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Need help on a simple question - tempeture change

  1. Dec 7, 2005 #1
    The question is:
    A 1000 g sample of lead shot, at 300°C, is dropped into 100 g of water at a temperature of 5.6°C. The specific heats of lead and water are 0.129 and 4.184 J/g°C, respectively. What is the final temperature of the mixture, in degrees Celsius?

    I was doing my chemistry pratice, then I saw this question, I don't have a clue at all. It looks very similar to the high school physics, but I can not remember anything. Can you just give me some hints?

    I just remember the change of Tw and Tl is the same.
    but how does it relate to the heat capacity?...

    I solve for the energy of the water: 100 X 4.184 X 5.6 = 2343.04 J
    .......................................lead: 1000 X 0.129 X 300 = 38700 J

    then what should I do?:frown:

    please help,
     
  2. jcsd
  3. Dec 7, 2005 #2
    nevermind, i finially remember what to do....
     
  4. Dec 7, 2005 #3

    Astronuc

    User Avatar

    Staff: Mentor

    And if the final temperature exceeds 100°C, there is a phase change of the water at 100°C, which would also have to be considered.

    Heat of vaporization of water = 2.26 J/kg or 539 cal/g.

    Hopefully you remembered [itex]\Delta H[/itex]= m cp[itex]\Delta T[/itex]
     
  5. Dec 7, 2005 #4
    I remembered it after I post, but I really appriciated it, thank you
     
    Last edited: Dec 7, 2005
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