In explaining $C_v = \left(\frac{\partial u}{\partial T}\right)_V$, my textbook first writes that $\delta q = du$ for a constant volume process, and then that that $dq=C_v dT$ 'from the definition of $C_v$' which they have given as "the energy required to raise the temperature of a unit mass of substance by one degree as the volume is maintained constant."

Then the two are equated to get $C_v = \left(\frac{\partial u}{\partial T}\right)_V$

However, what's not explained is why can dQ be expressed as the product of some quantity $C_v$ times $dT$. Intuitively it makes sense that heating a system would increase its temperature. But why would this necessarily be depending on the product of temperature change $dT$ and some $C_v$. Is this based on some empirical observation that hasn't been stated? I don't see how it follows directly from the definition


2 Answers 2


Your book has it backwards. The definition of $C_v$ is $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$From Gibbs phase rule, if you have a single phase, then U is a function of two parameters, in this case T and V. But if V is constant, then $$dU=C_vdT$$. From the first law of thermodynamics, at constant volume, no P-V work is done and so, in this case, $$dU=dQ=C_vdT$$

  • $\begingroup$ Okay. Makes perfect sense when you do it this way. But I want to understand how the textbook got $dQ=C_vdT$ doing it their way. I believe their intention in doing it "the other way around" is to explain the motivation behind defining $C_v$ that way. I just don't get how what they wrote for dQ follows from definition. Is it from some empirical observation I'm not aware about? $\endgroup$
    – xasthor
    Commented Jan 14, 2023 at 8:48
  • $\begingroup$ The alternative to it being an empirical result is: An answer here(now they've deleted it) pointed out that it follows from the fact that for a function f(x,y,z), and constant y,z that $df= \frac{\partial y}{\partial x} dx$. Is there any validity to this? IMO, if we have $C(T)$ and some antiderivative G, then $dG = C(T)dT$ but it's not clear to me that $dG$ represents heat transfer. And in this case $G(T)$ representing some sort of net heat transfer seems nonsensical too and seems to imply heat transfer is a state function $\endgroup$
    – xasthor
    Commented Jan 14, 2023 at 9:22
  • $\begingroup$ The reason this seems so confusing is that it was taught to us incorrectly in Freshman Physics. In Freshman Physics we learned that dQ=CdT, even though Q is a function of path while C is supposed to be a function of state. In thermodynamics, they surreptitiously try to correct this without telling us that this is what they are doing. $\endgroup$ Commented Jan 14, 2023 at 11:58
  • $\begingroup$ So what that answer was implying, i.e that it follows from the fact that for a function $f(x,y,z)$ that $df= \frac{\partial f}{\partial x} dx$ if $y,z$ are constant is incorrect? Also, they haven't written dQ = CdT in the textbook by itself. They specified that it's for constant volume, so Q being a path function is accounted for. $\endgroup$
    – xasthor
    Commented Jan 14, 2023 at 12:08
  • $\begingroup$ No. The math is right. Also, in freshman physics, they never bothered to include the constraint on v being constant. Also, in a case like an ideal gas where there is no dependence of U on v and in a case where work is done, $\Delta U=\int{C_vdT}$ but $Q\neq\int{C_vdT}$ $\endgroup$ Commented Jan 14, 2023 at 12:22

A thermal (entropy) reservoir, e.g., a large metal block, a "calorimeter", by definition, is a device whose interaction with its environment is characterizable by its experimental temperature change. Now starting at a given temperature one can measure the effects of a small (ideally infinitesimal) temperature change in the calorimeter and use that to define the experimental "heat", (experimental entropy), communicated by $\delta Q = \alpha TdS = \alpha C(T) dT$ in some arbitrary units defined by the coefficient $\alpha$. The convention is to have $\alpha =1 $ and then the differential coefficient, so-called "heat capacity", $C=C(T)$ characterizing the "calorimeter" is defined in units of energy per temperature. If the calorimetric measurement is done at fixed volume then $C=C_v(T)$. The important point is that this definition is based on temperature measurement.


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