# How does the Carnot Cycle work?

I am learning the Carnot cycle, which consists of four transformations for a fluid.

In the first two transformations the volume of the fluid expands, generating work.

This happens in two steps:

1. First, we approach a hot source and the gas expands and work is generated (okay, temperature increases and fluid wants to expand → the generated work comes 100% from the consumed heat, because the internal energy did not change during the isothermal process)
2. Then we remove the hot source and the gas is supposed to keep expanding (Question A: Not really sure how the gas keeps expanding after we remove the hot source. If no heat source is close, it seems logic to expect that the expansion will just stop. Why would it want to continue? Expansion means work done. Since there is no heat flow, the gas would be losing internal energy. Why would the gas want to lose internal energy to create work?).

In the second two transformations the volume of the fluid decreases.

This happens in two steps:

1. First, we approach a cold source, so the fluid naturally flushes some heat away to the cold source and hence decreases its temperature and hence decreases its volume. (That is quite reasonable).
2. Finally, the fluid keeps compressing, automatically increasing its temperature and pressure, coming back to state number 1 (Question B: How can expect the fluid to keep being compressed automatically? Being this an adiabatic process and an increase in temperature, this means work must be done on the system. Who is doing this work?).

1 and 3 are based on putting a thermal source close to the fluid so nature will do what it does, according to the second law.

But 2 and 4 seem like magic to me, since I don't understand what goes on behind the scenes.

There is a similar question but I still don't get it. Thought it was better to create a new question, rather than commenting on an old one. Also, this way I can expose my complete point of view on the case.

I would also like to understand how this relates to the efficiency of the carnot cycle.

Which comes from something like:

$$\dfrac{GeneratedWork}{GivenEnergy}=\dfrac{HeatIn-HeatOut}{HeatIn}$$

This is understandable but then it goes on like:

$$\dfrac{HeatIn-HeatOut}{HeatIn} = 1 - \dfrac{T_{cold}}{T_{hot}}$$

There are two confusions with this:

Question C.

How is it possible to mix heat and temperature in such an easy way?

This is almost like saying that heat = temperature.

Also, I understood heat is never absolute, but a variation. While temperature is actually an absolute measure.

Question D.

If 4 actually requires external work for the compression, how come is this applied work not reflected in the efficiency? (maybe it is, but I do not see it)

UPDATE:

This video is very helpful: https://youtu.be/d6eJ8mccvu0

• In (adiabatic) step 2, imagine that you have a vertical cylinder with a piston, and a bunch of small weights is sitting on top of the piston. As you gradually remove weights from the piston, the gas expands and the piston rises. This results in work being done to raise the remainder of the weights that are still on the piston. Step 4 is the opposite of this. In both cases, work is done to raise the weights, and the gas expands without heat being added. – Chet Miller Aug 19 at 12:38
• @ChetMiller Thanks, but, "In both cases, work is done to raise the weights" --> Who or what is doing that work? So the cycle is just the idea, but someone must find a way for that work to be done. How then, is that work applied to the system considered in the efficiency? I only see net work, but not the applied work. And... I think it is okay to think it will be more or less work depending on how you create the real machine. This is considering an ideal process, but still, it is not stating where that work comes from. The "As you gradually remove weights from the piston". Who is doing this? – Alvaro Franz Aug 19 at 13:25
• Oops. I mis-stated. In step 4, adding the weights does work on the gas, as the piston get lower. Also, please understand that the work done by the gas in step 2 exactly matches the work done on the gas in step 4. So,, there is no net work done is steps 2 and 4. – Chet Miller Aug 19 at 13:38
• @ChetMiller I cannot see that work being done is the same since we take the same amount of stuff from state 2 to 3, and then from 4 to 1. All of them have different internal energy, so... how can they be the same? Also, I don't understand why at step 2 the fluid would want to expand. If there is no heat transfer, why would the gas expand at all? I expect it to just stay there. What makes it expand? – Alvaro Franz Aug 19 at 13:49
• Taking the weight off the piston lowers the applied pressure from the outside and causes it to expand. The net work is done in steps 1 and 3. Actually, just see @Bob D's answer. – Chet Miller Aug 19 at 15:03

The typical depiction of the Carnot Cycle is with the use of a closed cylinder containing an ideal gas fitted with a piston with a shaft that extends outside the cylinder to interact with the surroundings. In the following I will refer to the fluid as an ideal gas.

First, we approach a hot source (okay, temperature increases and fluid wants to expand)

The first process is reversible isothermal (constant temperature) expansion. The temperature does not increase. For an ideal gas that means the product of the pressure and volume is constant. As the volume expands the pressure resisting the expansion by the surroundings must proportionately decrease. The work done in the expansion exactly equals the heat added and the change in internal energy is zero. By the first law $$\Delta U=Q-W$$. Since $$\Delta U=0$$, we have $$Q=W$$ and $$Q=T_{HOT}\Delta S$$ where $$\Delta S$$ is the change in entropy.

Then we remove the hot source (Question A: Not really sure how the fluid keeps expanding after we remove the hot source. If no heat source is close, it seems logic to expect that the expansion will just stop.).

The second process is a reversible adiabatic (no heat transfer) process. The answer to question A is the gas keeps expanding because the external pressure is gradually intentionally reduced to allow the expansion to continue. Since $$Q=0$$ the work done in the expansion is at the expense of the internal energy, $$\Delta U=- W$$. Note that $$W$$ is considered positive when the gas does work on the surroundings (expands).

First, we approach a cold source, so the fluid naturally flushes some heat away to the cold source and hence decreases it's temperature and hence decreases its volume.

The third process is a reversible isothermal (constant temperature) compression. This is the same as the first process in reverse. In this case surroundings does work on the gas equal to the heat transfer out of the out of the gas to the surroundings.

Finally, the fluid keeps compressing, automatically increasing its temperature and pressure, coming back to state number 1 (Question B: How can expect the fluid to keep being compressed automatically?).

The fourth and final process is a reversible adiabatic compression. It is the same as the second process only in reverse such that work is done on the gas by the surroundings as opposed to the gas doing work on the surroundings. The answer to question B is the gas is not compressed "automatically". An external agent in the surroundings exerts pressure slightly greater than the gas pressure in order to compress the gas.

But 2 and 4 seem like magic to me, since I don't understand what goes on behind the scenes.

Think about the cycle being performed by a piston and cylinder with the piston shaft connected to something outside the cylinder. Basically, what goes on behind the scenes, which is outside the cylinder, is an external agent (the surroundings) is either exerting a pressure on the gas slightly greater than the pressure of the gas in order to compress the gas (process 4) or exerts a pressure slightly less than the pressure of the gas allowing it to expand (process 2). It is important to note that the positive work done by the gas in process 2 exactly equals the negative work done on the gas in process 4, so the net work done for the two processes is zero.

For processes 1 and 3 the pressure exerted by the external agent also accommodates the expansion or compression of the gas. But in these cases it does so in such a way that the temperature of the gas remains constant.

Question C.

How is it possible to mix heat and temperature in such an easy way?

This is almost like saying that heat = temperature.

In equating the efficiency on the left with the relationship of temperatures on the right you are leaving out the important in between steps.

The net work generated is

$$W_{net}=Q_{IN}-Q_{OUT}$$

This equates energy terms. Then for the two isothermal processes, where $$\Delta S$$ is the change in entropy, we have the following:

$$Q_{IN}=T_{HOT}\Delta S$$

$$Q_{OUT}=T_{COLD}\Delta S$$

Efficiency is give by

$$ϒ=\frac{Q_{IN}-Q_{OUT}}{Q_{IN}}$$

Substituting we have

$$ϒ=\frac{T_{HOT}\Delta S-T_{COLD}\Delta S}{T_{HOT}\Delta S}$$

Which simplifies to

$$ϒ=1-\frac{T_{COLD}}{T_{HOT}}$$

Question D.

If 4 actually requires external work for the compression, how come is this applied work not reflected in the efficiency? (maybe it is, but I do not see it)

As indicated above, the adiabatic compression work done on the gas in process 4 exactly equals the adiabatic expansion work done by the gas in process 2. That means they cancel out and are not part of the efficiency calculation.

Hope this helps.

• Bob D. This is just perfect. Really appreciate the fact that a stranger teaches clearer than paid teachers at Universities. Sure this helps, not only me, but other future readers too. Thanks. I have been thinking about "the positive work done by the gas in process 2 exactly equals the negative work done on the gas in process 4" but I do not understand why this is true. We are working with the same temperature changes in both processes, but the volume and pressure involved are different. Being work the total pdV changes, how can I clearly see that both amounts are equal? – Alvaro Franz Aug 19 at 15:01
• @AlvaroFranz Glad it was helpful. Regarding 2 and 4 you are right we are dealing with the same temperature changes. But for an ideal gas, the change in internal energy depends ONLY on the temperature change and is given by, for any process, $\Delta U=C_{v}\Delta T$. Since for the adiabatic process $\Delta U=-W$ that means the magnitude of $W$ will be the same for both processes, but the sign will be different and therefore the two works cancel. Hope this clarifies for you. – Bob D Aug 19 at 15:12
• @Nico Brenner I approved your edit, but I certainly didn't intend to "belittle" the OP. You can see from the OP's comments it wasn't taken that way. – Bob D Aug 19 at 20:19
• @JMac Thanks and I agree with you. Just didn't want to make a big deal about it. Regards, – Bob D Aug 19 at 20:22
• @NicoBrenner No problem. No harm done. – Bob D Aug 19 at 20:29

I think you misunderstood the Carnot cycle.

The Carnot cycle goes as follows: During an isothermal expansion, heat is taken in from a hot reservoir to the working substance. Then, an adiabatic expansion occurs: the substance's temperature falls with no heat exchanged. Then, an isothermal compression sees heat "flowing" from the working substance to a colder reservoir. Finally, an adiabatic compression sees the working substance contract and increase its temperature to complete the cycle.

A potential source of your misunderstanding:

• in a Carnot cycle, heat is exchanged between the substance and the reservoir during isothermal processes (i.e. temperature is constant throughout). If you are not familiar with these, I'd advise you to do some research. Essentially, the answer to "how can it expand if no hot source is nearby" boils down to the fact that, when heat is supplied to a system, it can be used to a variety of effects. The most logical one is increase the particles' speeds, thus increasing the temperature. Another potential use of heat is to do work against the environment. An isothermal expansion is essentially this: all the heat supplied is used to expand the gas, i.e. to do work against the environment. I think this answers question A.

For question C, note that both sides of the equation are dimensionless. You're comparing two dimensionless ratios. You can do that for anything -- for example, $$\text{time}_A/\text{time}_B=\text{mass}_A/\text{mass}_B$$ is dimensionally valid. Clearly, mass and time are (kinda) different concepts -- nonetheless, this equation could come up in the description of some physical system or process.

• Thanks for your time, but I still do not understand it. For the adiabatic expansion, you say "Then, an adiabatic expansion occurs: the substance's temperature falls with no heat exchanged.". I can see that temperature falls if no heat is echanged but simultaneously the volumen increases. But I do not understand what thing makes the volume increase. If you leave a fluid there in an adiabatic box, it won't just say: "hey, i need to expand". It just stays there. – Alvaro Franz Aug 19 at 13:22