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Internal energy change

  1. Aug 2, 2014 #1
    Hi Guys! I have a confusion which I hope you can help clear up. The mathematical expression of the first law of thermodynamics can be stated as δu=Q+W where u is the internal energy of the system, Q is the heat added(or taken from) to the system and W is the work done by or on the system. If I lift a glass of water to a certain height, the internal energy is unaffected correct? But if we refer to the equation before, work is being done on the system (changing its gravitational potential ), so why doesn't this form of work done qualify to appear in the system? ( My hunch is that for this form, the system has to include the cup and the zero gravitational potential reference ).
    Next question, If we would freeze a cup of water at a constant temperature, does the internal energy change? From the equation, heat is taken away from the system, but I am not too certain as to any work is done by the system. So, δu most likely <0
    Alternatively, u = sum of molecular kinetic and potential energy. since temperature is constant, kinetic energy is constant, but I am sure that potential energy has increased since water molecules form more bonds with and come close to each other in the solid state. Here, I would think that δu>0. Then, I seem to contradict myself LOL. Perhaps someone can point out where I got my assumptions wrong?

    Thank you for your time, have a great day.
  2. jcsd
  3. Aug 3, 2014 #2

    Simon Bridge

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    The thermodynamic equation does not include every possible way that work can happen nor every possible way that energy can be manifest.

    (1) When you lift the glass of water, at a constant speed, the work you do manifests as a change in the height of the glass. The equation is just not appropriate here. The acceleration and deceleration stages sets up a pressure gradient in the water that would have an effect on the internal energy. The thermodynamic equation would tell you what portion of the work did this.

    (2) When you freeze a glass of water, the exact form the equation takes depends on how you freeze it. As far as the system is concerned, if there is no mechanical work on or by the water, then you have ΔU=ΔQ.

    This would suggest that the internal energy of a solid is less than the internal energy of a liquid, at the same temperature. What you need to do is work out if the relations you are thinking about are valid for solids.

    The energy situation in a change of state is usually better handled from the POV of entropy.

    I think what you are discovering here is that the rather simple models you have been given are not the whole story.
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