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Thermodynamics-Conceptual problem

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data

    When we heat an object,it expands.Is work done by the object in this process? Is heat given to the object equal to the increase in its internal energy ?

    2. Relevant equations



    3. The attempt at a solution

    Had it been a gas ,the work done would be pΔv,but I am unclear how do we see the work done in general while an object expands.I am assuming,the object may be a solid or a liquid.

    I am unsure whether the problem assumes the object is a gas.

    Kindly help me in clearing the doubt.
     
  2. jcsd
  3. Jul 3, 2013 #2

    ehild

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    When we heat an object, it is not sure that expands. Think about ice. You heat,. it becomes water, and occupies less volume.

    Or think about gas heated at constant volume. You can realize a set-up when the volume does not change.

    If the object exerts force on its surroundings and that force causes some displacement then work is done by the object.

    ehild
     
  4. Jul 3, 2013 #3
    Generally use First Law Q = U + W, for these types of problems. Do follow the correct sign conventions carefully









    The general formula for Work Done by gas or solid is ∫P*dv.
     
  5. Jul 4, 2013 #4
    Suppose the object is a round metal plate which expands on heating.The radius increases from R1 to R2.

    Do we say that the plate exerts a force on the surroundings ? Will the work done be given by pΔv ?
     
  6. Jul 4, 2013 #5

    ehild

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    No work in vacuum. In air, the surface of the expanding disk exerts force on the air, compressing it. So it does work on the air. If the expansion is so slow that the pressure of the air stays constant during the expansion, the disk does pdV work on the surrounding air.

    If there is a box touching the disk and you heat the disk, there will be a normal force between the box and disk. Supposing, the disk is fixed at the centre, the box is shifted by distance R2-R1. The work done by the disk on the box is the integral of the normal force between R1 and R2.

    ehild
     
  7. Jul 4, 2013 #6
    Very nice...Thanks a lot :biggrin:
     
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