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!

Thermal Equilibrium of a system

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A combination of .25kg of H2O at 20C, .4kg of Al at 26C and .1kg of Cu at 100C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final T of the mixture.


    2. Relevant equations



    3. The attempt at a solution

    At first this seemed really simple to me I just figured out if the T would be below or above 26 to determine which parts will lose energy and which will gain.
    So it turns out that Tf will be less than 26 because the energy required to heat up water to 26 is greater than the copper can supply.

    Setting up the equation is where I get lost.

    If the energy lost by the Al and Cu must equal the energy gained by the water then shouldn't the equation be:

    Q(Cu)+Q(Al)=Q(H2O)

    where Q= mCΔT

    The book has it set up as

    Q(Cu)+Q(Al)+Q(H2O)=0

    which is really confusing me.

    Ok so if I let the lost energy be negative I get the same thing as the book does but then it seems to me like I'd get the same answer in every case.

    The book's way:
    -Q(Cu)-Q(Al)=Q(H2O)
    Q(Cu)+Q(Al)+Q(H2O)=0

    But then let's say Tf > 26 then we get

    -Q(Cu)=Q(Al)+Q(H2O)

    Which simplifies to

    Q(Cu)+Q(Al)+Q(H2O)=0

    exactly the same thing! That doesn't make any sense to me.
     
  2. jcsd
  3. Feb 28, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    One material's loss is another material's gain. And a negative gain is a loss. You have a tendency to want to use positive numbers only, but the book allows negative Q too. Which is good if you consider that a ΔT can be < 0 !
     
  4. Feb 28, 2014 #3
    Let Q(Cu) be the energy gained by copper.
    Let Q(Al) be the energy gained by aluminum.
    Let Q(H2O) be the energy gained by water.
    If any of these turn out to be negative, it just means that that particular species lost energy. So,
    Q(Cu)=0.1CCu(T-100) is the energy gained by Cu, where T is the final temperature, even if T is less than 100.
    The same kind of thing for Al and H2O in terms of T.

    If you do it this way, then Q(Cu)+Q(Al)+Q(H2O)=0 makes sense, since the sum of the three energy "gains" is zero.
     
  5. Feb 28, 2014 #4
    Ah that makes sense. Thx.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thermal Equilibrium of a system
  1. Thermal Equilibrium (Replies: 1)

Loading...