 000 21
To find out how much energy is produced per gram we use the formula…
Energy per gram of fuel = Energy evolved x Mass of fuel burnt
Energy per gram of fuel = 21kj x ?
Below is a table showing how much energy is produced per gram when burning the alcohols in question…
Type of alcohol Energy per gram (kj)
Methanol 9.63
Ethanol 5.36
Propanol 4.16
Butanol 2.76
Pentanol 2.19 wwca caw escacas ayca caba nca kcca cauk.
As you can see the energy per gram decreases as the length of the molecule increases. This is because more fuel is burnt so there is more of it to be filled with the energy. This is shown in graph 1.
To find out how much energy is produced per mole you have to use this formula..
Energy per mole = Energy per gram x Formula mass
Here is a list comprising all the formula masses of the chosen alcohols…
· Methanol 32g
· Ethanol 46g
· Propanol 60g
· Butanol 74g
· Pentanol 88g
Below is a table showing the energy produced per mole…
Type of alcohol Energy per mole(kj/mol)
Methanol 308.16
Ethanol 245.56
Propanol 249.60
Butanol 206.46
Pentanol 192.72
Again the results decrease as the molecule length increases suggesting that during the experiment more energy was last in the longer alcohols, this is shown in graph 2.
As you can see, graph 1 has a nice smooth curve suggesting that the results are reliable. But graph 2 has an anomalous result (x), this does not allow the curve to follow a smooth pattern. I think this is because the energy per gram in ethanol and propanol was only slightly different and when multiplied with the formula mass, propanol comes out larger. from www.essaybank.co.uk
My prediction was not clear enough because to vary the amount of heat evolved, time would have to come into it. But I did predict that the energy per gram would decrease as the molecular length increases. I think this because the alcohols with more carbon atoms in like pentanol burnt more fuel so there would be less energy per gram because more fuel has been burnt. The reason why more fuel has been burnt is because of the large number of carbon atoms and large molecular length, hence the surface area is large allowing more energy to be released. wwba baw esbabas ayba baba nba kcba bauk!
EVALUATION
I believe my results were very inaccurate. Below is a list of reasons why it was an inaccurate experiment.
1. Energy given off through sound and light.
2. Heat conducted and convected away through the air.
3. Radiation of heat out into the atmosphere.
4. The fact that the beaker gets hot.
5. The rubber clamp transferred heat way.
6. Heat may be take away through guts of wind. wwgc gcw esgcgcs aygc gcba ngc kcgc gcuk:
7. The fact that at higher temperatures, heat is lost faster to the air and out of the beaker, due to the bigger heat difference, making the higher temperatures more inaccurate, and making a shallower gradient on the graph. wwce cew esceces ayce ceba nce kcce ceuk;
8. By incomplete combustion
9. The amount of energy you give the alcohol originally.
10. 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.
11. Evaporation of water so there will be less water to heat, making the water hotter.
12. The size of the wick.
13. Not all of the water was the same temperature.
14. The flame size changed due to the type of alcohol, hence it was a different distance away from the beaker each time.
Numbers 1 – 10 would decrease the reading and Numbers 11 – 14 would increase the reading.
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. On the other hand, 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. wYCA7 from wYCA7 essay wYCA7 bank wYCA7 co wYCA7 uk
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). wwbf bfw esbfbfs aybf bfba nbf kcbf bfuk:
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.
