1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Q=m*c*deltaT


    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook