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I have learned in thermodynamics that it is impossible for a cyclic system that is receiving energy as heat from one and only one thermal reservoir to produce a net positive amount of work to the surroundings.

I understand the other version of this statement that it is impossible for a cyclic system to convert all the energy input to work (100% efficiency) easily, but I don't understand how a cyclic system can produce a net negative amount of work (from surrounding to the system)

The convention I use is work(system->surrounding)>0.

Please explain with a simple practical example how a cycling system which only communicates with one thermal reservoir can be subject to work done on it, and not able to to do work on the surrounding.

Thanks in advance for your answers

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  • $\begingroup$ Imagine an object on the floor surrounded by air at some temperature $T_0$. You move the object (i.e. you supply work to counter friction) and let the object stop. The object will have raised their temperature to a temperature $T>T_0$. The object will transfer energy to the environement until it reaches thermal equilibrium $T=T_0$. Does that answer your question or I misunderstood? $\endgroup$ – MarcoCiafa Jan 31 at 1:52
  • $\begingroup$ Thanks, but that was not my concern. $\endgroup$ – Michael Tamajong Feb 1 at 7:24
  • $\begingroup$ See update to my answer with the example you are looking for. $\endgroup$ – Bob D Feb 9 at 15:03
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I have learned in thermodynamics that it is impossible for a cyclic system that is receiving energy as heat from one and only one thermal reservoir to produce a net positive amount of work to the surroundings.

That is correct. The applicable statement of the second law is the following Kelvin-Planck statement:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir

I understand the other version of this statement that it is impossible for a cyclic system to convert all the energy input to work (100% efficiency) easily, but I don't understand how a cyclic system can produce a net negative amount of work (from surrounding to the system)

Not sure what you mean by the "other version", but net negative work (i.e., net work input instead of net work output) is done when you reverse the heat engine cycle making it a refrigeration or heat pump cycle. For the refrigeration and heat pump cycle the following Clausius's statement of the second law applies:

No refrigeration or heat pump cycle can operate without a net work input

The refrigeration or heat pump cycle moves heat from a low temperature reservoir and transfers it to a high temperature reservoir. Since, as I'm sure you are aware, heat does not flow naturally from cold to hot, in order to do this work must be done on the working fluid. Work done on a system is negative.

For a refrigeration or heat pump cycle we don't talk about efficiency. We talk about the Coefficient of Performance, or COP. This equals the desired heat transfer divided by the work input required, or

For a refrigerator:

$$COP_{R}=\frac {Q_L}{W}$$

For a heat pump:

$$COP_{HP}=\frac {Q_H}{W}$$

Where

$Q_L$ = the heat transfer from the low temperature environment

$Q_H$ = the heat transfer to the high temperature environment

$W$ = the work input (negative work) required to cause the transfer.

Please explain with a simple practical example how a cycling system which only communicates with one thermal reservoir can be subject to work done on it, and not able to to do work on the surrounding.

An example is an ideal gas that undergoes a free adiabatic expansion followed by a reversible isothermal compression.

We have an insulated rigid chamber that is divided into two parts, one containing an ideal gas and the other a vacuum. An opening is created in the barrier between the two parts allowing the gas in one part to spontaneously and rapidly expand into the evacuated part. This is an irreversible process (You would not expect the gas to spontaneously return to its original part leaving a vacuum behind).

Since the chamber walls are rigid, there is no boundary work done ($W=0$) and since it is insulated there is no heat transfer ($Q=0$). From the first law, closed system, $\Delta U=Q-W=0$. If the gas is an ideal gas, then there would also be no temperature change since internal energy of an ideal gas is a function of temperature only.

Now in order to return the gas to its original state we remove the insulation and insert a frictionless piston into the chamber. We then perform a reversible isothermal (constant temperature) compression returning the gas to its original volume and pressure. Since the process is isothermal $\Delta T=0$, $\Delta U=0$ and $Q=W$ where $Q$ is the heat rejected to the surroundings and $W$ is the compression work done on the gas.

The end result is we have a cycle in which work is done exchanging heat with a single reservoir. But the key is that the work done is negative work. The surroundings does work on the gas and not the other way around. This does not violate the Kelvin-Planck statement of the second law because it applies to a heat engine and a heat engine is one that takes in heat from the surroundings and produces net work on the surroundings (positive work).

Hope this helps.

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  • $\begingroup$ I see you thought I was talking about a refrigeration system when I wrote that net work is negative. But thats not what I meant. It's true that the net work is negative for a refrigeration system, but in the statement of the law, it is written that no cyclic system operating with a single reservoir can deliver a net work to the surrounding. The author of the book (Fundamentals of thermodynamics, Moran, Shapiro pg 246) emphasizes that it is possible for the system to communicate with one reservoir and receive a net amount of work done on it, that is negative net work $\endgroup$ – Michael Tamajong Feb 1 at 7:36
  • $\begingroup$ According to my textbook Fundamentals of Engineering Thermodynamics, Moran, Shapiro, pg 246): "According to the Kelvin–Planck statement, a system undergoing a cycle while communicating thermally with a single reservoir cannot deliver a net amount of work to its surroundings: The net work of the cycle cannot be positive. However, the Kelvin–Planck statement does not rule out the possibility that there is a net work transfer of energy to the system during the cycle or that the net work is zero." This means he is talking about any cyclic system: power or refrigeration. $\endgroup$ – Michael Tamajong Feb 1 at 7:39
  • $\begingroup$ @MichaelTamajong I can see how the net work could be zero when exchanging heat with a single reservoir. All you have to do is reverse a single process. For example, perform a reversible isothermal expansion and then reverse it, i.e, a reversible isothermal compression, both with the same reservoir. Net work is zero. But I don't see how you can get net negative work in a cycle with a single reservoir. Does Moran provide an example? $\endgroup$ – Bob D Feb 1 at 13:26
  • $\begingroup$ That is exactly what I am looking for. Thanks for this example, however. $\endgroup$ – Michael Tamajong Feb 2 at 14:08
  • $\begingroup$ @MichaelTamajong Is there more context in the book related to the statement "However, the Kelvin–Planck statement does not rule out the possibility that there is a net work transfer of energy to the system during the cycle"? Seems strange to make such a statement without providing something to back it up. $\endgroup$ – Bob D Feb 2 at 14:17

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