Heat of Fusion For Ice

Purpose:

The purpose is to find the experiment heat fusion of ice using a makeshift calorimeter and compare it to the actual heat fusion, 6.01kJ.

Procedure:

1. Start with 100 mL water at 50°C

2. Place in calorimeter and record the temperature

3. Drop 2-3 ice cubes in and stir until temp. is constant, but do not run out of ice

4. Measure the new volume

Data:

Volume of water: 100mL

Temperature of water in calorimeter: 50°C

Constant temp after addition of ice: 1.3°C

New volume of water: 163 mL

Calculations:

1. Mass of water used:

100mL=100g

2. Heat given by water:

100g x -48.7C° x 4.18 J/gxC° = -20,356.6J

3. Heat gained by ice:

-20,356.6J--> 20,356.6J

4. Mass of melted ice:

163g - 100g = 63g

5. Moles of melted ice:

63g --> 3.5mol

6. Heat Fusion of Ice:

20.4kJ / 3.5mol = 5.8kJ

Conclusion: 

In this lab ice was added to 50°C water until the temperature reached a constant temperature around 0°C. Once the data was collected, a series of calculations--such as penguins, heat gained by the ice, heat lost by the water, and the amount of melted ice in grams and moles--were made to establish the final heat fusion of the ice. The outcome was 5.8kJ, which means we had a 3.5% error factor, the correct answer being 6.01kJ.