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Finding entropy involving metal in water

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A piece of metal at 80C is placed into 1.2L of water at 72C. This thermally isolated system reaches a final temperature of 76C. Estimate the overall change of entropy for this system.

    2. Relevant equations
    deltaS = Q/T

    3. The attempt at a solution
    I think that i have to find Q for metal and water separately. correct? and if so i dont know how to find Q for the metal because i dont have its mass or what kind of metal it is to know its heat capacity.
    does anyone have any suggestions?
  2. jcsd
  3. Apr 14, 2010 #2
    Please, can anybody help me with this question?
  4. Apr 14, 2010 #3
    The amount of heat that leaves the metal is equal to the amount of heat that goes into the water.
    Last edited: Apr 14, 2010
  5. Apr 14, 2010 #4
    I don't think it's possible without the metal :|
  6. Apr 14, 2010 #5
    But don't I still need to know the type of metal or mass of metal to find the Q in order to find the deltaS for the metal?

    And I dont even know how to find out the water part because what is 1.2L in kg?
  7. Apr 14, 2010 #6
    Well you can find the mass of the water using the density of the water. But I just noticed this process isn't isothermal so I do think you are lacking information here.
  8. Apr 15, 2010 #7
    That's what I thought too.
  9. Apr 18, 2010 #8
    there has to be a way to answer this question .. it is part of an assignment that I have to hand in, so any other suggestions?
  10. Apr 18, 2010 #9
    Thinking about it some more there is a solution.

    You should be able to show using a formula from your book that

    [tex] \Delta S = m_w c_w ln(\frac{349}{345}) + m_m c_m ln(\frac{349}{353})[/tex]

    Also, since heat in equals heat out

    [tex] m_w c_w (349-345) = -m_m c_m (349-353)[/tex]

    That's all you need to solve this problem.
  11. Apr 18, 2010 #10
    thank you for the help!
  12. Jun 14, 2010 #11
    I am having this same problem. What's the ln mean?
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