What stops entropy from forming if you extract 100% of heat flow and use it for work?

In a carnot engine you convert 100% of heat put inside the system during the isothermal phase into work. This is the basis of maximum efficiency and makes sure that it is reversible. If you don't convert 100% of the heat flowing into the system into work done, you have performed an irreversible cycle and thus $$\Delta S > 0$$.

What exactly is happening here, why is entropy created and why does that make it irreversible? I am specifically talking about 1 isothermal path as an example. Since not all heat went into work, isn't it still (adiabetically) accessibleand 'stored' into its internal temperature (since the heat not used for work went inside the temperature?)

• I think the whole premise is incorrect. It is not true that basis of maximum efficiency of a Carnot cycle is 100% conversion of heat into work in isothermal stages. That 100% conversion holds only for ideal gas, and makes it easy to calculate the process. But Carnot engine can work with any fluid. The only important property is that the whole cycle is reversible, and then efficiency is the standard Carnot efficiency. By the way, the vertical and horizontal lines need not be irreversible. It is common to assume that when you see such a line in the pV diagram, it is reversible. May 12, 2022 at 0:36
• Whether a process is reversible or not depends on details of the process, and cannot be completely determined just from the pV diagram. May 12, 2022 at 0:38
• Is the question assuming that the path labeled with arrows is isothermal? Why??? May 12, 2022 at 0:46

What exactly is happening here, why is entropy created and why does that make it irreversible?

For the irreversible isothermal expansion less work is done because less heat is taken from the hot reservoir. This also results in an increase in entropy of the surroundings for the cycle. However, in the case of an ideal gas, for both the reversible and irreversible expansions 100% of the heat that is taken from the hot reservoir is converted to work due to the first law.

The following is the explanation for the example of an ideal gas. Refer to the figure below. In the figure the irreversible process consists of the following:

At equilibrium state 1, visualize a vertically oriented piston/cylinder with a weight on top. That weight plus atmospheric pressure constitutes the initial external pressure on the system at state 1. It also equals the equilibrium pressure of the gas.

Now suddenly the weight is removed. The external pressure suddenly drops to the final external pressure $$P_2$$, but the internal pressure of the gas is not in equilibrium with the external pressure. Also, since the reduction in external pressure happens so quickly (non quasi-statically) it is an irreversible process and there is not enough time for heat to transfer to the gas, i.,e in effect it happens adiabatically.

Now after the weight is removed, the gas is allowed to expand (irreversibly, since the gas is not in equilibrium with the external pressure) until the gas comes into thermal equilibrium with the thermal reservoir of the surroundings, i.e., in equilibrium at state 2. We call this path isothermal because even though the gas is not in internal equilibrium, the temperature of the gas at the boundary with the surroundings equals the temperature of the thermal reservoir

For simplicity we will assume the three other processes in the cycle (2 adiabatic and isothermal compression) are reversible.

For both the reversible and irreversible expansions both the first and second laws must be satisfied.

First Law:

We start with the first law applied between equilibrium states 1 and 2, that is

$$\Delta U_{12}=Q_{in}-W_{out}$$

For an ideal gas, any process

$$\Delta U_{12}=nc_{v}\Delta T$$

Since both the reversible and irreversible paths begin and end at the same temperature, $$\Delta T=0$$ or $$T_{1}=T_{2}=T_{h}$$ where $$T_h$$ is the temperature of the hot reservoir. Therefor $$\Delta U_{12}=0$$ and for both paths

$$Q_{in}=W_{out}$$

From the figure we see that less work is done for the irreversible path (dark gray area) than the reversible path (light+dark gray areas), or

$$W_{out}(irr)\lt W_{out}(rev)$$

from which follows

$$Q_{in}(irr)\lt Q_{in}(rev)$$

So for the irreversible path less work is done because less heat is added, but still all of the heat added is converted to work per the first law.

The thermal efficiency for any heat engine cycle is the net work done divided by the gross heat input, or

$$\eta=\frac{W_{net}}{Q_{in}}=\frac{Q_{in}-Q_{out}}{Q_{in}}=1-\frac{Q_{out}}{Q_{in}}$$

Since $$Q_{in}$$ for the irreversible cycle is less than $$Q_{in}$$ for the reversible cycle whereas $$Q_{out}$$ is the same for both cycles (the isothermal compression being reversible), it follows that the efficiency of the irreversible cycle is less, or

$$\eta_{irr}\lt \eta_{rev}$$

Second Law:

For both the reversible and irreversible cycle, the change in entropy of the system is zero. That's because entropy, like internal energy, is a state property and the system begins and ends at the same equilibrium state. The change in entropy of the surroundings is the sum of the changes in entropy of the reservoirs.

For the reversible cycle where $$T_c$$ and $$T_h$$ are the temperatures of the cold and hot reservoirs,

$$\Delta S_{surr}(rev)=\Delta S_{cold}+\Delta S_{hot}=0$$

$$\Delta S_{surr}(rev)=\frac{Q_{out}(rev)}{T_{c}}-\frac{Q_{in}(rev)}{T_h}=0$$

For the irreversible cycle, since $$Q_{in}(irr)\lt Q_{in}(rev)$$

$$\Delta S_{surr}(irr)=\frac{Q_{out}(rev)}{T_{c}}-\frac{Q_{in}(irr)}{T_h}\gt 0$$

The increase in the entropy of the surroundings for the irreversible cycle equals the entropy created during the irreversible isothermal expansion.

Hope this helps.

• I don't understand how it is the 'other way around', isn't that kind of redundant? I am probably misunderstanding here. But an isothermal expansion is reversible because you transfer all heat into work so the work is 'captured' and no energy is 'lost'. That is why it stays isothermal. If it didn't do that, the temperature would change wouldn't it? May 9, 2022 at 18:52
• Ahh I think I now understand it. For an isothermal reversible process: $Q_{in}=-W$ For an isothermal irreversible process $Q_{in}=T \Delta S-W$. Since $T$ is constant and not all heat is converted into work, this means $Q_{out} \neq 0 \implies T\Delta S \neq 0 \implies \Delta S \neq 0$. But this means taht an isothermal path can be irreversible if it doesn't reach max $W$ May 9, 2022 at 20:39
• @bananenheld After re reading my answer I realized it did not correctly respond to your original question. So I have completely revised it as shown above. Check it out and see if it helps. May 10, 2022 at 15:26
• @BobD why do you assume the horizontal and vertical lines are irreversible, and why do you assume they are isothermal? May 12, 2022 at 0:43
• @JánLalinský Because the path occurs as follows: At equilibrium state 1, visualize a vertically oriented piston/cylinder with a weight on top. That weight plus atmospheric pressure constitutes the initial external pressure on the system at state 1. It also equals the equilibrium pressure of the gas. May 12, 2022 at 1:06

Entropy is a measure of the number of microstates available to do work. If an engine converts all its heat to work, then the all the available microstates are occupied and entropy decreases to zero. This is a reversible process because if you empty all the occupied work states, you have enough work to recover all the heat. In an inefficient system, any heat not converted to work is an available state and the number of available microstates rises proportionally to the heat lost (entropy increases). Initially, the system was given enough heat to fill all the work states but some were used for "non work" purposes. This is now an irreversible process because if you empty all the filled work states and convert back to heat, the total heat recovered is less than the initial amount.

The most conventional view is that the inflowing "heat" is converted into "work" in an isothermal process. But there are some other views emphasized famously by Bronsted as regards to how "heat" can or cannot do work in an isothermal process, and I admit being more than sympathetic to these views... I do not think I can put it better than this argument from Sorensen[1]:

The basic spatial process consists of moving a piston corresponding to an infinitesimal volume $$\delta V$$ when the pressure on one side is $$p_1$$ and on the other side $$p_2$$. If the piston is moved towards the $$p_2$$ side, the process may be characterised as a transfer of $$\delta V$$ from $$p_2$$ to $$p_1$$. The work done is $$\delta W_{spatial} = (p_1-p_2)\delta V$$

One striking example of the fundamental manner in which Bronstedian energetics differ from traditional thermodynamics may be given at this stage. Consider an ideal gas confined in a cylinder with piston in thermal contact with an entropy reservoir. On the other side of the piston is vacuum, but a weight is placed on the piston. Now, traditional thermodynamics would maintain that the conversion of heat taken from a single reservoir into work is possible in a non-cyclical process. Heat flows into the gas and the piston with the weight goes up. As the internal energy of the gas remains constant in the isothermal expansion, the potential energy taken up by the weight corresponds to the heat absorbed. Bronsted finds this untenable. He maintains that the conversion of thermal energy at a single temperature into other kinds of energy is impossible under all circumstances. The two basic processes which are coupled together in the present example are the transfer of $$\delta V$$ from zero pressure to $$p$$ and the transfer of mass $$M$$ from the gravitational potential $$\phi$$ to $$\phi +d\phi$$. The transfer of entropy from the entropy reservoir to the ideal gas is just a neutral tramport without any energetic significance, because the transfer occurs between regions with the same temperature. The absorption of entropy is only a consequence of the special constitutive relation of the ideal gas (the ideal gas law). Consider that the material enclosed in the cylinder were water at 2 °C. Now, when heat is absorbed, the piston goes down! The "tragicomedy of thermodynamics" - as Truesdell has put it - has been the persisting tendency to entangle in special constitutive relations, so that the purely energetic content of the theory becomes invisible.

In other words isothermally absorbed "heat" cannot do work, this of course is in harmony with Kelvin's principle. All work done by the gas during isothermal expansion is done against a "work content" of the gas that is its free energy or rather "exergy". The absorbed entropy ("heat"=temperature $$\times$$ entropy transported) in a reversible isothermal process does not contribute to the work done by the gas, instead it will just pass through and will be rejected in the "other" isothermal stage at a lower temperature in the $$3^{rd}$$ stage of the reversible Carnot cycle.

[1] Torben Smith Sorensen: Bronstedian Energetics, Classical Thermodynamics and the Exergy. Towards a Rational Thermodynamics. I.; Acta Chemica Scandinavica A 30 (1976) 555-562

• "In other words isothermally absorbed "heat" cannot do work, this of course is in harmony with Kelvin's principle." What Kelvin's principle are you referring to? May 9, 2022 at 12:58
• @Bob_D there is no two-stage thermodynamic cycle whose one leg is isothermal and the other is adiabatic with positive work output. May 9, 2022 at 13:07
• But the Carnot Cycle is not a "two stage" cycle, so I'm not sure what that means May 9, 2022 at 13:22
• I tried Googling "Torben Smith Sorensen: Bronstedian Energetics, Classical Thermodynamics and the Exergy. Towards a Rational Thermodynamics" but got no hits. But if Bronsted says that a body has "TS heat content" I'm already suspect since any thermodynamicist know that bodies don't contain heat. Anyway, I personally found the quoted material confusing. And I'm not sure it will be more helpful to the OP than a more mainstream physics response to the OP's post. May 9, 2022 at 14:09
• I started to read the article but frankly find it too convoluted. I guess its above my ability. Regarding your last sentence, of course the high and low temperatures are not the same- I don't see the point. In order to complete the cycle you need to reject heat and therefore need a lower temperature process to reverse the cycle. The adiabatic processes simply serve to connect the isothermal processes. May 9, 2022 at 17:30

Since not all heat went into work, isn't it still (adiabetically) accesisible and 'stored' into its internal temperature (since the heat not used for work went inside the temperature?)

Consider the case where the isothermal expansion accelerates a (low-weight) piston, which then smashes into a mechanical stop outside the system or even is slightly braked by air drag. In such cases, thermal energy is generated irreversibly (and entropy generated) outside the system, heating the surroundings. Such energy is no longer available to do work for this particular engine, agreed?

• It makes sense that external heat lost is in a sense gone to the engine. But for an isothermal expansion $Q_{in} = -W$ but what if $Q_{in} = Q_{system} -W$ instead of heat lost to the surroundings. May 9, 2022 at 18:08
• What is $Q_\text{system}$? May 9, 2022 at 18:33