[kimmyc2008]

A 325 gram piece of gold at 427 C is dropped into 435 grams of water at 22.0 C. The Specific heat of gold is 0.131 J/gC. Calculate the final temperate of the mixture.

[ebaines]

The trick here is to understand that the gold and water will be at the same temperature once they reach equilibrium, that energy is conserved, and that the specific heat is a measure of how much heat energy is represented by a 1 degree change in a gram of material. Call the final equilibrium temperature T. Intuitively you can see that the equilibrium temp is somewhere between the initial temp of the hot gold and the cool water. The initial heat energy of the gold is 235 g * 427C * .131J/gC, and the initial heat energy of the water is 435g * 22 C * 1J/gC (I am assuming that the specific heat of water is 1 J/gC -check your text to verify). The sum of these initial energies must be equal to the heat energy at eqiulibrium, which is (235*.131+435*1)*T. Now you can calculate T.