The data shown in Table 1 is shown again in the following table, Table 2, but with some calculated values with their respective uncertainties (in bold) :-
Table 2
It is assumed that the initial temperature of ice is 0.00 °C without any uncertainty, and therefore, the change in temperature for ice has an uncertainty of 0.05 °C only.
Now, using the equations above and the values in Table 2, the latent heat of fusion of ice can be calculated as follows:-
Latent heat of fusion of ice based on Trial 1 data
Step 1 – Calculate the heat lost by the water in the copper calorimeter
Heat loss = (mass of water × specific heat capacity of water × temperature change of water and calorimeter) + (mass of copper calorimeter × specific heat capacity of copper ×
temperature change of water and calorimeter)
→ Heat loss = ((34.84 g ± 0.02 g) × (4.20 J g-1 °C-1) × (13.00 °C ± 0.10 °C)) + ((36.14 g ± 0.01g) × (0.385 J g-1 °C-1) × (13.00 °C ± 0.10 °C))
→ Heat loss = (1902.26 J ± ((0.02 ÷ 34.84 × 100) + (0.10 ÷ 13.00 × 100) %))
+ (180.88 J ± ((0.01 ÷ 36.14 × 100) + (0.10 ÷ 13.00 × 100) %))
→ Heat loss = (1902.26 J ± 0.83%) + (180.88 J ± 0.80%)
→ Heat loss = 2083.14 J ± ((0.83 ÷ 100 × 1902.26 J) + (0.80 ÷ 100 × 180.88 J))
Heat loss by water and calorimeter = 2083.14 J ± 17.24 J
Step 2 – Equate this with the total heat gained by the ice and hence, calculate the latent heat of fusion of ice
Here, the heat gained by ice constitutes the heat gained by ice at 0.00 °C to convert to water at 0.00 °C and then the heat gained by that water to reach the final equilibrium temperature with the water in the calorimeter:-
Heat loss by water and calorimeter = Heat gained by the ice
→2083.14 J ± 17.24 J = (mass of ice × latent heat of fusion of ice) + (mass of ice converted to water × specific heat capacity of water × change in temperature of the ice)
→ mass of ice × latent heat of fusion of ice = (2083.14 J ± 17.24 J) – ((10.34 g ± 0.02 g) × (4.20 J g-1 °C-1) × (4.00 °C ± 0.10 °C))
→ mass of ice × latent heat of fusion of ice = (2083.14 J ± 17.24 J) – (173.71 J ± ((0.02 ÷ 10.34 × 100) + (0.10 ÷ 4.00 × 100))
→ mass of ice × latent heat of fusion of ice = (2083.14 J ± 17.24 J) – (173.71 J ± (2.69 ÷ 100 × 173.71 J)
→ (10.34 g ± 0.02 g) × latent heat of fusion of ice = 1909.40 J ± 21.91 J
→ latent heat of fusion of ice = (1909.40 J ÷ 10.34 g) ÷ ((21.91 ÷ 1909.40 × 100) + (0.02 ÷ 10.34 × 100)%)
→latent heat of fusion of ice = 184.66 J g-1 ± (1.34 ÷ 100 × 184.66 J g-1)
Latent heat of fusion of ice based on trial 1 data = 185 J g-1 ± 3 J g-1
Latent heat of fusion of ice based on Trial 2 data
Using the method used to calculate the value for the latent heat of fusion of ice for trial 1 data, the latent heat of fusion of ice was calculated for trial 2 data and is as follows:-
Latent heat of fusion of ice based on trial 2 data = 239 J g-1 ± 4 J g-1
Hence, the average latent heat of fusion of ice is:-
Average latent heat of fusion of ice = (185 J g-1 ± 3 J g-1) + (239 J g-1 ± 4 J g-1) ÷ 2
= 212 J g-1 ± 4 J g-1
Conclusion and Evaluation
The literature value of the latent heat of ice of fusion is 334.40 J g-1, as obtained from the IB Diploma 5th edition book by K.A Tsokos on pg. 165.
The value obtained experimentally is 212 J g-1 ± 4 J g-1, and therefore, its percentage deviation from the literature value, or the total error in the experiment, is:-
((334.40 – 212) ÷ 334.40) × 100 = 36.60%
The random error is:-
(4 ÷ 212) × 100 = 1.89%
Hence, the systematic error is:-
Systematic error = Total error – Random error
Systematic error = 36.60 % - 1.89 %
Therefore, the systematic error is 34.71%
The possible sources of errors in the experiment are:-
- Extra heat generated within the system due to friction when the stirrer was shook up and down.
- The system wasn’t a perfect insulator and therefore, heat might have been lost to the surroundings.
- Not all the ice might have melted.
- The substance whose temperature was to be measured might not have been left with the thermometer in it for a long time before the measurement was made, causing inaccuracy in change in temperature values.
- Heat loss to surroundings when ice is transferred to the water in the calorimeter
Besides the errors, the following is why there is a deviation from the literature value of the experimentally obtained value:-
- The value of specific heat capacities of water and copper used in the calculation are not exact; they are published average values obtained from experiments performed earlier
Suggested improvements for the future include:-
- Slow or no movement of the stirrer as it generates unwanted heat even though it speeds up the reaction but if accuracy is desired, time has to be compensated for this.
- It would be made sure that all the ice has melted.
- The thermometer would be left for a longer time in the substance whose temperature is to be measured and only after that would a measurement be taken.
- Heat loss to the surroundings would be minimized by minimizing the time between the transferring of the ice to the water in the calorimeter.
An experiment for the future could be to determine the latent heat of vaporization of water.
