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Internal energy when stirring a liquid.

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A liquid is irregularly stirred in a well-insulated container and thereby undergoes a rise
    in temperature. Regard the liquid as the system. (a) Has heat been transferred? How
    can you tell? (b) Has work been done? How can you tell? Why is it important that the
    stirring is irregular? (c) What is the sign of ΔU? How can you tell?

    2. Relevant equations
    ΔU = Q-W
    ΔU = Q+W

    3. The attempt at a solution
    I know this is an isochoric process even though work is being done. In this case there is no change in volume.

    For (a) I said there is no heat transferred since we are just doing work and the system is well insulated.
    (b) Yes work has been done on the system because we are stirring it...
    (c) The sign of ΔU is positive since we are doing work on the system and the kinetic energies of the particles will increase. Thus there is an increase in temperature/internal energy.

    What do you guys think of my answers? I have an answer key but it seems to have mixed up the answers to a similar question and I do not trust the answers.
     
  2. jcsd
  3. Oct 9, 2011 #2
    Your answers are OK.
    Remember that the first law of thermodynamics assumes a convention re signs for increases/decreases in U,W and Q.
     
  4. Oct 9, 2011 #3
    What do you mean by a "convention re signs"?
     
  5. Oct 9, 2011 #4
    We used to learn at school that if
    [itex]\Delta[/itex]U = [itex]\Delta[/itex]Q + [itex]\Delta[/itex]W
    is used then one must follow the sign converntion. This is that whatever increase U must be entered as a positive number and whatever decreases U must be entered as negative.
    One must be careful because some books use an equally consistent but different convention.
     
  6. Oct 9, 2011 #5
    Oh yeah... Our books have it as ΔU=Q-W. Although at times I do not agree, but I will have to cope with it :P
     
  7. Oct 9, 2011 #6
    Then you enter work by system as a positive number because the minus is already built in in the equation used.
     
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