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Pre Lab Every solid requires a certain amount of energy to make it melt. The amount of energy required to change a solid into a liquid is known as the heat of fusion. This experiment will incorporate the Law of Conservation of Energy to help determine the heat of fusion of water. Aim/Purpose To calculate the heat of fusion for water by applying the law of conservation of energy Equipment See Lab Page 1 Materials See Lab Page 1 Safety Be careful when working with hot water Procedure 1. Record mass of calorimeter. 2. Add 150 mL of water to calorimeter and record mass. 3. Determine mass of the water. 4. Place thermometer in water to record initial temperature. 5. Add approximately 15g of ice to the water, record the total mass, calculate the mass of the ice. 6. Record the temperature the moment the ice has completely melted. 7. Calculate T. 8. Repeat steps 1 to 8 once more. 9. Calculate the Heat of Fusion for each trial. 10. Calculate the percent error for your experimental value. Observations Mass of Calorimeter Mass of Calorimeter and H¬2¬O Mass of H2O Mass of calorimeter and H2O and Ice Mass of Ice Temperature of H2O only Temperature of H2O and melted ice Change in temperature Heat of Fusion Trial 1 3.96 g 149.58g 145.62g 162.04g 12.46g 21.0oC 15.4oC 5.6oC 273.6 J/g Trial 2 3.96g 147.12g 143.16g 185.37g 38.25g 31.0oC 12.1oC 18.9oC 245.1 J/g Calculations

Questions 1. Percent Error for trial 1 = 18.1%

2. 360 g of 20oC cold water is needed to cool 200 g of 90oC warm water to 45oC.

3. If 400 Joules are added to a sample of water, raising the temperature from 10oC to 50oC, the mass of the water had to be 2.39 g.

Conclusion After determining the formula for calculating the heat of fusion of water which was-

And plugging in the appropriate numbers and units the Heat of fusion was calculated to be 273.6 J/g. This had a 18.1 percent error of the actual heat of fusion of 334 J/g. Some sources of error could be inaccurate measurements or the quantity of ice and water used.