# Why can we express the heat added as the product of temperature change and some $C$?

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

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, $$dU=dQ=C_vdT$$
• 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? Jan 14 at 8:48
• 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 Jan 14 at 9:22
• 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. Jan 14 at 12:08
• 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 $Q\neq\int{C_vdT}$ Jan 14 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.