Reversible heat exchange with surroundings

1. Apr 10, 2005

Willa

why is it acceptable to assume any heat exchanged with the surroundings of a system is a reversible heat exchange? The only explanation I can find is that it's because the surroundings are essentially unchanged by the heat they absorb...but I dont understand that as an explanation.

I take the example of a hot cup of water in cool surroundings...surely the heat transferred to the surroundings is not reversible...it wont suddenly go back into the cup will it!? so why do they say it's ok to say it is a reversible heat exchange?

2. Apr 10, 2005

Andrew Mason

If a change in a system can be reversed (ie the original state restored) without going outside the system, it is reversible. So, for example, a heat exchange in which the work done during the hot-cold exchange is available (ie it is stored) and is sufficient to enable the cold-hot exchange, is a reversible exchange.

For example, an adiabatic expansion of a gas causes a weight to be lifted. The lifted weight is then used to compress the gas adiabatically back to its original state. So we can say that the expansion is reversible.

AM

3. Apr 10, 2005

Iraides Belandria

Willa : I don´t know if the following opinion may help you : In general, a reversible process is constituted by a sequence of states that deviates infinitesimally from equilibrium, and in each state , it can be turned into the reverse direction leaving the surroundings unmodified. The overall work and heat of this cyclic process is zero.Therefore during a reversible process the system and the surroundings could be restored to their original states without producing any change in the rest of the universe.This demands that the driving forces of the process should be differential in size. In the case of heat exchange , it requires that the temperature difference between the system and the surroundings must be infinitesimal. Otherwise, the process is non reversible. Therefore, it is not acceptable to assume that any heat exchanged with the surroundings is a reversible heat exchange.
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4. Apr 10, 2005

Andrew Mason

This is not quite correct. In a reversible process, the system and its surroundings can have any temperature difference you like. The Carnot engine is the example. The Carnot cycle is one in which the work done by the gas in the expansion cycle is equal to the work required in the compression cycle (ie. to bring the system back to its original state).

AM

5. Apr 11, 2005

Willa

nah sorry i just dont understand how a hot object cooling in cold surroundings can be reversible. You're saying that an object that was once hot but is now cold can become hot once again without any energy being supplied to the system which consists of the object and the surroundings...what gives...i just dont get that?

i get reversible gas expansions because an infintesimal increase in external pressure causes the gas to compress again. But i just dont understand the situation with the surroundings!

6. Apr 11, 2005

Andrew Mason

The ideal Carnot cycle is a reversible cycle. It operates between two different temperatures.

Consider this: if the hot object is a gas and expands adiabatically it gets cold as a result (but since it is adiabatic, the energy has not been lost to the system). If the work done by the gas is stored (say by moving a ratchet with a weight attached) and is then released (by releasing the ratchet, one click at a time) so the gas compresses (adiabatically), one ends up with the gas in its original state (or arbitrarily close to it) with no energy being added to the system.

AM

7. Apr 11, 2005

Iraides Belandria

Andrew Mason: I think you have a wrong interpretation of reversibility in a heat exchange process, which is the case I am explaining. Under this circunstance reversibility requires that the temperature difference between the system and surroundings must be infinitesimal. But, if there is not exchange of heat between the system and surroundings, like in an adiabatic process, it is evident that the system and the surroundings can have any temperature, because there is not interaction between them, but the pressure differences of the system and surrondings and any other gradient should be diifferential in size. Because the Carnot cycle presents two adiabatic stages, it presents, the behavior you suggest. However, in the two isothermal stages of this cycle, where there is heat exchange, the temperature differences of the system and the corresponding surroundings ( hot and cold reservoir) are infinitesimal. Finally, you mention that the expansion work is equal to the compression work in a Carnot Cycle. This is not correct. They are different.
In relation, to Willa question, I still believe that you cannot assume that any heat exchange with the surroundings can be reversible. To attain such condition it is neccesary infinitesimal temperature differences between system and surroundings.
Now, suppose the initial example where a hot cup of water at 100 °F( system) is inmersed in cold place at 32°F( surroundings). What happen if they are in contact for an infinite time?. We can guess that they will reach the same equilibrium temperature. This final temperature is a value between 32 and 100 °F. ¿ can we restore the original temperature of the hot water and the cold place without affecting the system and the surroundings or the universe ?. Evidently, the answer is not, unless the temperature differences are infinitesimal.

8. Apr 12, 2005

Andrew Mason

That is certainly the case during reversible processes in which heat flows between the system and the surroundings. But it is not the case, obviously, in processes where heat does not flow out of or into the system (adiabatic).

So if you combine the following steps:

1. connect a gas at pressure P and temperature T (P is higher than ambient pressure) to a hot reservoir at temp. T

2. let the gas expand to a volume at which further (adiabatic) expansion (unconnected to the hot reservoir) will bring it to the temperature of the cold reservoir

2. save the work done during the isothermal expansion (eg. a weight is lifted)

3. disconnect the hot reservoir

4. let the gas expand adiabatically until it reaches the temperature of the cold reservoir.

5. save the work done during the adiabatic expansion (eg. a weight is lifted further)

6. connect the gas to the cold reservoir

7. apply the saved energy (raised weight) to the expanded gas and compress it isothermally (heat flows out of the gas to the cold reservoir) to the point at which further (adiabatic) compression will bring it up to its original temperature and pressure

8. disconnect the cold reservoir.

9. compress gas further, adiabatically (using stored energy) so the gas will return to its original pressure at the temperature of the hot reservoir.

Now the final result is, of course, that net work has been done (the weight is higher than it was when the process started. And heat has flowed from the hot to the cold reservoir.

But it is reversible:

Use the remaining weight height (stored work) to run a Carnot refrigerator cycle on the hot compressed gas: Adiabatic expansion to temperature of cold reservoir, connect cold reservoir, Isothermal expansion at cold temp, followed by adiabatic compression to temperature of hot reservoir, followed by Isothermal compression connected to hot reservoir.

After running the Carnot cycle backward, all the heat that flowed to the cold reservoir in the heat engine cycle has flowed back from the cold to the hot reservoir using the work generated in the engine cycle.

Quite right. I should have said: the work done in the forward cycle (heat flows from hot to cold) is equal to the work required to reverse the cycle (heat flows from cold to hot).

I agree with you that during the heat exchange process the two bodies exchanging heat must do so at the same temperature (infinitesimal temp differences). But if the system temperature changes (during adiabatic expansion/compression) heat can be made to flow from the hot to cold reservoir reversibly (as I have shown above).

AM

9. Apr 12, 2005

Clausius2

I am sorry Andrew, but this time I agree with Iraes Belandria. I think his explanation is correct, or almost correct. The only possibility of exchanging reversibly heat is by means of an infinitesimal difference of temperature. Any heat flux enhanced by a non infinitesimal difference of temperatures causes an increasing of the universal entropy. The proper mechanism of heat flux is caused by means of thermal boundary layers in the case of fluid flows, inside of which there are sources of entropy relationed with heat dissipation. In fact, the gradient of temperatures is a source term in the transport equation of entropy of fluid flow.

In your example, a carnot engine works between two temperatures well differenced. BUT the gas or fluid inside engine works with a temperature such that the difference between surroundings and the proper fluid temperature is infinitesimally low. I mean, there is no temperature gradient across the "wall" of the engine, and so no net heat flux. We know there is heat exchange, but this heat is not caused by a gradient of temperature, in fact this heat exchange doesn't exist, because if it really exists then the process will be reversible and it is naturally impossible.

It is impossible to exchange heat reversibly, and this statement comes directly of the heat transport mechanism which originates irreversibilities. The unique form of exchanging heat reversibly is by means of an infinitesimally slow process, but for engineers it is useless and by the way impossible.

10. Apr 12, 2005

Andrew Mason

See my last post. I agree with you. I should have said "The Carnot cycle is one in which the work done by the gas in flowing from hot to cold is equal to the work required in making it flow from cold to hot (ie. to bring the system back to its original state)."

I agree that the Carnot cycle this is an ideal and is of little practical use, but it can be achieved (ie. to within arbitrarily close limits). The original question asked whether heat can be made to flow reversibly between two significantly different temperatures and suggested that it can't be. The answer is that it can be, although as you point out, it is of little practical use.

AM

11. Apr 13, 2005

Clausius2

Sorry, I didn't pay too much attention . I only read some lines....Anyway we agree, all of us are happy now and this problem seems solved....

12. Apr 13, 2005

Willa

so can i just establish then, are we saying that the heat given out to the surroundings by, say, a hot object closed system cooling (say a cup of water or something...no evaporation for simplicity), that heat is NOT reversible?

Well that cant be the case because I can do calculations on, say, water freezing to ice...and in that case the equation works out fine by assuming the heat transfer is reversible! I'm so confused, someone clear it up!

by this I mean:

$$\Delta S_{universe} = \Delta S _{system} + \Delta S _{surroundings}$$
$$\Delta S_{universe} = \frac{\delta q_{rev-sys}}T_{sys} + \frac{\delta q_{rev-surr}}T_{surr}$$
$$\Delta S_{universe} = \frac{\delta q_{rev-sys}}T_{sys} + \frac{ -q_{sys}}T_{surr}$$

using that I get good answers...all I have assumed is that $$\delta q_{rev-surr} = -q_{sys}$$

I dont need to assume the temperatures are the same

Last edited: Apr 13, 2005
13. Apr 13, 2005

Iraides Belandria

Dear Willa, If Tsys is different of Tsurr, the heat under the integral sign should be the irreversible heat. Depending on the process , the irreversible heat may be easy, difficult or impossible to estimate for the actual trajectoy of the process. However, if you know that entropy is a function of state , independent of trajectory, then you can calculate the entropy change of the irreversible process by considering an imaginary reversible path by which the process can be carried out. The heat estimated for the hypothetical reversible process does not neccesarily coincide with the irreversible heat of the actual process, because heat is not a function of state. I dont know, if these ideas may help you. I would like to know your numerical example and calculations if you continue to be confused.

14. Apr 14, 2005

Andrew Mason

Willa: Thermodynamics is a difficult and confusing subject! Your determination to understand it is commendable (and I am sorry if some of my answers may have confused you even more).
A hot cup of coffee in a picnic cooler can give up its heat
to a cold tank of water reversibly but it will take a very long time. If you just plunk the cup in the water, the process is not reversible. You have to connect a Carnot engine to the coffee cup and transfer the heat from the cup to the water (slowly). If you do that, the heat flow is reversible (ie. it can be reversed without adding energy to the system).
The heat transfer has to occur at the same temperature in order to have 0 entropy change:

If you transfer the heat from a hot cup to cold surroundings by putting the cup in a cold surroundings, the heat is lost to the cup at a high temperature and gained by the water at a lower temperature. Since the heat lost, Q, is equal to the heat gained by the surroundings, $S_h = Q/T_h < S_c = Q/T_c$ so $\Delta S = Q/T_c - Q/T_h > 0$

If you add ice at 0 deg. C to water at 0 deg. C, there is no temperature difference - the heat lost by the water is gained by the ice at the same temperature so there is no entropy change. This is a good example of reversible heat flow.

AM

15. Apr 14, 2005

Willa

surely this is provided that Q is reversible heat?

Look, I'm not very good at thermodynamics at all...so I'm just after a non-fussy explanation. I have written in my notes that because the surroundings tend to be large, it can be assumed that heat exchanged with the surroundings is reversible from the point of view of the surroundings. The only reason I can think this to be true is that the surroundings are essentially unaffected by a gain or loss in heat: it is an infintesimal change to the surroundings. Such a change must be reversible. Is that at all correct?
The bit which throws me off in my notes is that it says "an infintesimal change in, say, the temperature, will cause the direction of heat flow to be reversed"....i just dont get that - you can change the temperate of the surroundings and the heat flow direction wont change (unless the exchange is to a system at the same temp as the surroundings). But I'm pretty sure we've done calculations where the surroundings arent the same as the system, i.e. water freezing to ice. To do that we used: $$\Delta S = \Delta S _{sys} - \frac{q _{sys}}{T _{surr}}$$ and things work fine.

So can someone explain whether or not my simple explanation is correct or not...and why (that the surroundings are unaffected by the heat exchange)

this is all assuming the surroundings are large btw

16. Apr 14, 2005

Andrew Mason

There is no such thing as reversible heat. There are only reversible paths or heat flows.

Whether heat is exchanged with the surroundings in a reversible (or nearly reversible) manner depends on how it is done. The only way to transfer heat from one reservoir to another (at different temperatures) is to employ a Carnot cycle.

A reversible flow of heat can only take place when the matter absorbing heat is at the same temperature as the matter losing heat. To make a hot cup of water lose heat reversibly to a cold reservoir you have to do it with an intermediate material - a gas. (The reason you can use a gas is because a gas can, by compression/expansion, change its temperature without heat flowing into or out of it). You make the gas the same temperature as the hot reservoir by adiabatic expansion/compression, put it in thermal contact with the hot reservoir and let the gas absorb heat by slow isothermal expansion (ie. the gas expands an infintesimal amount so it cools an infinitesimal amount, so it absorbs an infinitesimal amount of heat and warms back up an infinitesimal amount. Thus it keeps expanding, cooling, absorbing, heating all at essentially the same temperature). You then expand the gas adiabatically (no heat flow) so it is at the same temperature as the COLD reservoir. You then connect it to the cold reservoir and start slow isothermal compression to transfer the heat to the cold reservoir (ie. the gas compresses an infintesimal amount so it warms up an infinitesimal amount and releases an infinitesimal amount of heat to the cold reservoir; the loss of heat makes it cool down an infinitesimal amount; further compression makes it warm back up an infinitesimal amount. Thus it keeps compressing, heating, releasing heat, and cooling all at essentially the same temperature).

Again, reversible heat flow can only occur between two materials in contact with each other at the same temperature. To make heat flow reversibly between two reservoirs at different temperatures, you have to use an intermediate material (that can change its temperature without adding/losing heat) to 'carry' the heat from the hot to the cold reservoir.

AM

17. Apr 15, 2005

Clausius2

Maybe Willa will understand it when he progress in his studies about fluid flow or heat transfer.

Anyway Andrew, you have made a great effort to explain it accurately. :tongue:

18. Apr 15, 2005

Andrew Mason

Or perhaps, 'she'.

I notice that in post 14 there should be a correction which is apparent from my explanation. The following is the corrected sentence:

AM

19. Apr 15, 2005

Willa

but suppose the surroundings are taken to be a large expandable gas. Absorbing some amount of heat will effectively be a near infintesimal change if the surroundings are sufficiently large. So surely the heat can go straight into the surroundings (i.e. air) in a reversible manner like you described.

I'm talking about very idealised situations here - i.e. the surroundings are infinitely bigger than the system in question, and it is a gas. But i think you're just saying that my notes are wrong, and they should say that it is only valid to consider it reversible provided the surroundings are the same temperature as the system.

I'm sorry, you're being too technical for me...thanks for trying but I just don't think I'm getting an answer to my problem. I understand this whole "you need a carnot engine" thing but what i was rather after was a simple "yes it is correct to assume heat transfered to a sufficiently large surrounding is seen as reversible from the point of view of the surroundings" or a "no that is wrong". I really wasnt after an in depth discussion because I really am not expected to go into it in my course that much...it was just a tiny point which my notes mention which I just dont understand, and I think it's being overanalysed here

thanks anyway, if you can't come up with a super simple explanation then it's ok, i'll just have to accept my notes

20. Apr 15, 2005

Andrew Mason

No it can't. If the object losing heat is at a different temperature than the surroundings while the heat is being transferred, there will be a net increase in entropy and therefore the process is irreversible.

A system can have different temperatures at different times. Heat can flow from a hot to a cold reservoir reversibly so long as the heat exchanges occur when the part of the system gaining/losing heat is at the same temperature as the reservoir it is losing/gaining heat to/from.

The answer is "no that is wrong". Direct heat flow from a small hot object to a large cool reservoir is not reversible. Indirect heat flow from a hot object to a large cool reservoir via the Carnot cycle, is reversible.

AM