In my test I will be burning 5 fuels to calculating their enthalpy changes,
I will be doing this spirit burners to heat 50cm3 of water. Then work out
the mass of fuel burnt.

Heatproof mat,
Clamp stand and clamps,
Flame Shield/draft excluder,
Copper calorimeter,
Spirit Burner and different fuels
There are some variables or measurements we have decided to use in our
experiment and these are the reasons why.

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. Temperature increase by only 15(C because when the water is heated it
will be close to the room temperature and not too much heat will be
lost to the surroundings, also this will still give us good results.

Will also save time.

. Only use 50cm3 of water - this is plenty of water to heat and again
get a good set of results, we don't wont to take forever to heat the
water and use large amounts of fuel to do this.

. Use a draft excluder - this will stop the flames energy being lost to
its surroundings, it will reflect the heat inwards and back towards
the copper calorimeter and will help reduce heat/energy loses. Making
our experiment have less percentage error.

. Make sure all the wicks of the spirit burners are all 1cm long, this
will keep my test fair,
. Also keep the copper calorimeter 5cm about the flame, again to keep
the test fair,
. Keep stirring the water, to make sure the heat is distributed evenly
though the water.

1.Set up apparatus ready for experiment.

2. Add 50cm3 of cold water to a copper calorimeter,
3. Take the temperature of the water,
4.Get our chosen fuel and measure the wick, so that its 1cm,
5. Hold the copper calorimeter 5cm above the spirit burner,
6. Put the flame shield around the spirit burner,
7. Light the spirit burner and heat the water above,
8. Stir the water constantly while being heated,
9. Watch the thermometer, and when the rise in temperature of the water has
risen by 15(C. Put out the spirit burners flame.

10. Keep watching the thermometer until it has stopped rising and record
the final temperature of the water.

I will then do the test for the other fuels.


|Experiment|Fuel |Start |End|Weight |Start|End |Temperature|
|||Weight|Weight| |Temperature|Temperatur|change|
|||||Change ||e||
|1. |Methanol|11.90 |11.36g|-0.54g |23|39 |+16 |
|||g|| ||||
|2|Methanol|13.65 |12.98g|-0.67g |20|35 |+16 |
|||g|| ||||
|3|Methanol|12.82 |12.08g|-0.74g |24|39 |+15 |
|||g|| ||||
|1|Ethanol|11.88 |11.60g|-0.28g |23|39 |+15 |
|||g|| ||||
|2|Ethanol|13.48 |12.05g|-1.43g |21|37 |+15 |
|||g|| ||||
|3|Ethanol|12.30 |12.00g|-0.30g |23|38 |+15 |
|||g|| ||||
|1|Propan-1-ol|13.03 |12.73g|-0.30g |24|39 |+15 |
|||g|| ||||
|2|Propan-1-ol|13.00 |12.70g|-0.30g |19|34 |+15 |
|||g|| ||||
|3|Propan-1-ol|12.89 |12.54g|-0.35g |26|41 |+15 |
|||g|| ||||
|1|Propan-2-ol|12.91 |12.63g|-0.28g |20|35 |+15 |
|||g|| ||||
|2|Propan-2-ol|13.60 |13.27g|-0.33g |21|36 |+15 |
|||g|| ||||
|3|Propan-2-ol|12.41 |11.86g|-0.55g |26|41 |+15 |
|||g|| ||||
|1|Butan-1-ol |12.33 |11.93g|-0.40g |21|39 |+18 |
|||g|| ||||
|2|Butan-1-ol |13.61 |13.22g|-0.39g |20|37 |+17 |
|||g|| ||||
For all of my experiments I have done 3 tests, except for Butane-1-ol as
the two tests I did do where very close together and show little error. So
for all the other I repeated the tests 3 times, making sure the results
where in close proximity to each other. There are a few errors however;
these have been tagged with a star (*) these errors may have accord through
human error or mechanical error.

There are many reasons why errors could have occurred in the experiment;
1.Human error could have occurred I my experiment, a measurement could
have been read wrong or written down incorrectly.

2.Another error, which could have occurred, would have been the fact a
set of scales was used which only goes to 2 decimal places. This could
have affected our results but only very slightly. This error is known as
percentage/precision errors. The precision error of a measurement of 5.46
g would be 0.005 because the true measurement could be 5.455g or 5.465g.

. Percentage errors are calculated using the following formula:
Percentage error = error x 100
So I will calculate the percentage errors for all of my fuels. Using the
formula I stated above and the results I got I have put in a table.

|Fuel|Percentage error in|
| |g's|
|Methanol|0.80g |
|Ethanol|1.70g |
|Propan-1-ol |1.56 g |
|Propan-2-ol |1.61 g |
|Butan-1-ol|1.22 g |
3.Another reason for percentage errors could have been the fact the draught
excluder wasunable to stop the draughts from tampering with the flam
and the heat it was giving off. A further error could have been a certain
amount of fuel was spilt when moving the spirit burner around. This would
make the mass fuel burnt seem higher than it should be. Giving us errors.

Working out enthalpy changes
Propane-1-ol: moles used = mass ( molecular mass
Moles used = - 0.32 ( 60 = - 5.33(10-3
Energy released by Number of moles used = (Hc ( number of moles
= -
2021 kjmol-1 ( 5.33 ( 10-3= 10.77 kJ
10.77kj of energy are required to raise 200ml of water by 15(C
This amount of energy is the same amount required for each fuel to raise 50
ml of water by 15 in temperature. We can then use this energy value to
calculate the ?HC (standard enthalpy change of combustion) of the other
fuels using the formula:
?HC = Energy released
No. Of moles
However first I need to calculate the number of moles using the formula:
No. Of moles = Mass
Molecular mass
Methanol: No. Of moles= Mass
Molecular mass
No. Of moles = -0.65 = -0.02
?HC = Energy released
No. Of moles
?HC = 10.77 = -538.5 KJmol -1
Ethanol: No. Of moles = Mass
Molecular mass
No. Of moles= -0.30 g =-6.52 x 10
?HC = Energy released
No. Of moles
?HC = 10.77= -1651.8 KJmol -1
-6.52 x 10
Propan-1-ol: No. Of moles = Mass
Molecular mass
No. Of moles= -0.32 =-5.33 x 10
?HC = Energy released
No. Of moles
?HC = 10.77 =-2020.6 KJmol -1
-5.33 X 10
Propan-2-ol: No. Of moles= Mass
Molecular mass
No. Of moles = -0.31 =-5.20 x 10
?HC = Energy released
No. Of moles
?HC = 10.77 = -2071.15 KJmol -1
-5.20 x 10
Butan-1-ol: No. Of moles= Mass
Molecular Mass
No. Of moles= -0.40 =-5.41 x 10
?HC = Energy released
No. Of moles
?HC = 10.77 =-1.990.76 KJmol-1
-5.41 X 10
|Fuel|?HC |Actual ?HC|Difference|
||||(KJmol-1) |
|Methanol |-538.5 KJmol-1 |-726 KJmol-1|-187.5|
|Ethanol|-1651.8 KJmol-1 |-1367KJmol-1|+284.8|
|Propan-1-ol|-2020.6 KJmol-1 |-2021KJmol-1|-0.4|
|Propan-2-ol|-2071.17 KJmol-1|-2006KJmol-1|+65.17|
|Butan-1-ol|-1.990.76 |-2676KJmol-1|-685.24|
The standard enthalpy change of combustion values which I calculated from
the results obtained during my experiment were slightly inaccurate as they
were slightly different form those given in the data book.


Although I did my best to minimize the risk of errors occurring by carrying
out the following procedures it is evident some errors occurred.

The reasons the enthalpy change in my experiment as different to the
enthalpy changes in the data book are because of heat lose. When heat is
produced, heat rises and there for will be lost, as the draught shield as
not really high enough it only covered the sprit burner and not the spirit
burner and copper calorimeter. Heat would also have been lost though the
clamp stand and heat mat, which the spit burner was sat on. Heat lose is
inevitable, as you will always loose heat to the surrounds, but I tried as
hard as possible to limit this, as I said in my method section.

Another reason the results from my test, aren't the same as the data
book is the fact the scales only read to 2 decimal places and as I
explained in my method this will affect my results. Another possibility
would be the fact the thermometer is not decimalised, so all the readings
are whole numbers, so there can be an inaccuracy there. Another problem
with the thermometers is the fact the numbers and lines are very small so
human error is more than likely to occur. Again affecting our test.

I've also plotted a graph which is attached after this page, but I've
plotted both the data books enthalpy changes and the enthalpy changes from
my experiment. The trend of this graph is that as the number of carbon
atoms in the fuel increases as the chain gets bigger) the standard enthalpy
change of combustion increases. This means more energy is released in the
combustion of alcohols with longer carbon chains. This is because the
longer the hydrocarbon chain the weaker the bonds between the molecules and
therefore less energy is required for combustion to take place.