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Heat flow

  1. Apr 17, 2009 #1
    If an object receives heat from a heat reservoir, is it necessary for the heat receiver to release heat?

    Can the heat receiver increase its internal energy without heat flow out of the receiver?

    And what happens when the system between heat provider and heat receiver reaches a steady state equilibrium where the infinitesimal difference in temperature between the receiver and provider cycles the heat flow? Does that mean that entropy is always increasing in an isolated insulated system between the receiver and the provider of heat??? or should we consider a system at steady state equilibrium as a perfectly quasistatic reversible process where the cyclic entropy equates to zero???
     
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  3. Apr 17, 2009 #2

    russ_watters

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    Not for a very long time, no. Ultimately, you'll run into problems keeping the object insulated and eventually you'll reach an equilibrium, but in theory you can make an object gain heat virtually forever if you pile on more and more insulation.
    That sounds like the same question as the first one...
    Absolutely nothing. Heat stops flowing (on a macroscopic level).
    A closed system at an internal thermal equilibrium isn't doing anything from a thermodynamic sense, and that includes entropy: the entropy does not continually increase if there is an equilibrium (that's what equilibrium means!).
     
    Last edited: Apr 18, 2009
  4. Apr 17, 2009 #3
    take the example where the heat provider and heat receiver are at 2 different temperatures where the difference is significant.

    if the receiver of heat begins to cycle the heat in, it must let the heat flow back out towards the provider into the section of the provider that lost the heat to the receiver. this cycle increases the internal energy of the receiver slowly until it is closer to the higher temperature of the provider of heat. this obeys the first 2 laws of thermodynamics. Eventually, the internal energy of the receiver cycles all the way up near the temperature of the heat provider. The total sum of heat received from the provider is the step by step cycling up of the internal temperature of the receiver while releasing heat into the provider minus the Work. this total sum is the Heat received from a heat source.

    steady state equilibrium is reached after all the heat from the source is received to raise the internal energy of the receiver to the heat source. (heat source is an infinite heat reservoir.)

    is everything correct????
     
    Last edited: Apr 17, 2009
  5. Apr 17, 2009 #4
    please give me some feedback?
     
  6. Apr 17, 2009 #5

    russ_watters

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    You asked something similar in a previous thread and got this answer: it isn't typically useful to think of heat transfer in that way. In particular, the second sentence contradicts the first and isn't necessary except in a convective loop.

    These are very odd questions - bordering on nonsensical even - where are they going? What bigger issue is motivating them?
     
    Last edited: Apr 17, 2009
  7. Apr 17, 2009 #6
    I'm visualizing the amount of energy lost when heat flows from provider to receiver. Particularly, I'd like to follow the energy and entropy in the heat flow from provider at T1 to receiver at T2.

    It's well known that receiver of heat ultimately obtains more entropy due to the lower Temp, and if you follow the infinitesimal amount of heat flow...you can see that it is a cycle between provider and receiver. And, if you assume reversible, quasistatic cycles that connect the movement from lower temperature to higher temperature of the receiver...it's clear why the entropy must increase until the temperatures reach the same Temp. between provider and receiver.

    I guess I'm just trying to iron out any wrinkles in this new way of thinking about heat flow using the Carnot Cycle model.

    is the total amount of work obtained from the heat flow into the receiver equal to the total change in internal energy of the receiver???????? after all [tex] Q_{in} - Q_{out} = Work_{net} [/tex]

    that doesn't make very much sense because that would imply an Adiabatic process....

    how would you calculate the new temperature of the receiver if you knew how much Work was obtained from the cyclic heat flow? or if you knew how much heat was retained from the cyclic heat flow?

    THIS IS A GOOD QUESTION!!!!
     
    Last edited: Apr 17, 2009
  8. Apr 17, 2009 #7

    Mapes

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    These questions, as phrased, are indeed nonsensical. Thermal reservoirs do not cycle. (Engines cycle, but I don't see an engine mentioned in your first few posts.) Thermal energy does not flow spontaneously from a reservoir at lower temperature to one at higher temperature. Total entropy does not increase after one or more reversible cycles.

    And when people point out these errors, you usually move on to a new set of questions. If you're really looking for answers, please describe your thought experiment fully, including all the components of the system and your assumptions, try asking one or two questions at a time, and stay focused on these questions. This approach has worked beautifully for the majority of members asking a question on these forums.
     
  9. Apr 17, 2009 #8

    Mapes

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    Ah.. are you asking about two thermal reservoirs connected by a Carnot engine? If so, the system does cycle due to the presence of the engine. The Carnot cycle is reversible and does not produce additional entropy. The thermal reservoirs are often assumed to be limitless so that their temperature does not change. If the thermal reservoirs were not limitless, the Carnot engine would eventually bring them to the same temperature (again, with no increase in entropy). Is this on track with what you're thinking about?

    EDIT: I'm really trying to understand your posts, but I'm getting lost. The terms "heat provider" and "heat receiver" are ambiguous; a Carnot engine serves as both. "High temperature thermal reservoir," "low temperature thermal reservoir," and "Carnot engine" would be much better.

    EDIT 2: Sorry, I screwed up myself regarding reversibility. The current post should be correct.
     
    Last edited: Apr 17, 2009
  10. Apr 17, 2009 #9
    can we not see every type of heat flow as a type of engine???
     
  11. Apr 17, 2009 #10
    I'm trying to figure out what exactly is entropy without the statistical part yet.

    for instance, y can't 2 adiabatic processes cross?? Why must heat flow in order to get work out of an engine?? Adiabatic processes can produce work without heat flow. yet, they obey the second law due to its internal energy change being equal to the isochoric energy change.
     
    Last edited: Apr 17, 2009
  12. Apr 17, 2009 #11
    yes!
     
  13. Apr 17, 2009 #12

    Mapes

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    I'm sorry, after a little more thought I realized I was wrong about entropy increasing in the second case, and I corrected my post above.
     
  14. Apr 17, 2009 #13
    you can connect the carnot engine to 2 heat reservoirs. One is at high Temperature and the other is at Low Temperature. When you have 2 objects at different temperatures, the two objects can function as the heat reservoirs. Thus, heat flows from the hot object to the low object.

    Now, if you follow what happens in the Carnot cycle, you'll notice that entropy describes the direction of heat flow, and I'm following the energy and entropy changes between the 2 objects.
     
  15. Apr 17, 2009 #14

    Mapes

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    Count Iblis wrote a great post last month https://www.physicsforums.com/showpost.php?p=2117035&postcount=6" about reservoirs connected together by reversible engines. If there are only two equal-size reservoirs at temperatures [itex]T_1[/itex] and [itex]T_2[/itex] and they are connected by a Carnot engine, the final temperature of the reservoirs is [itex]\sqrt{T_1 T_2}[/itex], there is a positive work output, and the total entropy is unchanged because of the reversibility of the Carnot cycle. If the two reservoirs are simply hooked together, the final temperature is [itex]\frac{T_1T_2}{2}[/itex], there is no work output, and the total entropy increases because the process is irreversible.
     
    Last edited by a moderator: Apr 24, 2017
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