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!

Thermodynamics Entropy question

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A 10-kg block of copper is initially 100 degrees celsius. It is thrown into a very large lake that is 280K. What is the entropy change of the piece of copper. What is the entropy change of the universe?

    2. Relevant equations



    3. The attempt at a solution
    I know that the entropy change of the universe is the change of the system and the change of the surroundings. I Don't really know how to go about this problem. I know for say two copper blocks next to each other I can do ncdeltaT + ncdeltaT and b/c they have the same mass and C it is easy. however I'm assuming this won't work for a few reasons in this problem because the lake I'm pretty sure I am supposed to assume is infinite compared to the copper Can anyone help pleaseee?
     
  2. jcsd
  3. Mar 4, 2009 #2

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you need to think about the energy associated with heating the copper, and about finding the resulting entropy change by integrating an equation like [itex]q=T\,dS[/itex].
     
  4. Mar 4, 2009 #3
    I worked out a solution let me see what you think(i was thinking something sorta like what you said)

    I said change in internal = Q and changein internal is the integral from 373 to 280 of the C for copper which is 2.723/8.314(which is R) in my book dT (my book gives C's in Cp/R for some reason) Anyway after this integral is found I multiplied by 8.314 to get rid of the R then I said change in entropy for system is this value over 373 and for surr is the negative Q over 280 does this seem correct?
     
  5. Mar 4, 2009 #4

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Just dividing the energy by 373 K isn't going to work for the copper; some of this energy is transferred when the copper is at 373 K, some at 303 K (for example), and some at 280 K. Try integrating the dS equation as I mentioned earlier.
     
  6. Mar 4, 2009 #5
    so to get Q i do use the integral of C from 373 to 280 and then i take that and put it over T and do an integral from 373to 280 dT right?
     
  7. Mar 5, 2009 #6

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Exactly. And if you assume that the heat capacity is temperature independent, the first integral is quite simple.
     
  8. Mar 5, 2009 #7
    thanks a lot !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Thermodynamics Entropy question
Loading...