An example of a common calorimetry problem asks the student to determine the experimental value for the specific heat of fusion of ice. In this scenario, 25.8 grams of ice are placed in a Styrofoam cup with 100 grams of water at 34.5 degrees Celsius.
Continue ReadingAfter leaving the cup alone for a few minutes, the ice melts completely, leaving the water temperature at 18.1 degrees Celsius. To solve this problem, the student needs to recognize that the amount of energy that the water loses while cooling is equal to the amount of energy necessary to melt ice. This can be written as Qice = -Qcalorimeter, where Qcalorimeter=m*C*T. This is further reduced to Qcalorimeter=(100g)*(4.18J/g/C)*(18.1-35.4), or -7231.4 J. Finally, the student uses Qice=-7231.4 J to determine the heat fusion.
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