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Homework Help: Thermodynamics when work being negative or positive

  1. Oct 3, 2013 #1

    How are you

    I have a question how we can know the work if - or +

    also heat energy

    please could see this question why in answer
    take work as -
    and heat as +

  2. jcsd
  3. Oct 3, 2013 #2


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    hi manal950! :smile:
    it's really english rather than physics …

    work done by a system is minus

    work done on a system (or to a system) is plus :wink:

    the question says "400 J of work is done by the system", so the work comes out of the system, so it's minus

    (if the question said "400 J of work is done on the system", then the work would come into the system, so it would be plus)​
  4. Oct 3, 2013 #3


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    Tim got it backwards for the sign convention you're using. You have the first law written as ##\Delta U = Q-W##. Assume for a moment that Q=0. If work is done by the system, the system is expending energy on its surroundings, so the system's energy U must decrease. For this to happen, you need W>0. Similarly, if the surroundings do work on the system, energy is being added to the system, so you want W<0 so that ##\Delta U>0##.
  5. Oct 4, 2013 #4
    thanks so much

    thanks tiny-tim
    thanks vela
  6. Oct 4, 2013 #5
    If the work was negative, that is -400J, then the answer would have been

    ΔU = Q - W
    = 1000 J - (-400 J)
    = 1000 + 400
    = 1400 J

    So negative work means that the system gained energy or an external force did work on the system.
    In this case the problem states that the work is positive 400 J. Which means that the system did work
    and therefore lost energy indicated by a lower change in its internal energy.
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