In most textbooks, the definition of entropy in reversible processes on a system $S$ is given simply as $$d S=\delta Q/T.$$ It seems to me this definition is insufficient since it does not specify what this $\delta Q $ is; this creates often confusion since in irreversible processes there are other sources of heat, for instance that due to the dissipation of turbulences or friction, which in fact should not be counted in the $\delta Q$. I wonder whether one should instead write something like $$d S_S= \delta Q_{E\to S}/T_S,$$ where $E$ stands for environment and $S$ for the system under consideration.

This would make it easier to explain the idea that entropy increases in isolated irreversible processes. In fact, since in any irreversible process there are other sources of heat (beside the one coming from the environment) that are dissipated into a system, the total heat absorbed by the system from the environment (eg, a heat bath) needs to be smaller than in reversible processes. Then clearly $$ T_S d S_{S}=\delta Q_{E\to S,REV} \ge \delta Q_{E\to S, IRR} .$$ If the irreversible process is such that there is no overall exchange of heat between $E$ and $S$, i.e. $S$ is thermally isolated, then one has $d S_S>0$ as required from the second Law. Since any real process will have almost always some heat dissipation into the system this statement is almost always true.

So my question is: do you think this almost trivial addition to the standard textbook definition of entropy is pedagogically useful or not?

  • You are asking about entropy change. You can go to this site . – user36790 Sep 22 '14 at 14:07

Heat is defined as a transfer from the environment to the system. No subscripts are needed for that purpose.

I think you would have to be very clear about the specific situation for the other processes that you mention before there can be any discussion about them. Friction where? Turbulence implies a non-equilibrium condition...

  • in any irreversible process (so, non equilibrium) there will be some friction and/or some turbulence or other source of heat not coming from E, this is all I am saying. I agree that heat is defined as transfer from E to S, I just wonder if it would be more clear to write this into the definition since any discussion of irreversible processes include other forms of heat. In Wikipedia for instance it says that dQ is the transfer of heat into the system, which could include also the heat induced on the system by dissipation of turbulence if you don't say where this transfer comes from. – user16397 Sep 22 '14 at 15:15

When textbooks write $dS=dQ/T$, quasi-static process with no friction is considered. $S$ is entropy of the system, $dQ$ is heat accepted by the system from the environment and $T$ is temperature of the system. You can add indices to $dQ$, but they are not necessary since $dQ$ already means total heat accepted from the environment. There are no other kinds or components of heat in equilibrium thermodynamics.

One should never write $d S_S= \delta Q_{E\to S}/T_S,$ unless it is a reversible process in the system, as well, because the temperature $T$ is whatever temperature the heat is supplied and that is the temperature of the environment (heat reservoir), presumably behaving independently of the amount of heat it supplies. The reservoir is always assumed to perform a reversible process. For the heat transfer to be reversible the system must have the same temperature as its environment, and then this distinction does not matter. So, if you wish, you should always write $$d S_S= \delta Q_{E\to S}/T_E,$$

  • the expression $dS=\delta Q/T$ is only valid for reversible processes, so $T_S=T_E$. – user16397 Sep 22 '14 at 15:24

In fact, $\delta Q/T$ is the entropy of $\delta Q$, and if $\delta Q$ is considered as the heat energy in transfer, it follows that $\delta Q/T$ is the entropy in transfer. For the inequality \begin{align}\oint dS \ge \oint \frac{\delta Q}{T}.\end{align} Whether reversible or not does not need to be considered for $\delta Q/T$ in that the entropy production in heat transport process is that \begin{align} d_iS=\Delta \left(\frac{1}{T}\right){\delta Q}.\end{align} From non-equilibrium thermodynamics, we have \begin{align} dS =d_eS+d_iS.\end{align} Where $d_eS $ denotes the entropy flux, and $d_iS $ denotes the entropy production. A well known fact is that $\delta Q/T$ is $d_eS$, such that the real meaning of Clausius inequality is that \begin{align} dS =d_eS+d_iS \geq d_eS=\frac{\delta Q}{T}.\end{align} Only when $d_iS=0$, we can get that \begin{align} dS = d_eS=\frac{\delta Q}{T}.\end{align} Why don't we define the entropy not by $dS$ for an arbitrary path but by $d_eS=\delta Q/T$?

“this creates often confusion since in irreversible processes there are other sources of heat, for instance that due to the dissipation of turbulences or friction, which in fact should not be counted in the δQ.”

If we consider that the heat energy is a non-conserved quantity, the changes in the heat energy within a system come from the two sources, one comes from the heat transfer $\delta Q$, another one comes from the heat conversion. Using $d_eq$ denotes the heat transfer $\delta Q$, and $d_iq$ denotes the heat production, a new function can be defined by \begin{align}dq=d_eq+d_iq\end{align} Where, $q$ is the heat energy within the system, then the fist law can be expressed as \begin{align}dU=dq+Ydx+\sum_j \mu_jdN_j\end{align} The entropy $ S$ for the system under consideration \begin{align}dS=\frac{dU}{T}-\frac{Ydx}{T}-\sum_j\frac{\mu_jdN_j}{T}+\frac{pdV}{T}=\frac{dq}{T}+\frac{p}{T}dV.\end{align} In thermodynamics, we have defined the heat in transfer $Q$, but failed to take into account a function to describe the energy of thermal motion within the system. It means that there is a state function, the internal heat energy that was lost.

Suppose that a system is subjected to an irreversible process between two thermodynamic equilibrium states. To determine the change in entropy of the system $\Delta S$ between these two equilibrium states, you execute the following sequence of steps:

  1. Forget about the irreversible process path. It cannot be used to determine the change in entropy. Focus only on the initial and final thermodynamic equilibrium states. (Entropy is a function of state).

  2. Identify a convenient reversible path between the same two thermodynamic equilibrium states. The reversible path does not need to bear any resemblance to the actual irreversible path. There are an infinite number of reversible paths to choose from, so make sure you select one for which it is easy to apply step 3 to determine the entropy change. The entropy change is exactly the same for all reversible paths.

  3. For the reversible path you have chosen, evaluate the integral of dQ/T from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state. This will give you the $\Delta S$

The change in entropy that you calculate in this way will automatically include all the dissipative effects that occurred during the irreversible process.

For an irreversible process, the form of the Clausius Inequality that applies is:

$$\Delta S \geqslant \int\frac{dQ}{T_B}\tag{1}$$where $T_B$ is the temperature at the boundary interface between the system and surroundings (the location where the heat transfer takes place). Frequently, the specification that $T_B$ must be used in the inequality is omitted in many literature references, most of which simply feature a T in the equation. This has been a source of confusion to students for many decades. Obviously, in an irreversible process, the temperature in the system is not uniform spatially if dissipative processes are occurring within the system. So certainly, a unique system temperature is not well defined for use in the inequality.

The integral on the right hand side of Eqn. 1 represents the net amount of entropy entering the system during the irreversible process. The remainder of the entropy change is the result of entropy generation from irreversible dissipation in the system. So, the Clausius inequality can be changed to an equality by writing:$$\Delta S = \int\frac{dQ}{T_B}+S_G$$where $S_G$ is the entropy generated.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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