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I Change in state variables

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  1. Aug 28, 2016 #1
    consider an irreversible cycle. There will be no changes in the state variables (like U, S) of the system undergoing the cycle, but for the surroundings, there will be change in the state variables, in fact, entropy increases.

    my question is, what happens to the internal energy of the surroundings?.. according to what i had calculated, it does change, it decreases...but does this mean its violating the conservation of energy?!...
     
  2. jcsd
  3. Aug 28, 2016 #2

    jfizzix

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    The conservation of energy would say that the total energy of the system plus surroundings is a constant.
    This means that if the total energy of the system remains constant, the total energy of the surroundings must as well.

    That being said, in this engine cycle, there is a net flow of heat into the system from the surroundings since some of that heat flow is being converted into work. As such, there is work being done on the surroundings by the system, whose total energy exactly cancels out with the net heat loss of the surroundings.

    In order for the system to do work, its volume must change. With the volume of the system-plus-surroundings being a constant, the volume of the surroundings must change by an equal an opposite amount. Without accounting for the change of volume of (and the work done on) the surroundings, it would seem like the internal energy of the surroundings decreases, but it actually remains constant.

    The conservation of total energy is sufficiently universal that it serves as a good sanity check on problems like these. Energy unaccounted for was a driving force in research into thermodynamics research back in the day.
     
  4. Aug 28, 2016 #3
    i don't understand how it exactly cancels out..
    case 1: during the expansion of the gas, dq amount of heat goes into the system and dw amount of work is done by the system. now if the cycle is isothermal, then from the view point of the surroundings, we have : dw <= - dq (the inequality is due to 2nd law since 100% conversion of heat is not allowed)

    case 2: during contraction, surroundings does the work and dq amount of heat is liberated from the system, then we have...dw = -dq (from 2nd law, 100% conversion of work into heat is allowed )

    adding, case 1: dw + dq <=0 implies ΔU<=0
    case 2: dw+dq=0 implies ΔU=0....therefore there is a net decrease, how does this get cancelled?
    sry if anything is not clear.
     
  5. Aug 30, 2016 #4

    jfizzix

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    There are two heat flows in the system.
    There's the heat flow from the high temperature reservoir into the system given by [itex]Q_{H}[/itex]..
    .. and there's the heat flow from the system into the low temperature reservoir given by [itex]Q_{L}[/itex].

    Since the total energy of the system remains the same over one cycle, the work must equal the net heat flow:
    [itex]W=Q_{H}-Q_{L}[/itex].
    The second law does forbid 100 percent conversion of heat into work, but this just means that [itex]W \leq Q_{H}[/itex]. The heat not converted is [itex]Q_{L}[/itex].

    The second law of thermodynamics says that change in entropy of the system plus surroundings after one cycle is greater than or equal to zero. Since it is a cycle, the system's entropy also remains constant at the end of the cycle, so the entropy of the surroundings must increase.

    Where a general infinitesimal change in entropy is given by:
    [itex]dS=\frac{dQ}{T}[/itex].
    We know that entropy leaves the surroundings through heat flow from the high temperature reservoir into the system...
    and entropy enters the surroundings through heat flow from the system into the low temperature reservoir.

    Since we may take these reservoirs to be large enough to have constant temperature, the net change in entropy in the surroundings in one cycle is given by:
    [itex]\Delta S =\frac{Q_{L}}{T_{L}} - \frac{Q_{H}}{T_{H}} \geq 0[/itex].


    As one extra bit, this means that where the efficiency [itex]\epsilon[/itex] of the cycle is given by the ratio of work gotten out over heat put in:
    [itex]\epsilon\equiv\frac{W}{Q_{H}}[/itex],
    the second law of thermodynamics (i.e., [itex]\Delta S\geq 0[/itex]) gives a tighter upper limit (than just 100%) to the efficiency of these sorts of heat engines:
    [itex]\epsilon \leq 1-\frac{T_{L}}{T_{H}}[/itex].
    This is also known as the Carnot efficiency limit, because it's also the efficiency of an ideal Carnot engine.
     
    Last edited: Aug 30, 2016
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