I take the temperature of ice as 0°C because the ice was melting when I started to use it in my experiment. I take this temperature theoretically and do not include uncertainty to this measurement as it is in theoretical level.
After having determined the raw data, I fill it into the table.
Further I provide constants that will be used during my determination. These are taken from Giancoli Physics 5th edition, page 421:
cwater = 4186 J*(kg*°C)-1
caliuminium = 900 J*(kg*°C)-1
Data processing:
Now I will provide the mass of the water. To do this I will subtract the mass of the calorimeter from the mass of calorimeter and water together:
mwater = (163.3 ±0.2) – (35.0 ±0.2) = (128.3 ±0.4) g
To determine the mass of the ice I will subtract the mass of water and calorimeter from the mass of calorimeter, water and ice together:
mice = (181.6 ±0.2) – (163.3 ±0.2) = (18.3 ±0.4) g
I fill the table with my findings.
Now I theoretically assume that there are no temperature losses while the water gets colder to the final temperature. I assume that Qlost = Qgained.
Now to determine the latent heat of fusion of ice I will fill that formula:
caliuminium*mcalorimeter*(tinitial – tfinal) + cwater*mwater*(tinitial – tfinal) = lice*mice + cwater*mice*(tfinal – tice)
lice*mice = caliuminium*mcalorimeter*(tinitial – tfinal) + cwater*mwater*(tinitial – tfinal) - cwater*mice*(tfinal – tice)
lice = [caliuminium*mcalorimeter*(tinitial – tfinal) + cwater*mwater*(tinitial – tfinal) - cwater*mice*(tfinal – tice)]/ mice
From here, lice = 277202.6011 ≈ 2.77 x 105 J*kg-1
To determine the absolute uncertainty of the measurement I will have to add relative uncertainties of different measurements and as there are a lot of them I will only provide the uncertainty itself without calculations.
Δl = 34173.68 J*kg-1 ≈ 30000 J*kg-1
(l ± Δl) = (2.77 ± 0.3) x 105 J*kg-1
Now I can compare my result with literature’s. In Giancoli Physics 5th edition page 425 it is provided that the latent heat of fusion of ice to 3.33 x 105 J*kg-1. Therefore, the percentage discrepancy is equal to 17%. Also, the theoretical percentage uncertainty is equal to 11%. I will discuss these finding in conclusion and evaluation part.
Conclusion and evaluation:
The percentage discrepancy of 17% shows that the determination was done quite precisely. Moreover, the percentage uncertainty of 11% suggests that the determination was done quite accurately as well. However, these are only theoretical assumptions as much energy and heat was transferred to air during the water cooling process. Further, I have to state more weaknesses and limitations of my determination.
Some errors were encountered despite the fact that I tried to be as accurate as possible. First of all, the main weakness of the determination was that all the determination was done theoretically and I could not measure how much heat was transferred to air during the water cooling process. I was not able to measure the heat loss. Moreover, I took the ice temperature theoretically as I also could not measure it. These were clearly the weakest parts of all determination. Of course, the results are good enough, as the percentage discrepancy shows, but still it was only a theoretical determination based on the assumption that Qlost = Qgained. However, as for the percentage uncertainty I should blame only the instruments as I had to use quite many of them, but the uncertainty they provided was relatively small. My percentage uncertainty does not include the uncertainty of theoretical assumptions.
Furthermore, some systematic errors have occurred as I had to do a lot of calculations and roundings during the data processing part. Also, the instruments may have been badly calibrated and this could have affected my determination. However, systematic errors are not important because even if they even were encountered, they were very small. Another thing is with random errors and heat loss as they were really significant because the percentage uncertainty shows relevantly high result.
I could provide several suggestions to improve the determination but obviously I am not able to make this determination not theoretically in school’s conditions. First of all, I would rather use more ice, bigger calorimeter and more water. Then, as I would still use the same equipment with same absolute uncertainties, the percentage uncertainty would be reduced significantly. The uncertainty would be less important and more accurate results would come. However, my suggestions would only lesser the uncertainties, but they would not totally cancel them and it would still be only a theoretical determination.
