The specific heat capacity $c_p = c_p(T)$ (J/kg/K) of an ideal gas depends on its temperature.

However, the gas has a static temperature ($T_s$) and a total temperature ($T_i$).

As we have for the "sensible" enthalpy ($h$) and for the total enthalpy ($h_i$) :

$h = c_p*T_s$

$h_i = c_p*T_i$

what temperature ($T_s$ or $T_i$) should I use to calculate the $c_p$ of the gas ?

I am not sure of my logic, but I thought this : as the heat capacity $c_p$ measures how much we can raise the gas temperature ($T_s$ I supposed ?), and as the "sensible" enthalpy $h$ is the energy involved in the change of the gas static temperature, I consider that $c_p$ is calculated with the static temperature $T_s$, and not with the total temperature $T_i$

Thank you

Edit : in order to clarify what I mean by calculating $c_p$. For instance, let's say I have a gas surrounding an object, moving at a certain speed $V$ in the object reference, and I have the temperature-dependent equation of $c_p(T)$ of that gas (e.g, NASA polynomials). Because the gas is moving in the reference of the object, we define two temperatures : $T_s$ (static temperature) and (total temperature) $T_i = T_s + \frac{V^2}{2c_p} > T_s$ . I want to know if I have to calculate $c_p(T)$ with $Ts$ or $T_i$, both temperatures do not give the same $c_p$.

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    $\begingroup$ What the heck is static temperature? $\endgroup$ Commented Jun 8, 2021 at 12:20
  • $\begingroup$ @ChetMiller Static temperature is the temperature of the gas with no velocity. The relation between static and total temperature is $T_i = T_s + \frac{V^{2}}{2c_p}$. Some people also know the following relation $T_i = T_s(1+\frac{\gamma - 1}{2}*M^2)$ which is valid for calorifically perfect gas if I am not wrong. Wikipedia link $\endgroup$
    – Jonses
    Commented Jun 8, 2021 at 12:38
  • 1
    $\begingroup$ Static temperature depends on the frame of reference of the observer. So, certainly an equilibrium physical property like heat capacity can't depend on that. $\endgroup$ Commented Jun 8, 2021 at 14:41
  • $\begingroup$ Now you say it, it makes sense about the dependence on the frame of reference. But I would have thought that it was the total temperature that dependds on the frame of reference, and that static temperature is intrinsic to the gas $\endgroup$
    – Jonses
    Commented Jun 9, 2021 at 6:20
  • $\begingroup$ Heat capacity of an ideal gas does not depend on temperature. Together with the comments above, I think the OP needs more clarity. $\endgroup$
    – Roger V.
    Commented Jun 17, 2022 at 13:05

2 Answers 2


An ideal gas is subset of ideal fluids. The specific heat of an ideal fluid depends only on temperature. The temperature of concern is in a static reference frame. One common form of the temperature dependence for specific heat capacity is the Shomate polynomial as on the NIST Chemistry Webbook site $C_p = A + BT^2 + CT^3 + \ldots$. A subsequent assumption is to state that $C_p$ is constant. This gives for example the formulation that molar specific heat $\bar{C}_p = 5R/2$ (J/mol K) for an ideal, monoatomic gas.

The "total" temperature as you have defined appears to be a representation to allow for changes in enthalpy $H$ and kinetic $E_K$ energy of an object. It would appear to provide a way to do this

$$\Delta \tilde{H} = \tilde{C}_p \Delta T_s$$ $$\Delta \tilde{E}_T = \Delta H + \Delta \tilde{E}_K = \tilde{C}_p \Delta T_s + \Delta (v^2/2)$$

Re-written, this becomes

$$\Delta \tilde{E}_T/\tilde{C}_p = \Delta T_i = \Delta T_s + \Delta (v^2/2\tilde{C}_p)$$

Note that energy and specific heat in these expressions are mass specific (J/kg and J/kg K).

You obtain the statement $T_i = T_s + (v^2/2C_p)$ when the difference is defined with respect to a zero point in absolute temperature and with reference to a static object. You can effectively say that the value $T_i$ is either the static enthalpy temperature equivalent $T_s = v^2/2\tilde{C}_p$ for a moving object at 0~K or the kinetic energy temperature equivalent $v^2/2\tilde{C}_p = T_s$ of a static object at a given temperature $T_s$.


In Thermodynamics and an Introduction to Thermostatistics, Callen writes

We [follow] the standard convention of restricting attention to systems that are macroscopically stationary, in which case the momentum and angular momentum arbitrarily are required to be zero and do not appear in the analysis.

The familiar material properties frequently used in thermodynamics (e.g., the isothermal and isentropic compressibilities, the isothermal and isentropic thermal expansion coefficients) are all considered to be taken for systems at rest in the frame of interest. The constant-pressure and constant-volume heat capacities are no exception; with no bulk motion, no ambiguity exists between the static and total temperatures you define, as these temperatures are identical.

  • $\begingroup$ I just want to clarify my mind, but as I understand, I must use in my specific problem what I define as static temperature to calculate the $c_p$ of the gas in motion ? $\endgroup$
    – Jonses
    Commented Jun 9, 2021 at 6:23
  • $\begingroup$ Maybe you could edit your question to describe what you mean by "calculate $c_P$." The value for a monatomic ideal gas is just $\frac{5}{2}R$, for example. $\endgroup$ Commented Jun 9, 2021 at 18:42
  • $\begingroup$ I have just edited my question. Yes, it is true that for an calorifically perfect monatomic ideal gas we have $c_p = \frac{5}{2}R$ , but this is not the case for all ideal gases. Many of them have a temperature-dependent $c_p(T)$ (e.g, NASA polynomials give the temperature-dependent relations of $c_p(T)$ for many gases). If the gas is motionless, total and static temperatures are equal, so there is no ambiguity to calculate $c_p$. When the gas is in motion (velocity $V$), its static and total temperature are different, so which should I use to calculate $c_p(T)$ $\endgroup$
    – Jonses
    Commented Jun 10, 2021 at 7:08
  • $\begingroup$ Thank you; that makes the issue much clearer. It looks like Jeffrey’s answer addresses this specific point. $\endgroup$ Commented Jun 10, 2021 at 15:17

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