If 150.0 grams of iron at 95.0 degrees Celcius, is placed in an insulated container containing 500.0 grams of water at 25.0 degrees Celcius, and both are allowed to come to the same temperature, what will that temperature be? The specific heat of water is 4.18 J/g degrees Celcius and the specific heat of iron is 0.444 J/g degrees Celcius.

Answer 1

heat lost by Fe + heat gained by H2O = 0

[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tnitial)] = 0
Solve for Tfinal.

Answer 2

21.6

Answer 3

The setup by DrBob222 is the most straightforward way. You could also set q of the metal as a negative (because it's losing heat) and set it equal to q of the water (which should be positive, as it's gaining heat). It's not possible for the water to end at a lower temperature than it started if the metal is at a higher initial temperature. You would plug in all known variables, such as masses, specific heat capacities, and initial temperatures, distribute the m•C product to (Tf-Ti) on both sides, then simplify. Final answer should be around 27.2ºC