Physics Lab Report

-Specific latent heat of fusion of ice

Aim

The aim of this experiment is to determine the specific latent heat of fusion of ice.

Data Collection

The following table, Table 1, shows all the recorded data that was collected along with the uncertainties (wherever applicable):-

Table 1

The following are some of the specific heat capacities which will be required in the calculation for finding the latent heat of fusion of ice:-

Water = 4200 J Kg-1 K-1 = 4.20 J g-1 °C-1

Copper = 385 J kg-1 K-1 = 0.385 J g-1 °C-1

Data Processing

The following equations are going to be used in the calculation:-

• Heat gained or lost by a substance = mass of the substance × specific heat capacity of the substance × change in temperature it undergoes

• Heat lost = Heat gained (for two substances in contact)

• Heat gained or lost by a substance while melting = mass of the substance × specific latent heat of fusion of the substance

34.84

47.02

Initial temperature of water in calorimeter (°C) ± 0.05 °C

17.00

16.50

Final temperature after adding ice to water in calorimeter (°C) ± 0.05 °C

4.00

7.00

Change in temperature for water and calorimeter (°C) ± 0.10 °C

13.00

9.50

Change in temperature for the ice (°C) ± 0.05 °C

4.00

7.00

Mass of calorimeter and water and ice (g) ± 0.01 g

81.32

91.08

Mass of ice that was added

(g) ± 0.02 g

10.34

7.50

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.

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.