# Why does the second law of thermodynamics prevent 100% efficiency?

So far in my thermodynamics lecture course, my understanding of the laws of thermodynamics is that the first law is about the conservation of energy, the second law says entropy must always increase or stay the same which apparently results in the fact you can never achieve 100% efficiency of heat engines, unless at $$T = 0\,\mathrm K$$, and the last law says that you can't get to $$T= 0\,\mathrm K$$.

I have never explicitly seen why the fact that entropy must always increase or stay the same results in the prevention of achieving 100% efficiency. The only proof I have is showing the Carnot cycle is the most efficient and that is only 100% efficient if the cold reservoir is at absolute zero, which it can not be at.

Is there any way to work from the statement: $$\Delta S \geq 0$$ (for any process in a closed system), to some result which says you can not achieve 100% efficiency?

• My wife and I used to paraphrase the 3 laws when we took thermo in college: You can't win; You can't break even; You can't get out of the game! – CrossRoads Jan 3 at 18:32

The fact that a heat engine cannot be 100% efficient is a consequence of the Kelvin-Plank statement of the second law, which can be summarized as

Kelvin-Plank Statement of Second Law "No heat engine can operate in a cycle while transferring heat with a single heat reservoir" (my emphasis on cycle)

COROLLARY to Kelvin-PLank: No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs.

You are probably aware that in the Carnot reversible isothermal expansion the heat provided by the high temperature reservoir equals the work done in the expansion. That process converts heat to work at 100% efficiency. But a process is not a cycle. To operate in a cycle the system has to return to its original state. After the reversible isothermal expansion there is no way to return to the original state and do net work without rejecting some heat to a lower temperature reservoir.

You could, for example, immediately follow the reversible isothermal expansion with a reversible isothermal compression and return to the original state. But the negative compression work would equal the positive expansion work for no net work done by the "cycle".

In order to produce net work in the Carnot cycle it is necessary to follow the isothermal expansion with an reversible adiabatic expansion. That reduces the temperature of the system to that of the lower temperature reservoir. Then a reversible isothermal compression then followed by a reversible adiabatic compression returns the system and surroundings to their original states and net work is done. But that work equals the heat into the system minus heat rejected, for an efficiency of less than 100%.

All heat engine cycles must reject some heat for an efficiency of less than 100%. The Carnot cycle does this in the most efficient manner.

I understand this, I guess what I am having trouble seeing is how the statement Δ𝑆≥0 is equivalent to the kelvin plank statement.

Actually we can show that the only way we can complete a cycle and have $$\Delta S_{system}=0$$ is to reject heat to a low temperature reservoir. By definition, a thermodynamic cycle is one where all the system properties (entropy, internal energy, pressure, temperature, etc.) are returned to their original state.

So let us transfer heat from the surroundings to a system by means of a reversible isothermal expansion of an ideal gas. Since for an ideal gas $$\Delta U$$ only depends on temperature, from the first law $$\Delta U=0$$ and $$W=Q$$, and theoretically we have the complete conversion of heat to work at 100% efficiency for the process. However, during the isothermal expansion the entropy of the system has increased by $$\Delta S_{sys}=+\frac{Q_H}{T_H}$$ where $$Q_H$$ is the heat transfer from the high temperature reservoir $$T_H$$.

In order to complete the cycle we must return the entropy of the system to its original state, meaning we must get rid of entropy of the amount $$\frac{Q_H}{T_H}$$. Now, here is the key point. The only way to transfer entropy is by heat transfer. That means to return the system to its original state we must transfer heat $$Q_L$$ to a low temperature reservoir. When we do this, the net work done in the cycle is

$$W_{net}=Q_{H}-Q_{L}$$

The efficiency $$ζ$$ of any cycle is the net work done divided by the gross heat added, or

$$ζ=\frac{W_{net}}{Q_H}=\frac{Q_{H}-Q_{L}}{Q_H}$$

Therefore, for any cycle,

$$ζ<1$$

The above applies to any cycle. It is whether or not a cycle is reversible that determines the maximum efficiency of the cycle. The maximum efficiency occurs when the cycle is reversible. For a reversible cycle both $$\Delta S_{sys}$$ and $$\Delta S_{sur}$$ are zero for $$\Delta S_{tot}=0$$. For an irreversible cycle there is additional entropy generated in the system which must be transferred to the surroundings as additional heat resulting in $$\Delta S_{sur}>0$$, and reducing the amount of work that can be done with the same heat input.

Finally, as stated in the Corollary to Kelvin-Plank, "No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs", which means the Carnot cycle is the most efficient of all reversible cycles.

I was looking for a more explicit mathematical result which shows you can not conserve energy and follow Δ𝑆≥0 at the same time.

You can conserve energy and have $$\Delta S_{tot}$$ ≥0 at the same time. It happens when the cycle is not reversible.

The First Law (energy conservation) and the Second law are completely independent laws. A cycle in which a process (or processes) is irreversible complies with the first law but generates entropy that must be transferred to the surroundings as additional (to the reversible cycle) heat. That in turn means less of the heat taken in does work in an irreversible cycle than the same cycle carried out reversibly. Since the entropy (and all other properties) of the system is unchanged for a complete cycle, the heat transferred to the surroundings results in $$\Delta S>0$$ for the surroundings.

Hope this helps.

• Just to clarify, it is not the second law which places the limit on how efficient a heat engine can be? So if we were to remove the second law completely it would have no effect on the upper limit of efficiency? I am having trouble understanding this since the argument is derived from Kelvin plank statement, which as you mention in your answer is a statement of the second law. – Vishal Jain Jan 4 at 8:53
• @Vishal Jain the second law does place an upper limit. First, Kelvin Planck says no cycle can be 100% efficient. Secondly the second law tells us maximum efficiency occurs when a cycle is reversible, I.e when $\Delta S_{sys}+\Delta S_{sure}=0$. And thirdly the Corollary to Kelvin Plank says the maximum efficiency of any reversible cycle operating between two reservoirs is the Carnot efficiency. – Bob D Jan 4 at 11:24
• @Vishal Jain I’m going to revise my answer to specifically show the connection between Kelvin Plank and that $\Delta S$ of the system has to be zero for a cycle. Please stand by – Bob D Jan 4 at 11:47
• @VishalJain I have revised my answer to hopefully make the connection between Kelvin Plank and entropy less ambiguous. The revision is to the response to the second to last highlight. Hope it helps. – Bob D Jan 4 at 12:40

If you can convert all of the heat to work, you're reducing entropy by definition ($$\Delta S = \frac{Q}{T}$$ , If $$Q<0$$ then $$\Delta S < 0$$).

If you allow yourself to let some heat flow into somewhere cold (heating something up instead of using all of the heat to work) you raise the entropy in the cold substance enough to let you not defy the second law, and the rest can go to useful work.

• "If you can convert all of the heat to work, you're reducing entropy by definition ($\Delta S = \frac{Q}{T}$ , If $Q<0$ then $\Delta S < 0$)." This implies something is wrong with converting all the heat to work. That doesn't violate the second law because it only describes a process (reversible isothermal expansion) and not a cycle. Only for a cycle it would violate the Kelvin Plank statement of the second law. – Bob D Jan 3 at 18:40
• Even if the process converted all the heat into work, like an isothermal expansion, the second law can still be true if the change in entropy of the surroundings exceeded that of the decrease in entropy of the working fluid. How can you show that something is being violated by converting all the heat into work? As @Bob D mentioned, it only seems to be a problem in a cycle, a single process can convert all the heat into work. Can you please edit your answer to describe why specifically in a cycle the second law breaks down if you convert all the heat into work. – Vishal Jain Jan 4 at 8:57
• @Ofek Gillon I should add that $\Delta S<0$ is only the change in entropy of the surroundings. During the reversible isothermal expansion $\Delta S>0$ Of the system by the same amount so that $\Delta S_{total}=0$ in compliance with the second law. The violation of the Kelvin Plank statement concerns 100% efficiency when operating in a cycle. – Bob D Jan 4 at 11:01
• @BobD , yup. Of course – Ofek Gillon Jan 4 at 11:14

The second law of thermodynamics proposed by Clausius, Kelvin, Carnot ..etc in its original form as T dS> dQ for irreversible process and Tds =dQ only for reversible thermodynamics process. The thermal efficiency of any cycle is defined as the work done or performed by the working gas (WD) divided by the inputs of heat Q. Applying the second law on a T-S or even a P-V chart. It is easy to show that reversible cycle is the most efficient cycle between T2 and T1. For Carnot cycle the heat input Q=T(hot ).(S2-S1) and, WD=T(hot or T2) .(S2-S1)-T(cold or T1) .(S2-S1)

The result is The thermal efficiency of Carnot cycle is :1- T1/T2

T1 is the temperature of the " cold " gas ejected by the exhaust tunnel obligatory hotter than the air sink nearly 300 K and therefore ejecting energy to the atmosphere . It is absolutely impossible for the T1 temperature to reach 0 K to obtain 100% thermal efficiency, moreover, the exhaust gas temperature must be some how greater than 300 K in order to sink in the open air of 300 K. The two main objectives of engineers and scientists involved in the automotive and other internal combustion engines (ICE) in increasing thermal efficiency are: i- Increase the theoretical or hypothetical efficiency of Carnot cycle by increasing the hot operating temperature T2 to maximum which is limited by the quality of the heat-resistant steel of the ICE and reducing the cold exhaust temperature T1 which is also limited to the exhaust tunnel efficiency. . ii- Approaching the Carnot's efficiency as closely as possible which is achieved by increasing the pressure ratio and the quality of the high-octane fuel. It means also a time to improve the quality of heat-resisting steels and improve the design of both the combustion chamber and the exhaust tunnel. As far as I know, this thermal efficiency has been somewhat improved to about 0.6-.7 but it must be multiplied by mechanical efficiency to get the overall efficiency. The mechanical efficiency itself must also be less than 1 due to mechanical losses in friction and unavoidable energy loss in forced cooling (water, air, or both) so as not to burn or distort the ICE steel combustion chamber. I assume the best modern efficiency that can be achieved from ICE is something near 0.5

In any irreversible process, heat is lost. in almost every case by the friction of whatever kind (resistance in a circuit, friction between gasses, friction between plates moving on top of each other, magnetic and electrical friction, or friction in waves e.g.).

This is wasted energy, so you can never be able to get 100% efficiency. Unless the process is a reversal one, but they're not. In the Carnot-cycle, the maximum is about 68%. Try looking up why this is the case!

But can't you use this heat again for doing work? Yes, but then you have to catch it first and store it. **If ** you succeed in doing this then the whole story repeats again... etc...etc...Until you come close to a peperpetuum mobile.