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According to Wikipedia, heat capacity or thermal capacity is a physical property of matter, defined as the amount of heat to be supplied to an object to produce a unit change in its temperature:$$C = \lim_{\Delta T \rightarrow 0} \frac{\Delta Q}{\Delta T}$$ According to another source (which is PDF only, and not in English), in the context of thermodynamic properties of gases due to the rotations of diatomic molecules, at very high temperatures:$$C_{V,\mathrm{rot}} \approx k_\mathrm{B} = \pu{1.38E-23 J/K}$$

Does this imply that adding $\pu{1 J}$ to this system will raise its temperature by $10^{23}$ kelvin?

I believe there is something in this definition I didn't get. Could you explain the sense of $C$ in a more intuitive way please?

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    $\begingroup$ Heat capacity [J/K] ]is property of a system/object, being an extensive property.. The property of matter/substance are the intensive parameters specific or molar heat capacities [J/K/kg] resp. [J/K/mol] $\endgroup$
    – Poutnik
    Apr 13 at 8:17
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    $\begingroup$ Are you aware that energy 1 J is for rotational energy of a single molecule an astronomically huge energy, available only at astronomicaly high temperature? (ignoring the fact any chemical bond would break at temperature many many orders lower). // The above C_V,rot is the heat capacity due rotational motion of a single molecule. $\endgroup$
    – Poutnik
    Apr 13 at 8:19
  • $\begingroup$ Oh, I see, I should have stayed at the $1 \ eV$ scale ^^ Thanks ! $\endgroup$ Apr 13 at 8:21
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    $\begingroup$ Typical thermal energy at 20 °C is $\frac{kT}{e} \approx \pu{25.2 meV}$ $\endgroup$
    – Poutnik
    Apr 13 at 8:26

2 Answers 2

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It is easier to think of the heat capacity at constant volume as the slope of the internal energy with temperature. This is effectively your first definition. The correct interpretation of the heat capacity is only possible via quantum theory, that is by realising that atoms and molecules have discrete energy levels.

Thermodynamics implicitly assumes that all energy levels are so closely spaced compared to temperature (at all temperatures) that they are effectively continuous. This is only true for translational energies thus the translational energy is the sum $\sum mv^2/2$ for $x,$ $y$ and $z$ directions giving $E=3k_\mathrm BT/2$ and so heat capacity $C_\mathrm{tr}=3k_\mathrm B/2$ for atoms and molecules. Doing the same calculation for rotation and vibrations assuming classical behaviour gives for a diatomic, $C_\mathrm{rot}=k_\mathrm B$ and $k_\mathrm B$ for the vibration, the total is $C_V=7k_\mathrm B/2$. These values overestimate the true experimental because by using classical arguments we have assumed all levels are fully occupied whereas we know that because levels are discrete, at any given temperature only some levels are populated according to the Boltzmann distribution.

This then returns to the calculation of the internal energy, which is the sum of every level's energy weighted according to its Boltzmann population. At very low temperatures only the zero point energies are populated and so the heat capacity tends to zero, a result impossible to obtain from classical thermodynamics. At very high temperatures the classical values are obtained for all molecules, but in many cases this temperature is far, far above room temperature.

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    $\begingroup$ You wrote: "Thermodynamics implicitly assumes that all energy levels are so closely spaced compared to temperature (at all temperatures) that they are effectively continuous." That's not quite right. Thermodynamics makes no assumptions about the microscopic character of the substance. For that reason, when we say, e.g., that Cv for a diatomic gas is 7/2 nR generally, we aren't using thermodynamics at all. Rather, we are using statistical mechanics uninformed by quantum mechanics. I.e., we are using classical statistical mechanics. It's the latter that also gives us the Dulong-Petit law. $\endgroup$
    – theorist
    Apr 14 at 5:07
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    $\begingroup$ In sum, values for heat capacity, like Cv=7/2 nR for a diatomic gas, are quantitites derived outside of thermodynamics that can be used as an input into thermodynamic expressions. But thermo is just as happy taking, as input, an expression for heat capacity as a function of T derived using quantum mechanics. Thermo has no opinion on the matter; it is a big tent that welcomes both. Thermo only complains if you attempt to use a negative heat capacity, since that causes a 2nd law violation. Take-home message: Don't blame thermodynamics for the limitations of classical statistical mechanics! ;) $\endgroup$
    – theorist
    Apr 14 at 5:31
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    $\begingroup$ @Theorist Thanks for you comments. I was trying to explain simply without being too technical. Of course thermo was developed before atomic theory was understood and treats matter as continuous undefined stuff but I think that it is easier nowadays to explain this in terms of energy levels as this is more familiar. Its great generality is its strength, but this I think, makes thermodynamics hard for students. $\endgroup$
    – porphyrin
    Apr 14 at 7:15
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As @Poutnik wrote in the comments, the heat capacity $C$ quantifies the influence of energy on the temperature of the system. There is no contradiction to write $C = k_\mathrm{B}$ since at the scale of a single molecule $\pu{1 J}$ is huge (energies are of the order of the electronvolt $\pu{eV}$ with $\pu{1 eV} \approx \pu{0.6E-19 J}$. Typical thermal energy at $\pu{20 ^\circ C}$ is $k_\mathrm{B}T/e \approx \pu{25.2 meV}$.

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    $\begingroup$ Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic // For more, see Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    Apr 13 at 8:31
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    $\begingroup$ Consider $\pu{1 eV} \approx \pu{1.6E-19 eV}$ as $\pu{1 eV} \approx \pu{1.6E-19 eV}$. // BTW $\pu{1 J} \approx \pu{1.6E19 eV}$ is wrong as $\pu{1.6E19} \ne \frac{1}{\pu{1.6E-19}}$. $\endgroup$
    – Poutnik
    Apr 13 at 8:38
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    $\begingroup$ P.S.: Be aware of e like \cdot versus E like \times: $\pu{6e-6}$ as $\pu{6e-6}$ versus $\pu{6.4E-6}$ as $\pu{6.4E-6}$. 10^6 in context of \pu{} can be written also as $\pu{e6 mg}$ seen as $\pu{e6 mg}$. $\endgroup$
    – Poutnik
    Apr 13 at 9:03
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    $\begingroup$ $C_{V\mathrm{,rot}}$ as $C_{V\mathrm{,rot}}$, a variant $C_{V\text{,rot}}$, with \text{} taking the string literally. $\mathrm{a b}$ displayed as $\mathrm{a b}$ , $\text{a b}$ as $\text{a b}$ $\endgroup$
    – Poutnik
    Apr 13 at 9:30
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    $\begingroup$ It is not about quantum vs classical physics, but simply the size of the system. 1 J, 1 mol, 1 g etc. -- these are scales used for macroscopic quantities, i.e. amounts of chemicals that we can see. 1 eV, k_B, etc. - these are things suitable for single molecules. $\endgroup$ Apr 13 at 11:30

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