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seanmcgowan
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Homework Statement
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)
Homework 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.
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?