Question

Several years ago a company invented plastic ice cubes. These were colored plastic cubes with frozen water (or another substances) inside. That way, as the solid melted, it could cool your drink, but did not dilute it with extra water (from the melted ice). If plastic cubes containing 86.3 g of frozen stuff are placed into 150.0 g of water at 38.0oC and the temperature drops to 29.4oC, what is the heat of fusion of the stuff in the cubes. Assume the plastic does not effect the heat transfer in any way.

Answer 1

heat absorbed by frozen stuff + heat lost by water in cooling = 0
86.3 x heat fusion + (mass H2O x specific heat water x (Tf-Ti) = 0
Tf = 29.4
Ti = 38.0
This calculation does not allow the melted stuff from cooling the water further.