I've recently started learning about entropy.

One possible definition is that it is the logarithm of number of microstates of particles.

Another possible definition is that
for a reversible (quasi-static + frictionless) process in a closed system (cannot transfer mass but can transfer thermal energy),
$ΔS=\dfrac Q {T}$

Where S is entropy, Q is thermal energy, T is absolute temperature

I have a few questions here:
1. Why must this formula be used in reversible processes and not irreversible processes?
2. Is heat transfer a reversible process? Is an object slowing down due to friction a reversible process?
3. Why does the temperature at which thermal energy is added affect the change in entropy?
4. Other than adding thermal energy increasing the average kinetic energy of the molecules (and distributing it over a wider range of speeds for a higher number of microstates), are there any other ways for added thermal energy to increase the entropy by increasing the number of microstates?

Here's my attempt to answer these questions.
1. Without the process being reversible (quasi-static), there will not be a defined and clear way to calculate the temperatures (state variable), so change in entropy cannot be calculated using the formula.
2. Heat transfer does not seem reversible because it is not quasi-static (reversible). Thermal energy does not seem reversible too, for the same reason.
Questions 3 and 4 were just ideas that I suddenly thought about.

I've been using $ΔS=\dfrac Q {T}$ quite a fair bit without putting much thought into whether the process is reversible or not.


4 Answers 4



The questions you post betray an underlying confusion about some of the terms in the language of thermodynamics. I prefer to start with this and then address each question specifically.

The System, Surroundings, and Boundary

A system is a region that we define. Anything outside the system is the surroundings. The system and surroundings are separated by a boundary. For simplicity, we will use a closed system. A closed system is one that does not have mass flow in or out across its boundary. We will also use a pure system, which is one that has only one chemical substance.

The combination of a system and surroundings is called the universe.

The State of a System, Surroundings, and Boundary

We can prove that the equilibrium state in a closed system of a pure substance of known mass can be defined exactly by any two of the three variables $T, p, V$. This means, we can define the absolute values of any of the thermodynamic parameters such as internal energy $U$, enthalpy $H$, entropy $S$, Helmholtz energy $A$, or Gibbs energy $G$ simply by defining any two of the three variables.

We are generally not interested in knowing the absolute equilibrium state of the surroundings as much as we are in specifying its temperature and pressure. A boundary has no state properties and no parameters because it has no finite dimension to it. It serves only as a separation across which the system and surroundings exchange energy in the form of heat or work (or “other work” such as light energy or electrical energy).

When we ask about the state of something, we are usually about the state of a system. Otherwise, we would be explicit (state of the surroundings or state of the universe).

Thermodynamic Properties

In the paradigm of statistical mechanics, the absolute internal energy of a system is related to the translation, rotation, and vibrational states of the molecules in the system. We generally ignore the electronic and nuclear states for reasons given below. Also in statistical mechanics, absolute entropy is related to the number of possible micro states of the system, stated as $S = k \ln \Omega$.

In the paradigm of the laws of thermodynamics, we are typically not as much interested in the absolute values of thermodynamic state properties of a system as we are interested in changes in state properties during a process with one exception. A change in a state property $Z$ is indicated as $dZ$ for an infinitesimal step or $\Delta Z$ for a difference between a final state minus an initial state. The one exception is the significance of absolute value of entropy $S$. At absolute zero temperature, the third law of thermodynamics defines $S$ as zero. In statistical mechanics, at absolute zero temperature, the state of the system is associated with only one micro state, that of a perfectly ordered single crystal of the material, and $S = k \ln \Omega = k \ln 1 = 0$ by definition as well.

The contributions of the electronic and nuclear states of a system in statistical mechanics are of interest to those who deal with processes that cause changes in the electronic or nuclear states of the molecules in a system. In the paradigm of the laws of thermodynamics for basic science or engineering applications, such processes are not considered. In these fields of study, when we talk about making a change to a system, we assume that the electronic and nuclear states stay the same.

The Relevance of the Laws

The first two laws of thermodynamics are concerned entirely with the universe. For any process, energy is conserved in the universe and entropy increases in the universe (for spontaneous processes). The relevance is that, for any process, we must always consult the changes in both the system and the surroundings to be able to determine whether we have violated a law during that process. To make this consideration, we analyze changes in the system and in the surroundings separately from each other, and then we add the results.

Types of Paths and Processes

A path is the course that is taken to change the parameters of the system or surroundings. A process is the entirety of the path from an initial state to a final state.

We can carry out a process along two different paths, reversible and irreversible. A reversible path is one where the system and the surroundings are in exact thermodynamic equilibrium at all points. In our pure, closed system, the statement that we will follow a reversible path is understood to mean that we will hold the temperature and pressure of the system and the surroundings exactly the same at all points.

Reversible processes that follow perfectly reversible paths throughout are hypothetical. They cannot be achieved in reality. In reality, all paths are irreversible. An irreversible process is one where the system and surroundings are not at perfect thermodynamic equilibrium at all points along the path.

Notice that, we cannot make a statement whether a path or process is reversible or irreversible based solely on what happens only in the system. We also cannot make a statement whether a path or process is reversible or irreversible based solely on what form of energy (heat, work, other work) flows across the boundary or which direction energy flows across the boundary. To determine whether a path or process is or is not reversible, we must always determine what is happening in the universe.

Implications for Thermodynamic State Properties

When we say that we will analyze a reversible process, we mean this.

$$\Delta U_{universe} = \Delta U_{sys} + \Delta U_{surr} = 0$$ $$\Delta S_{universe} = \Delta S_{sys} + \Delta S_{surr} = 0$$

When we say that we will analyze a (spontaneous) irreversible process, we mean this.

$$\Delta U_{universe} = \Delta U_{sys} + \Delta U_{surr} = 0$$ $$\Delta S_{universe} = \Delta S_{sys} + \Delta S_{surr} > 0$$

The change in the universe is always the sum of the changes in the system and the surroundings. How do we find the changes in entropy of the universe? We must always determine the changes in both the system and the surroundings separately! There is absolutely no equation standing alone to determine the entropy change of the universe that does not combine the changes of the system and the surroundings! Since $\Delta Z$ for any state property is path independent, we use a reversible process. Why? Because, even though they do not exist in reality, they are so much easier to envision and analyze. We can determine $\Delta S$ so by taking the final minus the initial state property. Alternatively, we can integrate the property or its state function definition $dS = \ldots$ over a reversible path.

Answers to Questions

1) The only way we will know whether a process is reversible or irreversible is to determine the entropy change of BOTH the system and the surroundings to obtain $\Delta S_{universe}$. Entropy is a state property of the system or the surroundings. Regardless of the actual path taken to complete a process (reversible or irreversible), a change $\Delta S$ in a system or in its surroundings only depends on the difference between the final and initial states. So, it is not that $\Delta S = \int \delta q_{rev}/T$ must be used only in reversible processes (and never in irreversible processes). It is that we always calculate $\Delta S$ for the system or the surroundings separately, we always use reversible processes to do so, and we only know whether the process is irreversible when we add the two values (system + surroundings).

2) In the real world, heat is only transported when we have a temperature difference. In this case, the system and surroundings are not at thermal equilibrium. Therefore, heat transfer in the real world only occurs along an irreversible path. We can presume that heat is transferred across the boundary during a reversible process, but only because reversible processes are presumptions anyway. The answer for friction is addressed using the Kelvin form of the second law. Work creates friction heat, and that heat cannot be converted back to the same amount of work.

3) It is how we define the entropy change during a process. Alternatively, it is the scaling metric by which heat input/output cause the system or the surroundings to increase/decrease the number of available micro states.

4) Molecules can have translational, rotational, and vibrational modes. The heat capacity of N$_2$ is higher than that of Ar due to the extra modes (vibration in this case).


The correct equation describing the change in entropy of a closed system experiencing either a reversible or an irreversible process between two thermodynamic equilibrium states 1 and 2 (initial and final) is $$\Delta S=\int_1^2{\frac{dq}{T_I}}+\sigma\tag{1}$$where dq is the differential amount of heat flowing into the system from its surroundings (through its interface with its surroundings), $T_I$ is the temperature at the interface through which the heat is flowing, and $\sigma$ is the amount of entropy generated within the system as a result of irreversibility. For a reversible process path, the amount of entropy generated within the system is zero, so the entropy change for such a path is $$\Delta S=\int_1^2{\frac{dq_{rev}}{T_{I,rev}}}\tag{2}$$Also, for a reversible process path, the temperature of the system T is the same as that at the interface: $T_{I,rev}=T$. So, the equation for a reversible process becomes:$$\Delta S=\int_1^2{\frac{dq_{rev}}{T}}\tag{3}$$So to get the change in entropy of a system that has experienced an irreversible process, you must first identify an alternate reversible process path between the same two end states, and then calculate the entropy change using Eqns. 3. If you want to find out how much entropy was generated within the system during an irreversible process, you just combine Eqns. 1 and 3 to obtain: $$\sigma=\int_1^2{\frac{dq_{rev}}{T}}-\int_1^2{\frac{dq_{irrev}}{T_{I,irrev}}}$$

A good way of thinking about entropy changes during a process, it is helpful to think of $\int_1^2{\frac{dq_{}}{T_I}}$ as the entropy change as a result of entropy transfer between the surroundings and the system and to think of $\sigma$ as the amount of entropy generated within the system during the process.

  • $\begingroup$ Hi Chet. I have a couple of questions relating to your answer and mine. In my answer we have a transfer of heat $Q$ between two thermal reservoirs, I’ll call H and L (high and low temperature). Question 1: Would the interface temperature be $\frac{T_{H}+T_{L}}{2}$? As long as there is a finite temperature difference between the reservoirs, there will be a finite temperature difference between each reservoir and the interface. The heat transfers are therefore irreversible and entropy is generated equal to the total entropy change of T plus L. $\endgroup$
    – Bob D
    Jul 11, 2019 at 13:29
  • 1
    $\begingroup$ In your discussion of your equation 1 the system would generate entropy σ is due to irreversible heat transfer into the system. Question 2: Isn’t it also true that entropy is generated in the surroundings as well since it, too, transfers heat across a finite temperature difference, albeit out of the higher temperature surroundings? $\endgroup$
    – Bob D
    Jul 11, 2019 at 13:30
  • $\begingroup$ In terms of my answer, which does not designate which reservoir is the “system”, we can determine each entropy generated in L and H by returning each to their original state. This involves transferring $Q$ out of L to an even lower temperature reservoir and transferring heat $Q$ into H from an even higher temperature reservoir. Then we can calculate the entropy generated in each reservoir. Thanks. $\endgroup$
    – Bob D
    Jul 11, 2019 at 13:30
  • $\begingroup$ @BobD Hi Bob. In the real world, the interface between two media (even baths) is going to be somewhere between the mean temperatures of the media. So there is going to be entropy generation in the thermal boundary layers. In the ideal world of two ideal constant temperature reservoirs, we are pretty much forced to imagine a thin layer of highly conductive material in between the two media, with a temperature gradient running from the high temperature to the low temperature. So each of the media experiences only entropy transfer at the boundary, and no entropy generation. All the entropy $\endgroup$ Jul 11, 2019 at 14:21
  • $\begingroup$ generation then occurs in the thin layer of conductive material between the media. So there is no entropy generation in each of the ideal reservoirs, as per the definition of an ideal reservoir. The thin layer of conductive material takes the place of the thermal boundary layers that would exist in real life. For two solids brought into contact, the interface temperature will be between those of the two solids, determined by the density, heat capacity, and thermal conductivity of both the two solids. $\endgroup$ Jul 11, 2019 at 14:25

I can respond to your first 3 questions from the perspective of macro thermodynamics. The fourth questions gets into the realm of statistical (micro) thermodynamics. Unfortunately I am not well enough versed in statistical thermodynamics. Perhaps someone else can fill that gap.

Why must this formula be used in reversible processes and not irreversible processes?

The equation applies for both reversible and irreversible processes because it only gives you the entropy change of the substance that is either receiving or losing heat at constant temperature. What determines the degree to which the process is reversible or irreversible is the temperature difference between the substances that are exchanging heat $Q$.

Heat is defined as energy transfer between two substances due solely to a temperature difference between them. The change in entropy of the higher temperature ($T_H$) substance that is transferring heat to the lower temperature ($T_L$) substance is negative because heat is transferred out, or

$$\Delta S=-\frac{Q}{T_H}$$

The change in entropy of the lower temperature ($T_L$) substance that is receiving heat from the higher temperature ($T_H$) substance is positive because heat is transferred in, or

$$\Delta S=+\frac{Q}{T_L}$$

The total entropy change of the two substances is

$$\Delta S_{tot}=+\frac{Q}{T_L}-\frac{Q}{T_H}$$

You can see that since $T_H$>$T_L$, then $\Delta S_{tot}>0$

The process approaches a reversible one when the difference between the temperatures equal zero. But it can never actually equal be zero because then there can be no heat transfer by definition.

Is heat transfer a reversible process? Is an object slowing down due to friction a reversible process?

As shown above all real heat transfer processes are irreversible. An object slowing down due to friction is another form of irreversibility. It is due to irreversible work, i.e., friction work, which is dissipative.

Why does the temperature at which thermal energy is added affect the change in entropy?

It has to with the reason for the necessity of the second law. The first law simply says energy is conserved. That means if heat $Q$ is transferred out of one substance equals the heat transfer $Q$ into another substance, the first law is satisfied. It doesn't matter if the heat is transferred from a high temperature substance to a low temperature substance or vice-versa. But we have never observed heat transfer from a low temperature substance to a high temperature naturally. After such a transfer occurs we would not expect that same heat to spontaneously (without some external work) transferring back from the lower temperature substance to the higher temperature substance. This means the process is irreversible. Yet that would not be violating the first law.

So the second law tells us that heat will only naturally transfer from a higher temperature substance to a lower temperature substance. In order to assure us that is the case, the law says that the total entropy change of the two substances must be greater than zero for all real processes, and zero for the ideal reversible process. So the second law needed to take into account not only the heat $Q$ transferred between the substances, but also the temperature of the substances at which the transfer occurs. Thus the equation for a differential change in entropy is


And if the heat transfer occurs at constant temperature

$$\Delta S=\frac{Q}{T}$$

And as indicated above, the total entropy change of two substances at different temperature exchanging the same heat between them must be greater than zero for any real (irreversible) process, and only approach zero for an ideal reversible process where the temperature difference is zero.

Hope this helps.


Let me start with a quick answer and then try to elaborate on it. Entropy, as you probably already know, is a state function. This is to say that that we'd like to define entropy in such a way that if you started with system in some state A (think some point on a P-V diagram) then move to another state B, the change in entropy shouldn't depend on the path taken as you move from A to B, $\Delta S$ should just be $S(B) - S(A)$. Now, if I moved the system in a cycle, start at A and end at A, we should expect the change in entropy to be zero, this can happen only if the the processes involved in the cycle where all reversible, otherwise going though one cycle would case a change in entropy and so much for us trying to define entropy as a state function.

Now let me elaborate on this, and in doing so I'll also try to answer question 2 since they are related. Suppose that as the system moves through the cycle, it takes some heat from the surrounding, call it $Q_{in}$, does some work, $W$, and lose some heat back to the surrounding, $Q_{out}$. Of course the first law would require that $W = Q_{in} - Q_{out}$. Generally, the change in entropy around the cycle would be $$\Delta S = \oint \frac{dQ}{T}, $$ where $dQ$ is the heat exchanged with the surrounding at every small step along the way of the cycle (it can be positive or negative). As I mentioned this integral need to be zero, and we know it's zero for reversible processes (to explain why this is true would make this answer too long for its own good). There are two kinds of irreversible process we need to look for going moving our system through the cycle, friction and the transfer of heat between two bodies of different temperatures. The second law of thermodynamics exists to prevent the reverse of these two things from happening, the transfer of heat form cold bodies to hot bodies, and the conversion of heat to work at the same temperature. The two processes are related in the sense that if you can do one you can do the other. So, I wanna say almost by design, a body slowing down due to friction is irreversible, since the reverse is just taking heat and doing work on a body to get it moving. As for heat transferring, it's only reversible if the two bodies are at the same temperature.

Roughly speaking, if the system had friction it would need more heat from the surrounding to do the same cycle and so we'd end up over estimating $Q_{in}$ and hence over estimating the above integral. If at any point of the process you had your system and the surrounding to be not at the same temperature, then there's some ambiguity about what temperature to use in the formula. This matters because as you mentioned, it matters at what temperature the system take the heat. Another answer has already hinted about why. In both cases, the solution to the problem is that no matter what path the system has taken, we can always imagine a reversible path, one which the system has no friction, and where the surroundings always have the same temperature as the system, and use this process to define the entropy for all states of the system. This is how you use the formula without thinking.

As for the last question, if you think of a harmonic oscillator, adding heat to the system would also allow it to explore more space (as well as momentum as you mentioned), and so increase of entropy would come from the oscillator moving faster and also further.

  • $\begingroup$ If a system goes through a cycle, which means that it is returned to its original state at the end of the cycle, the change in entropy of the system is zero irrespective of whether the process was reversible or irreversible. $\endgroup$ Jul 12, 2019 at 1:36
  • $\begingroup$ Well it depends on how you calculate this change in entropy. If you're already thinking that you have a formula for the entropy given the state, then yeah, returning the system to the initial state the formula gives the same number. The subtle point is how you can get such a formula for the entropy given the state, when we know only how entropy changes from one point to the other. Again, I think the friction example is a good one, comparing the same cycle, once with friction and one without would yield a different value for the integral for the change in entropy. $\endgroup$
    – A. Jahin
    Jul 12, 2019 at 1:53
  • $\begingroup$ The formula only applies to reversible paths. Entropy is a function of state, which means that it is physical property of the system, independent of any process. If it were not a function of state, we couldn't, for example, have entropy values listed in the steam tables. So, irrespective of any formula, if the state of the system in the end is the same as at the beginning, the change in entropy is zero. @Jeffrey J Weimer, how do you weigh in on this? $\endgroup$ Jul 12, 2019 at 2:30
  • $\begingroup$ @A.Jahin There are several things in your answer that you might consider editing. 1. As pointed out by Chet Miller, the change in entropy of a system in a cycle is zero whether or not the cycle is reversible since it is a state function. The difference is for a reversible cycle the change is zero for both the system and the surroundings. For an irreversible cycle the system needs to transfer the entropy it generated to the surroundings so that its entropy change is zero. But the entropy of the surroundings increases for a total entropy increase (system + surroundings). $\endgroup$
    – Bob D
    Jul 12, 2019 at 15:39
  • $\begingroup$ @A.Jahin 2. The second law does not “exist to prevent” irreversible processes from happening; such as prevent heat naturally flowing from low to high temperature. Nature does. The second law gives us a thermodynamic property, entropy, to account for what nature proscribes, just as the first law gives us a property, internal energy, to account for the conservation of energy that nature also proscribes. $\endgroup$
    – Bob D
    Jul 12, 2019 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.