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Specific heat mixture, final temp, etc.

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    The specific heat of lead is 0.03 cal/g*C. 300 grams of lead shot at 100* C is mixed with 100 grams of water at 70* C. What is the final temperature of the mixture if the container is insulated.

    2. Relevant equations



    3. The attempt at a solution

    Q1 = c1m1(Tf - 100) (lead)
    Q2 = c2m2(Tf - 70) (water)

    Q1 + Q2 = 0

    c1m1(Tf - 100) + c2m2(Tf - 70) = 0
    9(Tf - 100) + 100(Tf - 70) = 0
    9Tf - 900 + 100Tf - 700 = 0
    109Tf = 1600
    Tf = 14.7*C = no


    EDIT, nevermind... 100 x 70 = 700, you heard it here first!
     
    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    100 x 70 ≠ 700
     
  4. Nov 16, 2011 #3
    I caught it. :redface:

    Works out fine now. Thanks!
     
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