Question about heat transfer

Gweedz
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Topic 190755

Say you have a 1000 gallon tank that is thermally isolated from the environment. Inside you have a liquid at 100 deg. F. Over time, will heat rise in the tank making the liquid near the top warmer than the liquid near the bottom?

Common sense says heat rises, but that would violate the 2nd law of thermodynamics. I must be missing something.

nfortino
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Question about heat transfer

Quote:

Say you have a 1000 gallon tank that is thermally isolated from the environment. Inside you have a liquid at 100 deg. F. Over time, will heat rise in the tank making the liquid near the top warmer than the liquid near the bottom?

Common sense says heat rises, but that would violate the 2nd law of thermodynamics. I must be missing something.

While common sense does say heat rises, it's wrong when applied in the way you are using it. Heat is not actually a physical thing. Instead it is a property which basically describes how fast the molecules of a substance are moving. For fluids (liquid or gas), a warm fluid is typically less dense than a cold fluid. Thus, in a system of warm and cold fluid, the warm fluid will typically rise to the top due to gravity. While it does make sense to say that heat rises, it is the fluid that actually does the rising. The fluid carries energy with it, in the form of heat, and thus one could say that the heat rose. The common sense part comes from the observation that warm air rises, and cold air falls, and then making the (false) leap to say that heat rises. Heat will never rise, however, without a carrier of energy, as heat, just like energy, is not a physical thing.

In the case of your tank of fluid, assuming the fluid is all one temperature, it has uniform density throughout. Thus, there will be no net motion of any one section of fluid, meaning no heat movement.

Gweedz
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Perfect explanation! It's so

Perfect explanation!
It's so good I had a couple other questions to ask but you answered those also! Thank you.

Mike Hewson
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RE: While common sense does

Message 25075 in response to message 25073

Quote:

While common sense does say heat rises, it's wrong when applied in the way you are using it. Heat is not actually a physical thing. Instead it is a property which basically describes how fast the molecules of a substance are moving. For fluids (liquid or gas), a warm fluid is typically less dense than a cold fluid. Thus, in a system of warm and cold fluid, the warm fluid will typically rise to the top due to gravity. While it does make sense to say that heat rises, it is the fluid that actually does the rising. The fluid carries energy with it, in the form of heat, and thus one could say that the heat rose. The common sense part comes from the observation that warm air rises, and cold air falls, and then making the (false) leap to say that heat rises. Heat will never rise, however, without a carrier of energy, as heat, just like energy, is not a physical thing.

In the case of your tank of fluid, assuming the fluid is all one temperature, it has uniform density throughout. Thus, there will be no net motion of any one section of fluid, meaning no heat movement.

You're right, but not for quite the right reasons, alas! :-)
The fluid at the bottom of the tank is at a higher density and pressure as it is supporting the weight of the fluid above it. What's actually happening is that all the molecules are continuously interacting and we are averaging their behaviour over time or distance. If we consider some volume of space within the tank then particles will be entering and leaving that part of the universe, knocking their neighbours etc. There's an equation ( called the Navier-Stokes I think ) which exactly represents this. It is however almost impossible to solve rigourously, largely because it contains both integrals and derivatives! But heat, broadly, can move about by three mechanisms - radiation ( emit newly created particles ), conduction ( transfer energy to adjacent particles ), or convection ( bulk movement of the existing 'hot' particles ). These distinctions are OK macroscopically but are somewhat blurred microscopically with a full quantum mechanical treatment. The main thing is to not omit or double count in tallying the energy budget.
In any case, if the tank is thermally isolated and began at a uniform temperature throughout then it will remain that way. 'Heat rising' is a consequence of some breach of these conditions, energy or mass input from outside our system boundary, and doesn't thus apply to this tank scenario.
Heat is definitely physical, it causes burns for instance. It is however a 'hidden' variable meaning that to calculate it you have to take some other physical measurement and derive a number to quantify it - so it's not a directly measurable quantity in the usual simple sense.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

ADDMP
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[quote In any case, if the

Message 25076 in response to message 25075

[quote
In any case, if the tank is thermally isolated and began at a uniform temperature throughout then it will remain that way.

Mike Hewson
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RE: will be an equilibrium,

Message 25077 in response to message 25076

Quote:
will be an equilibrium, but there will be a temperature difference


equilibrium is defined as when there isn't a temperature difference after all the interacting bodies in an isolated system are left for a long enough time. You're describing what would be perhaps termed a quasi-steady state. Each mechanism you can think of has it's characteristic 'relaxation time' which is a measure of the speed that the system's temperature approaches uniformity if left to itself.

Quote:
If not, where is the flaw in the argument?


Other mechanisms, radiation etc... will eventually bring uniform temperature. For instance photons will transfer energy across the intervening cavity, or lattice vibrations will transmit energy around the walls of the tank.
Cheers, Mike. :-)

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

MarkF
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Mike is correct the

Mike is correct the temperature will be constant through out the environment. The relationship between the temperature and the mean velocity of the gas atoms will vary according to mass times mean velocity squared divided by two plus mass times gravitational acceleration time height equals some constant.
m*v*v/2+m*g*h=e

ADDMP
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RE: RE: will be an

Message 25079 in response to message 25077

Quote:
Quote:
will be an equilibrium, but there will be a temperature difference

equilibrium is defined as when there isn't a temperature difference after all the interacting bodies in an isolated system are left for a long enough time. You're describing what would be perhaps termed a quasi-steady state.

OK, good, let's call it a steady-state.

Quote:
If not, where is the flaw in the argument?

Other mechanisms, radiation etc... will eventually bring uniform temperature. For instance photons will transfer energy across the intervening cavity, or lattice vibrations will transmit energy around the walls of the tank.
Cheers, Mike. :-)

Certainly you can introduce other competing mechanisms. But I don't yet see how other mechanisms will STOP the mechanism I described.

If there are several competing mechanisms, and if those other mechanisms can be shown not to have any gravity effect, then will we not arrive at a steady state in which all the mechanisms come into a balance? Heat will be transferred downward by bouncing helium atoms & the temperature at the bottom will rise until the temperature difference is enough that all that heat can be transmitted upward by sidewall conduction or whatever. So far, that will not lead to a ZERO temperature difference. Only a REDUCED steady-state difference.

BTW, with regard to photons, I don't think the gravity effect on photons is PRECISELY ZERO. A photon falling downward into a gravitational well will gain a little energy. It will not gain in speed, but it will gain in energy by a very slight blue-shift. It will increase in frequency. Ie, it will become "hotter". Is that not right?

And what, exactly, do we know about the effect of gravity on the phonons that cause heat conduction in solids? Do we know that to be precisely zero?

There must be a flaw, but I don't think you have really nailed it yet.

But I thank you for taking a shot.

ADDMP

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RE: Mike is correct the

Message 25080 in response to message 25078

Quote:
Mike is correct the temperature will be constant through out the environment. The relationship between the temperature and the mean velocity of the gas atoms will vary according to mass times mean velocity squared divided by two plus mass times gravitational acceleration time height equals some constant.
m*v*v/2+m*g*h=e

Thanks. But I don't quite get you here. Isn't temperature associated with random kinetic energy? If e is constant, then does your equation say that
kinetic energy,

KE = m*v*v/2 = e - m*g*h

Does that mean as the height, h, increases, the KE decreases?

If the KE decreases as we go upward, does the temperature also decrease, or what?

ADDMP

MarkF
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ADDMP: No as Mike said the

ADDMP:
No as Mike said the temperature remains constant. That is an irreducible fact of thermal equilibrium.
Yes mean KE decreases with height. Remember the classic parabolic path of a cannon ball.
The same argument would apply to the walls of the container except that the KE term can not be as simply expressed.
The density distribution as a function of Total Knergy is proportional to Exp[-TE/k/T].
k is Boltzmann's constant and T is the temperature Kelvin.
In stated environment every movement upward/downward looses/gains energy to/from the gravity field. Side to side motions are not affected directly but reduce/increased by collisions.

ps where I said mean velocity I should have said root mean square of the velocity.

Mike Hewson
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RE: Isn't temperature

Message 25082 in response to message 25080

Quote:
Isn't temperature associated with random kinetic energy?


Not quite. The 'Zeroeth' law of thermodynamics defines the idea that if two bodies are isolated and in contact such that energy can be exchanged, then after a while the measurable properties of both cease to change. When that happens they are deemed to be at the same temperature. You can use this idea to create a thermoscope, which is a device that changes it's measurable properties in response to said temperatures. When calibrated in a reproducible fashion, to some agreed scale, then you have a practical thermometer. In everyday life that is generally associated with the average kinetic energy of massive particles, but one can speak of the temperature of a group of photons for instance.
As regards competing mechanisms, there is the 'equipartition' theorem which prescribes how energy is distributed between different modes of energy. The helium atom for instance has a structure to it's electronic energy levels, transitions between which can be stimulated by collisions, photon capture/radiation etc... these are inelastic processes which reduce the kinetic energy of the atom as a whole. The interior of the tank contains a 'bath' of 'thermal' photons which will interact with the helium atom. ( Some of those laser cooling traps for the Bose-Einstein condensate experiments use these sorts of mechanisms to cool a collection of atoms to near zero! ).
The gravity isn't really relevant to any of the above. When equilibrium is achieved then gravity will cause a density gradient, but not a temperature difference. In other words the helium atom will spend most of it's time near the bottom of the tank. If the whole caboodle was in free fall then the atom will be equally likely to be anywhere in the tank. Either way, it will have an average kinetic energy the same as the temperature of it's surroundings which is the matter of the tank, thermal photons etc...
Equilibrium implies that 'measurable properties are unchanging', but doesn't necessarily mean 'measurable properties everywhere the same'.
Cheers, Mike. :-)

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

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