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Heat can be considered as an inexact differential. Hence, we could conclude that: there exists a path $\gamma$,such that: $$ \oint\delta Q \ne 0 $$ Hence, $\int\delta Q\ne\Delta Q$.

i.e. $\Delta Q$ makes little to no sense.

Yet, I have seen equations where $\Delta Q$ is used. E.g. Latent Heat.

Moreover, I have seen situations where people treat this $\delta Q$ is treated as $d Q$ while writing some equations.

Is this justified? If so, how? Under what conditions?

Qmechanic
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Aditya Krishna Panickar
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    I don’t see any problem with writing dQ as an incremental amount of heat added along a path. – Chet Miller Feb 08 '24 at 12:52
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    @ChetMiller I would add provided we speak of a differential AMOUNT of heat as opposed to a differential CHANGE in heat – Bob D Feb 08 '24 at 12:58
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    In order to emphasise the inexactness of heat, it is better to use $\delta Q$ or some other notation other than dQ as the notation dF is generally used for an exact 1 form obtained by applying exterior derivative d to a function F. – user10001 Feb 08 '24 at 13:30
  • How do we denote Diffrential amount compared to differential change? – Aditya Krishna Panickar Feb 08 '24 at 14:33
  • @ Chet Miller The issue is that writing it as $dQ$ makes it seem as if heat is a state function. Which it definetly is not. The differential amount of heat added definetly depends upon the path taken – Aditya Krishna Panickar Feb 08 '24 at 14:38
  • A pertaining question is: What does it actually mean to define a function $Q()$. Is it even possible? I think it must be! But, what is its use? – Aditya Krishna Panickar Feb 08 '24 at 14:40
  • Basically, my question is, is it just a stupid convention, or is there any mathematical intricacy that I missed. Some contrived intergral theorem or manipulation perhaps? – Aditya Krishna Panickar Feb 08 '24 at 14:44

2 Answers2

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The notation used for heat in thermodynamic processes is one of the poorest in Physics. A source of confusion is that the total exchanges of heat (Q) and work (W) always sum to the internal energy difference between the final and the initial state. Therefore, one should write: $$ \Delta U = Q + W \tag{1} $$

A notation like $$ \Delta U = \Delta Q + \Delta W \tag{2} $$ It is incorrect but could be accepted, provided it is clear that the $\Delta$s on the right-hand side of the equation do not represent differences but increments of heat and work exchanged with the system. That is the meaning that should be assigned to the notation for latent heat in Clapeyron's equation. Altogether, I find the form ($1$) clearer.

For small changes in the state, the difference between $\Delta U$ and the differential $dU$ can be neglected. In many textbooks of the beginning of the twentieth century (Planck's textbook for sure), the differential form of equation $(1)$ was written as $$ d U = q + w \tag{3} $$ that I find coherent with the fact that the symbols on the right-hand side of the equation, in general, are just two real numbers and do not represent variation or differential of anything. For an irreversible process, they do not even depend on state variables. Of course, in the case of reversible processes, it is possible to give an expression of $q$ and $w$ as $TdS$ and $-pdV$, respectively, but this is a special case.

Unfortunately, such a terse notation has been forgotten, and all kinds of $d-$like symbols have been introduced to represent $q$ and $w$, like $\delta q$, or a barred $d$. Even worse, the term inexact differentials is often used for such quantities, introducing additional confusion between the genuine mathematical concept of non-exact differential forms and some quantities that are not a function of the state variables, in general.

To summarize, the only case where it is justified to use a notation like $dq$ is in the case of reversible processes, where it is a shortening for the differential $TdS$.

GiorgioP-DoomsdayClockIsAt-90
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  • Isnt inexact differential same as path dependence. – Aditya Krishna Panickar Feb 08 '24 at 17:01
  • Moreover, for a quasistatic process, T is just an integrating factor right( to change $\delta Q$ into path independent)? just like P is an integrating factor (to change $\delta W$ into path independent) – Aditya Krishna Panickar Feb 08 '24 at 17:04
  • @AdityaKrishnaPanickar Not really. Inexact differentials may be path-dependent, but in order to be such they must be functions of state variables. This is not the case when the process is not quasi-static. This is a situation quite different from path dependence. Also about P as integrating fac integrating factor, this is true only if $dW$ is a function of the state variables. – GiorgioP-DoomsdayClockIsAt-90 Feb 09 '24 at 00:40
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Since all infinitesimally transported heat $\delta Q$ is also some amount of entropy $dS$ transported at some temperature $T$, that is, $\delta Q = TdS$, you can always sum (integrate) their product over some path without any problem: $Q[\gamma]=\int_{\gamma} TdS$ where the integral is now the total heat $Q[\gamma]$ transported over the path. Here $Q[\gamma]$ depends on the path and not just on the endpoints and any "point" of the parametrized path $\gamma$. Because of their implied continuity $T,S$ is well defined by the interface through which the transport occurs.

hyportnex
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  • We definitely can integrate an inexact differential. However, writing the end result as $\Delta Q$ seems to be suspicious, as was my main question. – Aditya Krishna Panickar Feb 08 '24 at 14:35
  • there is nothing suspicious about this: every point in the path represents a certain amount of entropy transported at some temperature, even if the "heat" in question is latent: Latent heat is the product of some amount of entropy released at some temperature. – hyportnex Feb 08 '24 at 14:39
  • Is invoking entropy necessary? Can you justify without entropy? Moreover, i don't seem to understand the entropy explanation properly. Have you encountered a reading material which aims at this? – Aditya Krishna Panickar Feb 08 '24 at 15:11
  • If you want to understand it then yes, invoking entropy is necessary. There is no way to understand what "heat" is or is not without invoking entropy. In fact, entropy is the agent of thermal work, just as mass is the agent of gravitational work, electric charge is the agent of electric work, etc. $dS$ is a real differential of something $TdS$ is not, or if you wish in $\frac{dQ}{T}$ the role of $T$ is to make $\delta Q$ "path-independent integrable", an integrating factor. Historically, this is how it was introduced by Clausius but that is not necessary. – hyportnex Feb 08 '24 at 15:38
  • As is, your answer is wrong. Not all the infinitesimal exchanged heat $\delta Q$ can be written as $TdS$. It is possible only in the case of a reversible process. – GiorgioP-DoomsdayClockIsAt-90 Feb 08 '24 at 15:41
  • 20 years prior, Carnot already knew that what he called "calorique" (now called entropy) is the agent of thermal work, see this or this or this and chat. – hyportnex Feb 08 '24 at 15:42
  • @giorgiop-doomsdayclockisat-90 I did not write exchanged, I wrote transported; not the same. – hyportnex Feb 08 '24 at 15:43
  • Even using the term transported, it remains that $TdS$ is equal to the heat only in the case of reversible processes (it is in the definition of entropy). – GiorgioP-DoomsdayClockIsAt-90 Feb 08 '24 at 16:12
  • @giorgiop-doomsdayclockisat-90 It depends on what you mean by $dS$. If by $dS$ you mean the increase of entropy in a system that has absorbed heat, then you are right. But to avoid that confusion I wrote "transported". Thermodynamic entropy is meaningful for irreversible processes too, and there is no heat transport imaginable that is without entropy transport at some temperature and the ratio of heat to temperature is meaningful for any $T>0$. To prove your point, please, describe any calorimetric experiment that can distinguish heat transport from entropy transport. – hyportnex Feb 08 '24 at 16:32
  • It is not possible to refer to experiments for a definition. The definition of entropy is based on recognizing that $\sum q_{rev}/T$ is a function of state. It is Clausus' definition of thermodynamic entropy. – GiorgioP-DoomsdayClockIsAt-90 Feb 08 '24 at 16:43
  • @giorgiop-doomsdayclockisat-90 Clausius's is one way of defining entropy, but even if we use Clausius it is enough to say that $T$ being an integrating factor and whatever you mean by $\frac{q_{rev}(T)}{T}$ is a well defined differential for an infinitesimal $q_{rev}$ and one is justified to call it $dS$ the transported entropy, and all will be fine. Note too that whenever you measure $q_{rev}$ calorimetrically over an infinitesimal temperature drop to avoid irreversibility you also measure, a fortiori, $dS$ and now it is up to you which one you consider more fundamental. – hyportnex Feb 09 '24 at 02:08