Results:
Table 1. Uncertainties of the Equipments
Table 2. Measurement of the Equipments
Table 3. Raw Data
Calculations:
Amount of ethanol in the spirit burner can be calculated by;
Amount of Ethanol = Filled spirit burner – Empty spirit burner
= 164.22 – 127.27
≈ 37g ± 0.020g
Uncertainty = 164.22g±0.01g – 127.22g±0.01
= (164.22g – 127.22g) ± (0.01 + 0.01)
= 36.95g ± 0.020g
≈ 37g ± 0.020g
Table 5. Processing Raw Data
Uncertainties for change in mass of spirit burner
Change in mass of spirit burner = initial mass – final mass
= 164.22g±0.01g – 163.58g±0.01g
= (164.22 – 163.58) ± (0.01+0.01)
= 0.64g±0.02g
Uncertainties for change in temperature
Change in temperature = final reading – initial reading
= 54°C±0.2°C - 24°±0.2C°
= (54 – 24) ± (0.2+0.2)
= 30°C±0.4°C
Table 5. Average of the Data
As shown from above, the average change in the mass of the spirit burner is 0.68g±0.02g, which indicates that there was 0.68g±0.02g of ethanol was used to ...
This is a preview of the whole essay
Uncertainties for change in mass of spirit burner
Change in mass of spirit burner = initial mass – final mass
= 164.22g±0.01g – 163.58g±0.01g
= (164.22 – 163.58) ± (0.01+0.01)
= 0.64g±0.02g
Uncertainties for change in temperature
Change in temperature = final reading – initial reading
= 54°C±0.2°C - 24°±0.2C°
= (54 – 24) ± (0.2+0.2)
= 30°C±0.4°C
Table 5. Average of the Data
As shown from above, the average change in the mass of the spirit burner is 0.68g±0.02g, which indicates that there was 0.68g±0.02g of ethanol was used to heat water to have temperature difference of 30°C. This needs to be changed into Kelvin as the formula requires the temperature to be in Kelvin. As the 30°C is the temperature difference, therefore the temperature difference in Kelvin will be 30K.
With this information, it is possible to find out the energy density of ethanol. Energy density is amount of energy produced per kilogram.
To find the quantity of thermal energy of water, Q, the formula Q = mcΔt could be used where s is the constant which has 4.18JK-1g-1 value. In this formula, m is the mass and Δt is change in temperature. There was 50ml of water heated therefore m is 50g.
As m equals to 0.68g and Δt equals to 30K, if we substitute these values to the formula;
Q = mcΔt
= (0.050kg) x (4180JK-1kg-1) x (30K)
= 6270J
Uncertainty for the mass will be 1g as it was measured with the measuring cylinder with an uncertainty of ±1.0ml. As ‘c’ is the constant, the uncertainty is not able to be calculated. Uncertainty in change in temperature would be ±0.80, as it is ±0.40 for both initial and final temperature recorded. Hence the uncertainty for Q will be;
Uncertainty = (0.050kg±0.0010g) x (4180JK-1kg-1) x (30K±0.40K)
= (0.050g±2%) x (4180JK-1kg-1) x (30K±1.3%)
= 6270J±3.3%
= 6270J±206.91J
≈ 6300J ± 210J
Therefore, approximately 6300 joules with the uncertainty of 290J are produced by 0.68g of ethanol to heat water to have temperature difference of 30K.
To calculate the energy density in joules per kilogram, the amount of energy calculated needs to be converted into per kilogram.
It is calculated that 0.68g of ethanol produced 6270J.
J/kg = (1000/0.68) x 6300
= 9264705.882J
= 9.3 MJ/kg
Uncertainties
J/kg = (1000/0.68± 0.020) x 6300±210
= (1000/0.68±2.9%) x 6300±3.3%
= 1470.588±2.9% x 6300±3.3%
= 9.26x106 x 6.2%
= 9.26x106 x 5.7x105
= 9.3 MJ/kg ± 0.57MJ/kg
Discussion and Conclusion:
Throughout the experiment of heating water with ethanol, the energy density of -the ethanol was calculated. The energy density of ethanol is 9.3MJ/kg with the uncertainty of 0.57MJ/kg. Quoted from ‘Overview of Storage Development DOE Hydrogen Program’ by George Thomas, the theoretical value of energy density of ethanol is 26.8MJ/kg.
The difference between the theoretical value and data gained is 17.5MJ/kg.
To calculate the percentage difference between the data gained and the theoretical value;
Percentage Difference = (theoretical value – data gained)/(theoretical value) x 100
= (26.8MJ/kg – 9.3MJ/kg)/(26.8MJ/kg) x 100
= (17.5MJ/kg)/(26.8MJ/kg) x 100
= 0.65298 x 100
= 65.298%
= 65%
There is 65% difference between the data gained and theoretical data.
The uncertainty of gained data is ±0.7MJ/kg. As the theoretical value is 26.8MJ/kg, it does not lie in between the uncertainty range. This is because of random error carried out throughout the whole experiment. As there are heat loss during the process of heating the water, there are a lot of heat being used heating up the surroundings, hence there are so much difference between the theoretical value and the gained value.
A was graph was not required to be drawn for this experiment. Since there is no change in any of the variables, as the change in temperature was kept constant and the amount of water used was also constant, therefore there were no independent or dependent variables. Hence a graph cannot be drawn to aid in the calculation of the energy density of ethanol
There was three equipment used which affected the uncertainties. As weighed with electronic weight, there is systematic error in the weight itself. Also there is systematic error when water was measured with the 100ml measuring cylinder. Furthermore, in measuring the change in temperature, there is also a systematic error in reading the thermometer.
There were a lot of calculations been processed in order to calculate the energy density. All of those calculations had uncertainties included. These uncertainties built up throughout the calculations and ended up with ±0.7MJ/kg.
Since there is no graph needed in this experiment, it is unable to find out whether the best fit line lies in between the uncertainty marks in the graph. Random error was carried out throughout the experiment. When the spirit burner was heating water, before the heat reached water, the heat contacted with the beaker and the surrounding air. Therefore, a lot of heat was loss during the process of heating the water.
Improvements:
Bibliography:
Thomas, George. Overview of Storage Development DOE Hydrogen Program [pdf]. Hydrogen Program Review. San Ramon, CA. May9-11, 2000.