Heat capacity $C$ of an object is the proportionality constant between the heat $Q$ that the object absorbs or loses & the resulting temperature change $\delta T$ of the object. Entropy change is the amount of energy dispersed reversibly at a specific temperature. But they have the same unit joule/kelvin like work & energy. My conscience is saying these two are different as one concerns with temperature change and other only at a specific temperature. I cannot figure out any differences. Plz explain me their differences.


$$dQ = T \ dS \tag1$$ $$dQ = C \ dT \tag2$$

Interesting, right? In $(1)$, the whole $T$ multiplies the infinitesimal $\frac{\text{J}}{\text{K}}$. In $(2)$ it's the opposite: the whole $\frac{\text{J}}{\text{K}}$ multiplies the infinitesimal $T$.

But you hinted that you knew that yourself already. Let's cut to the chase: both are different beasts entirely, just like heat and torque are not related just because they carry the same unit (joules are newton-meters, right?).

However, if you still want a defining difference between them, other than "they're just different", I'd give you this:

Entropy by itself is not useful and cannot even be measured. What is useful are changes in entropy, or how it differs from one state to the other. In this sense, it's akin to internal energy and enthalpy, for which only relative values matter. Heat capacity, on the other hand, can have its absolute value determined experimentally, and it won't depend on a reference value like entropy does. Its absolute value is immediately useful, if you will. In this sense, it's akin to pressure and specific volume, for which absolute values make sense.

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    $\begingroup$ But from statistical physics it seems that entropy has really a lot similarity to heat capacity, not just the units. Entropy is logarithm of number of microstates $S=k ln(W)$ while number of microstates can be in simple case exponential of number of degrees of freedom (e.g. if each DOF has two possible states $W=2^N$). Than entropy is proportional to number of DOFs $S\propto N$, and heat capacity is also proportional to number of DOFs due to equipartiition theorem. $\endgroup$ – Prokop Hapala Aug 30 '16 at 8:37
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    $\begingroup$ Regarding the joules vs newton-meters: energy is scalar, torque is a vector quantity. So two very different things. But both heat capacity and entropy are scalar quantities. $\endgroup$ – Calmarius Jul 16 '20 at 20:24

If you consider a constant volume transformation, the corresponding specific heat will be defined as:

$C_v(T) \equiv \left( \frac{\partial U}{\partial T}\right)_{N,V}$

Now, it is not forbidden to use Leibniz rule for the decomposition of partial derivatives and for instance:

$\left( \frac{\partial U}{\partial T}\right)_{N,V} = \left( \frac{\partial U}{\partial S}\right)_{N,V} \cdot \left( \frac{\partial S}{\partial T}\right)_{N,V} = T \left( \frac{\partial S}{\partial T}\right)_{N,V}$

Which means that $C_v(T) = T \left( \frac{\partial S}{\partial T}\right)_{N,V}$

Hence, $C_v$ and $S$ are definitely two different things. In particular, the specific heat contains some (partial) information about the entropy of the system (and its possible variation under some constraints) but not all of it.

Hence in term of heat exchanged, we know that:

$\delta Q = TdS$ upon expanding $dS$ as total differential, we thus have one possible reading of the heat exchanged (at fixed number of particles):

$\delta Q = T \left( \frac{\partial S}{\partial T}\right)_{N,V} dT + T \left( \frac{\partial S}{\partial V}\right)_{N,T} dV = C_v(T) dT + T \left( \frac{\partial S}{\partial V}\right)_{N,T} dV $

The second term $T \left( \frac{\partial S}{\partial V}\right)_{N,T}$ is what any specific heat function (regardless whether we look at constant volume or pressure) will always miss and is ultimately related to the thermal expansion properties of the material.

  • $\begingroup$ I think I fail to understand the importance of the last sentence. The second term would only matter if you're using (in this example) Cv in a non-isometric process, right? Then the second term would compensate for that "inadequacy." Could you please elaborate on that sentence? $\endgroup$ – André Chalella Sep 8 '14 at 10:56
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    $\begingroup$ The point here stresses that heat capacity only contains partial information about how your system can exchange heat and that is the reason why this is not the same as calculating the entropy variation locally, integrated then over the thermodynamic trajectory followed by your system during the transformation. Now, in a system where you only consider transformations at fixed volume, then $C_v$ carries as much information about heat as the entropy itself. $\endgroup$ – gatsu Sep 8 '14 at 12:50

As a partial answer (a complete answer would require more time and more knowledge about classical thermodynamics on my part): As you thought, these are two entirely different objects.

The heat capacity is a material-dependent object that - as you say - measures the difference in temperature when energy is absorbed by the material. It's object and process independent: If you add this amount of heat, you'll get that increase in temperature. EDIT: By "process dependent", here, I mean that the processes can be either reversible or irreversible. The capacity DOES of course depend on overall constraints: Leaving the volume constant or the pressure gives you a different heat capacity.

The entropy change on the other hand is a material-independent quantity. The entropy change is proportional to the heat transfer in a reversible process (at constant temperature! Otherwise you have to account for the temperature difference). However, most processes are irreversible, so the quantity is process dependent.

If you look at the statistical interpretation, the difference becomes more clear: The entropy counts the number of specific states that the system can be found in, given the thermodynamic parameters known. Since temperature is something like the averaged kinetic energy, the heat capacity loosely counts the number of degrees of freedom in the atom (heat is randomized kinetic energy that needs to be transfered to the material: The more degrees of freedom, the less energy every degree needs to take on when energy transfered to the material, i.e.: more degrees of freedom, higher heat capacity).

  • $\begingroup$ -1, both are material- and process-dependent. Even though most heat transfer problems in solids and liquids will admit a single specific heat, it's definitely not true for gases. Also, entropy is a property of a pure substance, right? How can it not be material-dependent? $\endgroup$ – André Chalella Sep 8 '14 at 10:13
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    $\begingroup$ @AndréNeves I'm pretty sure he means the entropy change is material independent - any material heated by an amount $\delta Q$ at temperature $T$ will change its entropy by $\delta Q/T$. The total entropy is material independent, however, precisely because the heat capacity is material dependent. $\endgroup$ – Nathaniel Sep 8 '14 at 10:16
  • $\begingroup$ You're right, I didn't think that through. But the heat capacity being process independent bit is a bigger issue, I guess? $\endgroup$ – André Chalella Sep 8 '14 at 10:19
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    $\begingroup$ Well, yes, I meant the entropy change of course, which is material independent, I'll make this more clear. Also, heat capacity is not process dependent in the way I thought about it: It's true, there are different capacities for whether you keep volume/pressure fixed, but within these equations, the process doesn't matter - as long as you fix volume or pressure, the process can be both reversible or irreversible, because the definition doesn't depend on this. I'm maybe thinking from a not too much thermodynamic perspective here, so you might well critizise this, at least I have to clarify. $\endgroup$ – Martin Sep 8 '14 at 13:02

Taking the first law of thermodynamics you have $$dU = \delta Q + dE_{work}$$

I.e. the change of internal energy is the change of energy expended by work of the macroscopic parameters plus heat. It is a non-trivial result in thermodynamics that there exists a function $S$, entropy, and absolute temperature $T$ which can be used to write the first law as $$dU = dE_{work} + T dS$$

That is $\delta Q = T dS$. If we want to compute specific heat we want to know how much heat we need to increase the temperature - difference of heat over difference of temperature - this gives us a derivative $$C = \left( \frac{\delta Q}{\partial T}\right)_{constraints} = T \left( \frac{\partial S}{\partial T}\right)_{constraints}$$
So the specific heat is actually a derivative of the entropy and the units are a consequence of loosing a kelvin by the $T$ in the "denominator" of the derivative and then gained again by multiplying by $T$.

  • $\begingroup$ Are you talking about a reversible path? $\endgroup$ – Yasir Sadiq Dec 7 '20 at 8:10

It is bit difficult to see that entropy and heat capacities are entire different concepts. They are both related to the filling of energy "destinations" in a system as temperature increases. They are different ways of seeing the same thing.

As a starting point we can clearly see that the two are related:

$C=T\left(\frac{\partial S}{\partial T}\right)_\textrm{constraints}$

Here's an initial attempt to put the concepts into words to at least give some sense of the relationship:

Entropy is the cumulative filling of energy destinations between absolute zero (motionless) and a given temperature.

Heat capacity is the rate of change of entropy with temperature… scaled by temperature.



All comments miss a point, or maybe I didn't get it right. I see it this way, lets start with other forms of energy (kinetic and potential, for example) of an harmonic oscillator. We know for a non damped HO the relation between position "$s$" and velocity "$v$" is given by total energy $E_o = \frac{1}{2}kx^2 + \frac{1}{2}mv^2,$ where $k$ is spring constant and $m$ is mass. This relationship (since energy is conserved here) can be turned in many ways. Differentiating (focusing on small values) both sides yields: $\mathrm{d}E_o = kx \cdot \mathrm{d}x + mv \cdot \mathrm{d}v$ assuming conservative forces $\mathrm{d}E_o =$ and we can express $\dfrac{\mathrm{d}v}{\mathrm{d}x} = - k \dfrac{x}{mv}$. Does the velocity depend on position, NO! the fact that the conservation law allows to relate two (measurable) quantities does not mean those two are affecting each other. The same is true for $C$ and $T$. Although heat capacity is found by known relationships does not mean they are based on each other. If we for the same amount of heat quantity get one sample to have more temp than another sample it means that for the first one more energy is stored in the system than second. Is $T$ is the kinetic energy the $C$ relates to potential energy. I hope I make sense.


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