# What is the meaning of following expresion $C=\frac{\delta Q}{dT}$ mathematicly

Our professor raised the following question during our lecture in Statistical Physics (even so it's related to Thermodynamics):

Many text books (even wikipedia) writes wrong expressions (from mathematical point of view) for Heat Capacity Coefficient, and the right way to wright it is as following: $$C=\frac{\delta Q}{dT}$$ But as we see it is neither usual differential, nor a functional derivative, so the question what is this?

I couldn't find the answer in math books, and it is true that many text books writes it in very different ways mixing exact and inexact differentials, so anybody have a clue what is the right expression for c and why from mathematical point of view?

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## 5 Answers

I) The use of $\delta$ in the derivative

$$C~=~\frac{\delta Q}{dT}$$

is because in thermodynamics, heat $Q$ is not a state function. In particular, the differential $\delta Q$ is in-exact.

II) In detail, the heat capacity $C$ is not obtained by differentiation of some ordinary function wrt. temperature $T$. Rather it should be viewed as a ratio

$$C~=~\frac{Q}{\Delta T}$$

where $\Delta T$ is sufficiently small (as seen from all physically relevant purposes).

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+1 Your state function reference is good... – Killercam Sep 11 '12 at 13:29
Maybe your answer is the clearest one among others, but it doesn't address the main issue, why we don't use functional derivative? that Q is state function is known, why we not use $$\delta T$$ instead? and what is the meaning of dividing variation on infinitesimal differential , what is this Mathematically/Geometrically ? – TMS Sep 11 '12 at 14:24

You can take the expression $C=\frac{\delta Q}{\mathrm dT}$ as the infinitesimal version of $$C=\frac{Q}{\Delta T}$$ or a formal rewrite of $$\delta Q=C\mathrm dT$$ which, however, doesn't make sense in the language of differential forms as division by the form $\mathrm dT$ is not defined.

Let's take a look at the meaning of $\delta Q=C\mathrm dT$ assuming differential forms:

By the second law of thermodynamics, $\delta Q = T\mathrm dS$. The $\delta$ has no special meaning, it's just a reminder that we're dealing with a differential form and not a function (we can't write $\mathrm dQ$ here as the form is not exact, ie not the differential of some state function $Q$).

Thermodynamical systems are in general at least two-dimensional and allow different choices of coordinates, so assume $S$ is represented by a function of temperature and another variable, eg $S=S(V,T)$ or $S=S(P,T)$.

The definition of heat capacity from above assumes that $S$ is a function of $T$ alone as the right-hand side doesn't contain terms with $\mathrm dV$ or $\mathrm dP$. In general, we thus need a further restriction on permitted processes, like $V=\mathrm{const}$ or $P=\mathrm{const}$, which yields $C_V$ or $C_P$ respectively.

Under this assumption, we have $$\mathrm dS = \frac{\partial S}{\partial T} \mathrm dT$$ ie $$C\mathrm dT = \delta Q = T\frac{\partial S}{\partial T} \mathrm dT$$ and finally $$C = T\frac{\partial S}{\partial T}$$

A further note for the more mathematically inclined:

Geometrically, the restrictions $V=\mathrm{const}$ or $P=\mathrm{const}$ define a 1-dimensional submanifold where the pullback of $\delta Q$ via the natural embedding will be (locally) exact. In fact, this pullback needs to be included to make the equations above conform to the notation used in differential geometry:

Let $\nu$ be our embedding with $\mathrm d\tau = \nu^*\mathrm dT$ non-degenerate. There's a function $C_\nu$ and (as $\nu^*\delta Q$ is closed) another function $Q_\nu$ (or rather a family of locally defined functions) with $$\nu^*\delta Q = C_\nu \mathrm d\tau = \mathrm dQ_\nu$$ that is $$C_\nu = \frac{\partial Q_\nu}{\partial\tau}$$ In case of $V=\mathrm{const}$, $Q_\nu$ is the pullback of the internal energy $U$, whereas in case of $P=\mathrm{const}$, $Q_\nu$ is the pullback of the Enthalpy $H$.

In physicist's notation this reads $$C_V = \left(\frac{\partial U}{\partial T}\right)_V \\ C_P = \left(\frac{\partial H}{\partial T}\right)_P$$

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Following your logic, i see that one must write it as follows:$$C=\frac{\delta Q}{\partial T}\:,\: C_{p}=\left(\frac{\delta Q}{dT}\right)_{p}$$Do you agree? – TMS Sep 11 '12 at 15:30
(I wrote the second one in dT because this expression assumes that P=Const , while the first one because Q(T,P) , if true the question remains why not to write functional derivatives? – TMS Sep 11 '12 at 15:36
@TMS: you can find arguments for most of these notations and even other ones like $$C_P=\frac{\mathrm dQ_V}{\mathrm dT}$$ best find a bunch of physicists and take a vote ;) – Christoph Sep 11 '12 at 17:18

The heat capacity can change with $T$ making this a non-exact differential. This is also the case with other equations in thermodynamics. The heat capasity you reference here of course also varies with pressure and volume and this is what leads to the following definitions of heat capacity at constant pressure $C_{p}$ and constant volume $C_{v}$.

$C_{p} = (\frac{\partial Q}{\partial T})_{p}$

and

$C_{v} = (\frac{\partial Q}{\partial T})_{v}$

I would interpret your original equation simply as

$Q = C\,\Delta T$

That is $Q$ is the amount of heat required by a substance with heat capacity $C$, to change the substance's temperature by $\Delta T$.

I hope this helps.

Extension to Address Comments:

Of course having a partial derivative makes sense in this context. Let's take the constant volum case; when heat is added to a substance (fluid for example) at constant volume no work is done, so the heat added equals the increase in the internal energy of the fluid. Writing $Q_{v}$ for the heat added at constant volume (like in the equations above), we have

$Q_{v} = C_{v} \Delta T$

scince $W = 0$ (work), we can write

$Q_{v} = \Delta U + W = \Delta U$.

Thus,

$\Delta U = C_{v} \Delta T$.

Taking the limit as $\Delta T$ aproaches zero we find

$\mathrm{d} U = C_{v} \mathrm{d}T$

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+1: more or less the same things I was getting at – Christoph Sep 11 '12 at 13:35
I was a bit too quick with my upvote - using lower-case letters normally means specific heat capacities and your definitions are bogus - $Q$ is not a state function, thus taking partial derivatives makes no sense... – Christoph Sep 11 '12 at 13:54
See extension to answer. This does make sense in this context... – Killercam Sep 11 '12 at 14:15

1
thanks for the clarification, it was a thinko on my part (on restriction to a 1-dim submanifold, any form is locally exact, so of course there's a function $Q$ on the submanifold); however, one should note that in case of $V=\mathrm{const}$, $Q$ will be the pullback of the internal energy $U$, whereas in case of $P=\mathrm{const}$ it'll be the pullback of the Enthalpy $H$, so using different symbols $Q_V$ and $Q_P$ (as you did) is probably a good idea; I should add something about that to my own answer... – Christoph Sep 11 '12 at 15:21
Agreed. Thanks for highlighting my error. All the best... – Killercam Sep 11 '12 at 15:31

You see, as was stated $Q$ is not a characteristic of the system (not a state function). It depends on the process. The straightforward way to implement the notion of process is to view $Q$ as a function of time. Thus you could view $C$ as:

$$C(t) = \frac{dQ/dt}{dT/dt}$$

In the case of a reversible process with no exchange of matter with the environment:

$$\frac{dQ}{dt} = T \frac{dS}{dt}$$

Let's consider a monoatomic ideal gas, and a process with constant volume and speeds that allow us to use equilibrium relations and assume reversibility:

$$C_V = \frac{T \; dS/dt}{dT/dt} = \frac{dU/dt}{dT/dt} = \frac{\frac{3}{2} R NT \; dT/dt}{dT/dt} = \frac{3}{2} R NT$$

To sum up, you can always view $dQ$ as a full differential, but on time, treating $Q$ as a function of time. I learned this trick from the book "Modern Thermodynamics: From Heat Engines to Dissipative Structures" by Kondepudi and Prigogine.

Note, that's perfectly strict mathematically --- just the quotient of two derivatives, no differential forms or some slippery reasoning with infinitesimals.

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I don't have that book, but I suspect that making Q(t) makes it exact differential or Path function in general, because we still have infinity ways on how we heat up (for example) our system. – TMS Sep 11 '12 at 16:09
there isn't much about it in the book, it is just said how to get rid of $\delta$ ( "inexact differential" as you call them) by viewing functions of time (thus turning them into "exact differential"). – Yrogirg Sep 11 '12 at 16:13

Most of the given answers already describe how to reach the mathematical definition and I can only add a more phenomenological approach.

Experimentally specific heat is defined as the coefficient of thermal energy input to temperature rise of an adiabatic system (with either constant pressure or volume)

$$C = \frac{\Delta Q}{\Delta T},$$ and we also know from experiments that $C = C(T)$. So not to average over a large interval we have to reduce the thermal energy as much as possible and measure the response, the temperature increase: $$C(T) = \lim\limits_{\Delta Q\rightarrow 0} \frac{\Delta Q}{\Delta T}$$ If there was a unique function $Q(T,C)$ we would write this as the derivative $$C(T) = \lim\limits_{h\rightarrow 0} \frac{Q(T, C)-Q(T+h, C)}{h} =\frac{dQ}{dT},$$ but it turns out that it matters how the thermal energy is put into the system. This means that there is no unique function $$Q = \int dQ,$$ so we fall back on the inexact differential, denoted by $\delta Q$, and define the specific heat as $$C(T) =\frac{\delta Q}{dT}$$

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I will repeat my self, I know how to derive it, am asking what is the meaning of such a structure and if it is right mathematically, because I found no Math book that using such structures (variational derivative over exact deferential) – TMS Sep 12 '12 at 16:51
@TMS: Ok, then the question is more what is an inexact differential, as you can also express $\delta Q$ as $f(T,C)dT+g(T,C)dC$ but there is no function $Q(T,C)$, where $f=\partial Q/\partial T$ and $g=\partial Q/\partial C$? – Alexander Sep 12 '12 at 17:34
Sorry I didn't understood your comment. – TMS Sep 12 '12 at 17:57
– Alexander Sep 12 '12 at 18:13
Putting in mind that there are Huge math books on variational methods, forms, ect.. I suspect that it is as simple as you state.. – TMS Sep 12 '12 at 18:16
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