# The Carnot engine and entropy?

The Clausius statement of the second law of thermodynamics is as given below:

"Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."

Why is this statement true and exactly what does it have to do with entropy? In terms of the change in entropy of the system, why is it true that no heat engine can operate at 100% efficiency?

• en.wikipedia.org/wiki/Second_law_of_thermodynamics – user126422 Sep 10 '16 at 4:14
• I read the article but didn't understand, which is why I posted the question. – user106570 Sep 10 '16 at 4:21

# Second Law of Thermodynamics:

## Postulate of Kelvin:

A transformation whose only final result is to transform into work, heat extracted from a source which is at the same temperature throughout the process is impossible.

## Postulate of Clausius:

A transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible.

These statements are both equivalent to each other and states what we call the Second Law of Thermodynamics.

The Carnot efficiency is given by

$$\eta = 1-\frac{Q_1}{Q_2}$$

where

\begin{align}Q_2 &= \textrm{Thermal energy absorbed by the system from a source }~ t_2,\\ -Q_1 &= \textrm{Thermal energy absorbed by the system from a source }~ t_1,\\ t_2 &\gt t_1\;.\end{align}

For any other non-reversible engine ($'$) operating between the same temperatures $t_1$ and $t_2$, the efficiency of the same can never exceed the efficiency of a reversible cyclic (Carnot) engine:

$$1-\frac{Q_1}{Q_2}\geq 1-\frac{Q'_1}{Q'_2}$$

Since, $Q_2/Q_1$ depends only on the temperatures of the source, the ratio can be expressed as

$$\frac{Q_2}{Q_1}= f(t_1,t_2)\;.$$

Suppose, a reversible engine operates between $t_0$ and $t_1;$ then $$\frac{Q_1}{Q_0}= f(t_0,t_1)\tag{I.a}$$

Let another reversible cyclic engine operates between $t_0$ and $t_2;$ we have

$$\frac{Q_2}{Q_0}= f(t_0,t_2)\tag {I.b}$$

Dividing $\rm (I.b)$ by $\rm (I.a),$ we get $$\frac{Q_2}{Q_1}= f(t_1, t_2) = \frac{f(t_0,t_2)}{f(t_0,t_1)}\;.\tag I$$

$t_0$ is arbitrary and can be taken constant; therefore, $f(t_0,t)$ only depends on $t$. So,

$$f(t_0,t) \equiv \mathrm K'~\theta(t)\;.$$

So, $$\frac{Q_2}{Q_1}= f(t_1, t_2) = \frac{f(t_0,t_2)}{f(t_0,t_1)} = \frac{\theta(t_2)}{\theta(t_1)}\;.$$

Since, $\theta$ is not unique, we can freely choose its unit: the scale can be chosen based on the difference of boiling point and freezing point of water at one atmosphere pressure.

This new scale $\theta$ can be showed to coincide with the absolute temperature scale $T\;.$

So, we have $$\frac{Q_2}{Q_1}= \frac{T_2}{T_1}\tag{I.i}$$

For any system working between the sources $T_1, T_2, T_3,\ldots, T_n$ with the interacted thermal energy as $Q_1, Q_2, Q_3,\ldots,Q_n$, the following is always true during a cyclic transformation:

$$\sum_{i=1}^n \frac{Q_i}{T_i}\leq 0\tag{II}$$

In case of continuum of sources, the summation becomes

$$\oint \frac{đQ_\textrm{system}}{T_\textrm{source}}\leq 0\tag {II.a}$$

This is the celebrated Clauisus Inequality which is a re-statement of the Second Law of Thermodynamics.

Note, $T$ in the denominator is not that of system; it's of the source - the reservoir.

However, $$T_\textrm{source} = T_\textrm{system} ~\iff ~\textrm{reversible cycle}$$ and $$\oint \frac{đQ_\textrm{system}}{T_\textrm{system}} = 0\tag{II.a.i}$$

# Entropy and the Second Law:

Let $\rm A$ and $\rm B$ be two equilibrium states of a system.

Consider two reversible continuous curves connecting $\rm A$ and $\rm B$ viz. $\mathtt I$ and $\mathtt I';$ together they constitute a reversible cycle. So, applying $\rm(II.a.i)$ we have

\begin{align} \oint_{\mathrm A \mathtt I \mathrm B \mathtt{I'}\mathrm A } \frac{đQ_\textrm{sys}}{T_\textrm{sys}} & = 0 \\ \implies \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt I} + \left (\int_{\mathrm B}^{\mathrm A} \frac{đQ}{T}\right)_{\mathtt {I'}} & = 0\\ \implies \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt I}- \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt {I'}} & = 0,\end{align} which implies $$\left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt I} = \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt {I'}}\;.\tag{III}$$

Now, this make us conclude that the integral $S(\mathrm A) = \left(\displaystyle\int_{\mathrm O}^{\mathrm A} \frac{đQ}{T}\right)$ takes the same value for two equilibrium states and doesn't depend on which reversible path connect the states ($\rm O$ is the standard state). And this is entropy.

Now, in order to show the association of the Second law with entropy, we would utilise $\rm(II.a)\;.$

Consider now $\mathtt{ I''}$ as a non-reversible transformation replacing the reversible $\mathtt I$ while the others remaining the same as above.

Now, applying $\rm(II.a),$ we have,

\begin{align} \oint_{\mathrm A \mathtt{ I''} \mathrm B \mathtt{I'}\mathrm A } \frac{đQ}{T} & \leq 0 \\ \implies \left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt{I''}} + \left (\int_{\mathrm B}^{\mathrm A} \frac{đQ}{T}\right)_{\mathtt {I'}} & \leq 0\\ \implies\left (\int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\right)_{\mathtt{I''}} -[S(\mathrm B) -S(\mathrm A)] &\leq 0, \end{align}

Therefore, $$S(\mathrm B) -S(\mathrm A)\geq \int_{\mathrm A}^{\mathrm B} \frac{đQ}{T}\;.\tag{IV}$$

For an isolated system, $đQ = 0,$ hence, $$S(\mathrm B) -S(\mathrm A)\geq 0\tag{IV.a}$$

So, the Second law predicts that for an isolated system, for any transformation, the entropy of the final state must not be less than the initial state.

For a general system (not isolated), the Second Law states:

$$\Delta S_\textrm{universe} = \Delta S_\textrm{system} + \Delta S_\textrm{source/reservoir/surroundings} \geq 0\;.\tag V$$

Why is this statement true and exactly what does it have to do with entropy?

Because perpetual machine doesn't exist.

While the first law forbids the construction of perpetual machine from which emanates energy, it provides no limitation on the transfer of energy into one form or other.

There could be possibility of thermal energy being totally converted to work or vice-versa.

It is worthy to quote Fermi:

There are very definite limitations, however, to the possibility of transforming heat into work. If this were not the case, it would be possible to construct a machine which could, by cooling the surrounding bodies, transform heat, taken from its environment, into work.

Since, the supply of thermal energy contained in the soil, the water, and the atmosphere is practically unlimited, such a machine would, to all practical purposes, be equivalent to a perpetuum mobile, ....

The Second Law forbids that.

As for why it is related to entropy, let me re-iterate the fact that the very definition of entropy change comes from Clausius Inequality, which is explicitly shown above, that in-turn is a re-statement of the Second Law since, its deduction is based on the validity of Kelvin's postulate and thus Clausius' postulate.

Also, the very interpretation of Second Law in the light of entropy viz., the entropy of the universe(system plus sources) never decreases would get contradicted had thermal energy been exchanged from hotter source to cooler source.

In terms of the change in entropy of the system, why is it true that no heat engine can operate at 100% efficiency?

An engine can't use all the thermal energy it received from the hot source for work during a cycle.

Note the word cycle.

If a system has to operate on a cycle, then the entropy increase of the system caused due to the receiving of thermal energy from the hot reservoir must have to be nullified out by an entropy decrease.

The engine does so by giving up thermal energy to the cold reservoir so that there is no change in entropy of the engine during the cycle.

That's the reason why it has to lose heat energy and thus averting the case of having $100~\%;$ and that is the last nail in the coffin of perpetual motion of second kind.

It is worthy to re-state the statement of Second Law in the present context of an engine working in a cycle:

It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and cooling of a heat reservoir.

Note, as asserted at the outset of this post, the Carnot efficiency is the maximum efficiency for a given set of temperatures; all other non-reversible engines have efficiency lesser than the former.

However, even the Carnot engine can't have $100~\%$ efficiency for it would contradict the Second Law and it's impossible for perpetual machine doesn't exist.

## References:

$\bullet$ Thermodynamics by Enrico Fermi.

• This is absolutely perfect! Now I've received one answer that talks about the macroscopic details of entropy and the other relating to statistical mechanics. Thank you! :D – user106570 Sep 10 '16 at 6:26

To be honest Clausius(and alternatively Kelvin) gave their statement of Second law of Thermodynamics from a completely emperical understanding. They were not assumptions or theorems rather facts of nature. Why this law is true is not understood in Thermodynamics but rather easily explainable in Statistical Mechanics. Heat is a form of energy and Entropy there is given as the no. of microstates a system can be in. Now the tendency of a system is to acquire the maximum number of microstates possible. It always tries to randomise itself the most. Now absorbing heat makes the higher energy levels available for the system. Thus the available microstates also increases. So any system naturally tries to aquire heat and increase its entropy. But if heat would naturally flow from colder to hotter body this won't be possible as the colder body would cool down to absolute zero and the available microstate for this 2-body system at equilibrium will be just 1. Thus heat must flow from hotter to colder body. This is the origin of second law.

• Oh, wow, this is a brilliant explanation! Thanks. One more; when Carnot, Clausius and Lord Kelvin proposed these statements, they didn't have a deep understanding of why this is true? – user106570 Sep 10 '16 at 5:53
• No. They took it as a fact of nature. – Ari Sep 10 '16 at 6:10
• One more; won't the heat flowing from the hotter to the colder body decrease the number of microstates available to the hotter body? By this argument, heat shouldn't even be a thing. What's the correct way to think about this? – user106570 Sep 15 '16 at 10:23
• No, when heat flows from hotter to colder body the microstates of the colder one increases more and that for the hotter one decreases by lesser amount(you can show that). There is a net increase and that's what matters. – Ari Sep 15 '16 at 13:47

Here is an oversimplified presentation of Clausius' 1850 argument:

Premise: Heat can never pass from a colder to a warmer body SPONTANEOUSLY.

Conclusion: Heat can never pass from a colder to a warmer body REVERSIBLY.

The premise is true but the conclusion does not follow from it (the argument is invalid). Then Clausius made another mistake that fatally confused the issue - his formulation of the new result actually coincided with the premise (the formulation of the "Clausius statement of the second law" that you give in your question is essentially the same):

http://www.mdpi.org/lin/clausius/clausius.htm "Ueber die bewegende Kraft der Wärme", 1850, Rudolf Clausius: "Carnot assumed, as has already been mentioned, that the equivalent of the work done by heat is found in the mere transfer of heat from a hotter to a colder body, while the quantity of heat remains undiminished. The latter part of this assumption--namely, that the quantity of heat remains undiminished--contradicts our former principle, and must therefore be rejected... [...] It is this maximum of work which must be compared with the heat transferred. When this is done it appears that there is in fact ground for asserting, with Carnot, that it depends only on the quantity of the heat transferred and on the temperatures t and tau of the two bodies A and B, but not on the nature of the substance by means of which the work is done. [...] If we now suppose that there are two substances of which the one can produce more work than the other by the transfer of a given amount of heat, or, what comes to the same thing, needs to transfer less heat from A to B to produce a given quantity of work, we may use these two substances alternately by producing work with one of them in the above process. At the end of the operations both bodies are in their original condition; further, the work produced will have exactly counterbalanced the work done, and therefore, by our former principle, the quantity of heat can have neither increased nor diminished. THE ONLY CHANGE will occur in the distribution of the heat, since more heat will be transferred from B to A than from A to B, and so on the whole heat will be transferred from B to A. By repeating these two processes alternately it would be possible, WITHOUT ANY EXPENDITURE OF FORCE OR ANY OTHER CHANGE, to transfer as much heat as we please from a cold to a hot body, and this is not in accord with the other relations of heat, since it always shows a tendency to equalize temperature differences and therefore to pass from hotter to colder bodies."

It is easy to see that the two-substances process considered by Clausius presupposes the action of an OPERATOR; this operator constantly and unavoidably undergoes CHANGES, changes that are absent when heat spontaneously "shows a tendency to equalize temperature differences and therefore to pass from hotter to colder bodies". In other words, the trivial fact that, in a spontaneous process, in the absence of an operator, heat always flows from hot to cold by no means implies that heat will flow from hot to cold in a non-spontaneous operator-driven process as the one considered by Clausius.

## protected by Qmechanic♦Sep 10 '16 at 7:59

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