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Is this right deltaUA=-deltaUB?

  1. May 7, 2015 #1
    If we have 2 systems like water(A) and cold water(B) in contact, if we ignore the energy exchanges between air and those systems, could we consider that the deltaUB=-deltaUA??
  2. jcsd
  3. May 7, 2015 #2
    Are A and B kept separate, and just allowed to exchange heat, or are they mixed with one another?

  4. May 7, 2015 #3
    They are mixed
  5. May 7, 2015 #4
    If they are mixed, then it is not possible to determine the separate identities of A and B in the final mixture. So you you can't determine ΔUA and ΔUB separately. However, you can determine the difference in internal energy between the final mixture and the initial separate internal energies of A and B, and set this difference equal to zero.

  6. May 7, 2015 #5
    If deltaUA and deltaUB=weight . mass thermal capacity . delta temperature, I can determine the deltaU of both A and B even if they are mixed together
  7. May 7, 2015 #6
    Now I understand what you were saying, ok but isn't it the same measuring the deltaU with the final mixture?
  8. May 7, 2015 #7
    You can do that, and you would obtain the same result as if you used the method I described in post #4. However, conceptually, if the fluids are mixed, it is not really appropriate to identify separate changes for the internal energies of A and B. After all, at the molecular level, the fluids would be intimately mixed, and you could no longer identify either liquid.

  9. May 7, 2015 #8
    So This isnt right deltaUb=-deltaUa?
  10. May 7, 2015 #9
    Also, if you were mixing different fluids to form a solution, there might be a heat of mixing, and it would then be problematic to identify the separate contributions of each of the two fluids.
  11. May 7, 2015 #10
  12. May 7, 2015 #11
    That would be OK for mixing two bodies of the same liquid, but what would you do if they were two different liquids, and there was a heat of mixing?

  13. May 7, 2015 #12
    So if they were two different substances that expression wasn't apropriated. Thanks about your help
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