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Hot metal + water = Tf?

  1. Oct 16, 2005 #1


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    hot metal + water = Tf???

    Yet again I find i am stumped by having no clue as to what I did in this incredibly easy problem.

    Simple right?

    1 = water
    2 = copper

    m1c1 delta T1 = -m2c2 delta T2

    This simplifies to...

    m1= (-m2c2 delta T2)/(c1 delta T1)

    which is...

    (-(110*.2*(25.6-82.4)))/((4.18*(25.6-22.3)) = 95.8533g

    Computer says i'm wrong and it slapped me

    I then did the lazy way and calculated the energy required to lower the copper's temperature. With that, I manually figured out how much water there was and got the exact same answer.

    Where am I going wrong here?
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 16, 2005 #2
    Significant digits? I don't know... :confused:
  4. Oct 16, 2005 #3


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    Did you do your numerical calculation right? Isn't the answer 90.59 g?
  5. Oct 16, 2005 #4


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    I tried 3 and 4 sig. figures with +/- 1 each direction and nothings coming out correctly. God i wanna pound the writers of this textbook.
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