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Finding temperature after something falls

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the temperature in K of 1.0 kg of lead initially at 300 K after it falls 200 m? The specific heat of lead is 128 J/kg C.
    a) 307 b) 311 c) 315 d)275 e) 279
    the answer my teacher said was c but i keep getting a.


    2. Relevant equations
    I thought I would use mc (Tf-Ti) = mgy

    3. The attempt at a solution
    The masses cross out. so
    (300)(delta T) = (9.8)(200)
    I get delta T = 6.53
    300 +6.53 = 306.53
     
  2. jcsd
  3. Jun 22, 2008 #2

    tiny-tim

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    Hi crazyog! :smile:

    erm … why are you using 300 for c? :wink:
     
  4. Jun 22, 2008 #3
    (1kg)(128) (Tf - 300) = (1 kg) (9.8) (200 m)
    128Tf - 38400 = 1960
    Tf = 315.31
    thank you!! :)


    ehh, how do I mark this solved?
     
    Last edited: Jun 22, 2008
  5. Jun 23, 2008 #4

    tiny-tim

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    Hi crazyog! :smile:

    I think the "SOLVED" facility got lost in the recent upgrade.

    Maybe it'll come back! :rolleyes:
     
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