Yet anther strange question about entropy

Discussion in 'Physics & Math' started by Secret, Jul 24, 2014.

  1. Secret Registered Senior Member

    (NB for the purpose of the discussion, entropy as it is understood in chemistry is sufficient)

    Recall in physical chemsitry and 1st year chemsitry, we have came across Gibss free energy and some examples of endothermic process, e.g. solvation of NH4Cl
    Such process works despite having a positive enthalpy change is because it is offset by the increase in entropy as the ions get solvated

    Now recall the conceptual ideas of entropy in chemistry
    1. Molecules that are quite floppy (thus can wiggle more) have higher entropy
    2. Entropy measures the extent of spreading of energy among the molecule's various modes

    Now recall second law
    entropy of an isolated system increases over time until it maximize at thermodynamic equlibrium
    Now my question is, if all that is needed for second law to be happy is to increase entropy, and entropy (for most purpose in chemistry) is a measure of energy spread

    If we have a molecule which is incredibly porous and incredibly floppy (let's call this A), can we do the following?

    1. A has temperature T K (thus has energy U1)
    2. A piece of iron block has temperature T+1 K (and has energy U2)
    3. Now stick the iron to A. Because A is incredibly floppy, the entire system A+Iron is many orders of magnitude more floppy than the iron
    4. Heat (Q) will flow from hot to cold as expected, in order to increase the entropy of A+Iron by attaining equlibrium
    5. Meanwhile, we attached a battery (assume 100% efficient and no loss over time) in between to convert some of the heat into electricity and stored into it. Thus we basically make a heat engine with work (W) extracted from a reservoir

    a. Can Q be made arbitrarily small (NOT ZERO, as this is impossible), so it is in the order of 10^-30, due to the incredibly floppy nature of A+Iron will allow a small amount of energy to be distributed over (maybe) a gazillon of thermal modes, thus able to meet the entropy requirement of the 2nd law, or is this impossible because Carnot's Theorem also have taken account of this and Q can never met the requirement if it is too small, even if the final state has a lot more possible microstates (chem: ways to wiggle)?

    b. If a is no, how to modify the calculations in Carnot theorem in order to introduce the factor of microstates of A (which can be set to any value to analyse different A, let's call this factor B) and show that there is a limit to how small Q can get even if B tends to infinity?

    c. do the rate of heat transfer depends on the total number of microstates in the final state of the system (that is will the transfer be more rapid if the final state of the system has more microstates? (Analogy, like how gas expanding to a very large container is quicker than the small one, given same initial pressures each for both containers))
    I am kinda rusty because the stuff is a year away, thus apologies if I missed out some obvious theorems that I could have used and help and reminder would be appreciated if that's the case

    View attachment 7235
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  3. origin Trump is the best argument against a democracy. Valued Senior Member

    Batteries do not convert heat into electricity.

    I really do not understand the question. If there is a flow of heat you can convert some of the heat to work but of course it will not be 100% efficient so entropy increases.
    Last edited: Jul 24, 2014
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  5. Secret Registered Senior Member

    Well... I am just trying to brought out the point that heat is converted to work, but my memory failed me in I forgot the way energy storage cells

    work (e.g. those which store up the solar energy collected in solar cells, I think that's either electricity->electric potential energy (for capacitors),->chemical energy (for chemical cells?))

    Let me try rephrasing the question
    We all know how a heat engine works in that some of the heat has to be wasted in the cold reservoir to achieve the overall increase in entropy (2nd law)

    But can there exist a heat engine where because the cold reservoir has a lot more microstates than the hot reservoir, very little heat can be wasted into it to achieve overall entropy increase, and because of this, the portion of heat being converted to work can be larger (even slightly) than carnot's, and still less than 100%?

    *(if this engine can exist but cannot exceed carnot efficiency, then I am not sure how to show the final result be still bounded from above by the carnot efficiency, since (by intuition) Q5 will be very small as W_cyan->infinity, and thus by 1st law, Work will tends to Q4?)
    **(if this engine cannot exist, then how to incoporate the no. of microstates of the reservoir into the equation so that one can work out the same result predicted by carnot's theorem?)

    I know how to do the proof for carnot's theorem (assume there exist eta>eta_carnot->make eta drive eta_carnot->heat flow backwards=impossible), but I don't know how to take account of the no. of microstates available in the reservoir
    View attachment 7244

    PS Typo in pics: T2 should be TC
    EDIT 1: Did some preliminary calculations, it seems it will only make it worse, thus * seemed to be correct (not sure if my reasoning is correct)
    For a heat engine as shown in the leftmost figure of attached pics, 2nd law requires:

    |Q1/TH|<|Q2/TC| -------(1)

    So if the cold reservoir has more microstates, it is effectively the same as raising its temperature TC to TC'

    Then if Q2' remains =Q2 it would be a lot less likely for Q2'/TC' to be larger in magnitude than Q1/TH (since there will be less possible sets of (TH,TC',Q2') that can obey (1) for a given Q1)

    Thus to maintain the same difference as the case for TC, Q2' has to be > Q2, thus the efficiency is worsen
    In a physical picture, since there is a lot more microstates in the cold reservoir, the entire payload of intenral energy U in the heat engine got spread out all over the system, and most end up in the cold reservoir, thus much less is available to do work

    But then it brought out another question:
    If my above reasoning is correct, that means when TC'<TH reaches a certain point, Q2'->Q1, Work->0, and heat flow will stop despite there's still a temperature gradient (as otherwise 2nd law will be mad), but that's absurd, how can a system not in thermal equlibrium (hence not in thermodynamic equlibrium) can achieve maximum entropy?

    So it follows there is something wrong in my reasoning, but I am not sure how to fix it

    EDIT 2:
    In light of EDIT 1, what if we have a hot floppy reservoir and a cold rigid reservoir, (so that TH>TC, TH'=f(W_TH,TH) TC'=f(W_TC,TC) and TC'<<TH', thus |Q1/TH'|<|Q2/TC'| and thus Q2 can be made smaller?), but from a physical point of view, won't that mean in equilibrium most energy stays locked in the hot reservoir and little is available to do work?
    Last edited: Jul 25, 2014
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