1. The problem statement, all variables and given/known data A 0.10 kg piece of copper at an initial temperature of 95(degrees)C is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15(degrees)C. What is the final temperature of the system when it reaches equilibrium? (Cp of Copper= 387j/kg X (degrees)C, Cp of Aluminum= 899j/kg X (degrees)C, Cp of water= 4186j/kg X (degrees)C) 2. Relevant equations I am homeschooled and unfortunately, when I have problems figuring the physics questions out, such as this one, I have a very hard time understanding what exactly is being done (its not like I can ask the teacher for a better explanation). I try to use common sense but some times it doesn't work out. I have tried looking for examples on the internet and come here only as a last resort. Any help here would be appreciated. 3. The attempt at a solution Mass of Copper= 0.10 kg Initial Temp.= 95(degrees)C Cp= 387j/kg X 95= 36,765 Mass of Water= 0.20 kg Initial Temp.= 15(degrees)C Cp= 4186j/kg X 15= 62,790 Mass of AL.= 0.28 kg Initial Temp.= 15(degrees)C Cp= 899j/kg X 15= 13,485 Formula I used: [Cp,w X Mw(Tf-Tw)= Cp,c X Mc(Tf-Tc)= Cp,al - Mal(Tf-Tw)] The first 1/3 works out like this: 62,790 X .2(Tf-15) 62790 X .2Tf -3 (62,787 X .2Tf)= Second part: 36,765 X .1(Tf-95) 36,765 X .1Tf -9.5 (36,755.5 X .1Tf)= Final Part: 13,485 X .28(Tf -15) 13,485 X .28Tf -4.2 (13,480.8 X .28Tf)= 62,787 X .2Tf= 36,755.5 X .1Tf= 13,480.8 X .28Tf From there it was simply a matter of division, subtraction and isolating the Final Temp. or Tf. I worked it out to: (39,512.3= 14Tf) Tf=2822.31 This didn't make sense so I figured that since the answer was in Tf I should then divide it by the original temp., so I added all 3 (95 + 15 + 15) and divided (2822/125). The answer I came up with is 22.6(degrees)C Am I even close?