I conducted an experiment of the heat of combustion for three alcohol fuels, methyl, propyl and ethyl.
The results were:
Methyl -
mass before was 231.81g
temp of water before heating was 15 deg c.
temp of water after heating was 40 deg c.
mass of methyl after heating 229.92g
volume of water heated 112 ml
result 1481 cal/g
ethyl -
mass before was 183,01g
temp of water before heating was 11 deg c.
temp of water after heating was 41 deg c.
mass of methyl after heating 180.90g
volume of water heated 130 ml
result 1848 cal/g
Propyl -
mass before was 232.23g
temp of water before heating was 13 deg c.
temp of water after heating was 39 deg c.
mass of methyl after heating 231.08g
volume of water heated 116 ml
Result 2623 cal/g
I need to analyse my results and draw a conclusion from them can you help me please.
You've done the hard part; i.e., the calculation. I checked the first one and it's ok. I didn't check the other two.
Look at the numbers.
methyl alcohol = 1481 cal/g
ethyl alcohol = 1848 cal/g
propyl alcohol = 2623 cal/g
Per gram of alcohol, which produces more heat? Look at the cost per gram and see which is the least expensive. It would be good to put something in about the precision of your work but apparently you performed each experiment only once.
Of course, I can help you analyze the results of your experiment and draw a conclusion from them. Based on the data you provided, you determined the heat of combustion for three alcohol fuels: methyl, propyl, and ethyl.
To analyze the results, we need to understand what the heat of combustion represents. The heat of combustion is the amount of heat released when a substance undergoes complete combustion in an excess of oxygen. In this case, you measured the heat of combustion by burning the alcohol fuels and measuring the temperature change of the water.
Looking at the results, we can observe that the heat of combustion values for methyl, propyl, and ethyl are 1481 cal/g, 2623 cal/g, and 1848 cal/g, respectively.
From these results, we can draw a few conclusions:
1. Heat of combustion: The heat of combustion values indicate the energy released per gram of the respective alcohol fuel during combustion. A higher heat of combustion value suggests that the alcohol fuel released more energy when combusted. In this case, propyl has the highest heat of combustion value (2623 cal/g).
2. Molecular structure: The difference in heat of combustion values may be attributed to the structural differences between the three alcohol fuels. Methyl, propyl, and ethyl have different molecular structures, and these differences can affect the amount of energy released during combustion. The specific arrangement of atoms and bonds within each alcohol fuel molecule can contribute to variations in bond energies and subsequently impact the heat of combustion.
3. Efficiency: The efficiency of combustion can also be evaluated from these results. A higher heat of combustion value can indicate a more efficient combustion process, where more energy is released for the same amount of fuel consumed. In this case, propyl (2623 cal/g) appears to be the most efficient fuel among the three tested.
It's important to note that drawing conclusions from just three data points might not provide a complete understanding of the relationship between the variables. To strengthen your analysis, it would be beneficial to conduct multiple trials for each fuel and calculate the average heat of combustion. Additional data points could provide a more accurate representation of the heat of combustion for each alcohol fuel.
Overall, your results suggest that propyl has the highest heat of combustion value, followed by ethyl and then methyl. However, further investigation with more trials is recommended to confirm these conclusions and explore any potential sources of error or variability in the experimental setup.