ChemLab #25 - Heat Of Fusion of Ice
Date Submitted: Score:   
Lab Group Members:
Introduction: The amount of energy required to convert a solid to a liquid, at constant pressure and temperature, is called the heat of fusion of the substance.

In this experiment, the heat of fusion of ice will be determined. The ice will be melted by placing it in a known volume of hot water contained in a plastic cup. The system will be left undisturbed until all the ice has melted. The amount of heat lost by the hot water (absorbed by the melting ice) in this process can be calculated according to equation #1:

q (water) = s (water) X m (water) X ΔT (water)
The volume of ice that melts can be determined by equation #2:
volume of melted ice = volume of water (final) − volume of water (initial)
The mass of melted ice can be calculated with equation #3:
mass = density X volume
The heat of fusion of ice can be calculated with equation #4:
heat of fusion of ice = heat absorbed by ice/mass of ice melted
Materials:   400 ml beaker, wire gauze, plastic foam cup, gas burner, thermometer, 100 ml graduated cylinder, ring stand, beaker tongs, 250 ml beaker, ring support, spatula
Experimental Design:
  1. Add approximately 100 ml of water to a 250 ml beaker, and heat the water to 60 oC, using a gas burner and standard ring stand assembly. While the water is heating, fill a plastic foam cup halfway with ice cubes. Place the cup in a 400 ml beaker for support.
     
  2. When the temperature of the water has reached 60 oC, use two 20 ml portions of this hot water to preheat a 100 ml graduated cylinder. Rinse the cylinder with each of the hot water portions and discard the rinses.
     
  3. Pour 30.0 ml of the hot water into the graduated cylinder. Record the volume on the data table. Measure and record the temperature of the water to the nearest 0.5 oC, and record on the data table.
     
  4. Quickly drain any excess water from the ice cubes in the cup. Add the measured hot water to the ice in the cup. Stir the ice water rapidly, but with care, until its temperature falls to 2 oC. At this point, some unmelted ice should remain in the cup. If all the ice has melted, add a bit more so that some ice remains unmelted when the temperature is 2 oC or below. Record on the data table the lowest temperature of the mixture of ice and water to the nearest 0.5 oC.
     
  5. Using a spatula, quickly remove any unmelted ice from the cup. As you remove the ice, drain as much water as possible back into the cup.
     
  6. Carefully pour the cold water from the cup into the graduated cylinder and record the final volume to the nearest 0.1 ml on the data table.
Data Table
Initial volume of hot water.   _______ ml
Initial temperature of hot water.   _______ oC
Final temperature of water and melted ice.   _______ oC
Final volume of water and melted ice.   _______ ml
 
Calculations:
  1. Calculate the ΔT of the hot water. This is simply the initial temperature of hot water − final temperature of water and melted ice.
  2. Calculate the mass of the hot water. Don't forget that 1 ml of water has a mass of 1 g. This is because the density of water is 1 g/ml. So the volume in ml is the mass in grams.
  3. Calculate the heat lost by the hot water. (See equation 1 in the introduction.) The specific heat of water is 4.18 joule/g • Co. Your answer will have joules as the unit.
  4. Calculate the volume of ice melted. (See equation 2 in the introduction.)
  5. Calculate the mass of ice melted. (See equation 3 in the introduction.) As in step 8, 1 ml of water has a mass of 1 g. So its just the volume changed to grams.
  6. Calculate the heat of fusion of ice in joules/g. (See equation 4 in the introduction.) The heat absorbed by the ice is the same as the heat lost by the hot water, so it's the answer from step 9. The mass is from step 11.
Conclusion:     The accepted value for the heat of fusion of water (ice) is 334 joules/g. How did your results compare to this accepted value?