Object of the ExperimentEnergy is required to change water from a solid to a liquid, i.e.to melt ice. In this experiment you will try to measure thelatent heat of fusion ofice (LHice),theenergyneeded (per gram) to melt ice. The needed energy will come from acup of warm water. The amount of water and its temperature willbe measured before adding some ice and then again after the icehas been melted. These data will be used in an energy balanceequation to determine LHice.Conducting the ExperimentYou will first need to obtain some ice. Let the ice sit ina bowl at room temperature until you notice that it is starting tomelt. This will probably take 15 to 20 minutes. In afreezer, ice is kept at a temperature well below freezing (typicallynear 0o F). We need to allow the ice to sit outsidethe freezer long enough that is warms up to freezing (0oC or 32oF). The ice can"t get warmer than freezing; once it reachesfreezing it will start to melt.Add 200 to 250 mL of warm tap water (40oC or less so that there isn"t too big of a temperaturedifference between the water and air) to the styrofoamcup. We keep the warm water in a styrofoam cup tominimize the loss of heat to the cooler surroundings. Keep arecord of the actual amount of water added to the cup. Measureand record the temperature of the warm water just before you are readyto add some ice.Add one or two ice cubes (or about 1/4 cup of crushed ice) to the warmwater and stir gently with the thermometer until the ice is completelymelted. Try to be sure you just add ice and none of the waterin the bowl from the melted ice. Once the ice in the cup hasmelted, measure and record the water temperature. If thetemperature has dropped to near freezing or if you are unable to meltall of the ice, you will need to repeat the experiment. You willneed to use less ice or more water or both.Use the graduated cylinder to measure the new volume of the water inthe styrofoam cup. This new volume will be larger than theinitial volume because it contains the water from the melted ice.Because water has a density of 1gram/mL (1 gram/cm3) there is a one toone relationship between a measurement of water volume in mL and themass of the water in grams. The difference in the starting andending water volumes is the mass of the ice that was melted.Repeat the experiment at least one more time using a different amountof water, a different amount of ice, and perhaps a different initialwater temperature.Data AnalysisThe conditions at the start of an actual trial run of theexperiment are shown below. Tinitial is the temperature of thewater water before the ice was added, Mwater is the mass of the warmwater (note the mass is actually determined by measuring the intialvolume of the water). We assume that the ice has an initialtemperature of 0o C because we letthe ice sit at room temperature until it had started to melt.
In the figure below the ice hasbeen added to the warm water and melted. We are at the "midway"point of the experiment; the ice has turned from a solid to a liquidbut its temperature hasn"t changed, it is still 0oC.
This energy came from the warmwater and cause the water to cool (from Tinitial to Tmid). Wecould use the following equation to determine how much energy was takenfrom the water (we won"t do the calculation because we don"t know whatTmid is).
Heat will continue to flow from thewarm water into the 0o C ice water. Energy will flow until all of the waterhas the same temperature, Tfinal.
Now we will combine all theseparate terms. Energy was removed from the water to first meltthe ice and then to warm the ice water to Tfinal. The toal energylost by the warm water is: We perform the experiment in aninsulated cup and we assume that all of the energy lost by the warmwater was used to melt ice and warm up the ice water; no energy flowedfrom the cup into the cooler air in the room. We"ll add the 1stand 4th equations above together: We"ll set the energy lost andenergy used equations equal to each other. This gives us anenergy balance equation that can used to determine LHice:
The specific heat of water is 1cal/gm oC. You can plug this valuetogether with your measured values for Mwater, Mice, Tinitial andTfinaland solve for LHice. An example calculation using the data shownin the figures above is given below. The known value for thelatent heat of fusion of ice is 80 calories/gram so the measured valuebelow compares pretty well.