Rate of change of internal energy I have come across a text which, without proof or detailed explanation, states that the rate of change in internal energy $U$ of a system with constant volume $V$ is given by
\begin{equation}
\frac{\partial U}{\partial t} = \frac{\partial}{\partial t} \left( \rho V C T \right),
\end{equation}
where $\rho$ is density, $C$ is specific heat capacity and $T$ is temperature.
Intuitively, this seems reasonable, but I am struggling with deriving the equation from first principles.
So far, I have used the 1st law of thermodynamics to state that $\frac{d U}{d t} = \frac{d Q}{d t}$, since the system's volume is constant. Furthermore, from the definition of the heat capacity, I also know that $\Delta Q = \rho V C \Delta T$. Dividing by $\Delta t$ and taking the limit $\Delta t \rightarrow 0$ would then yield $\frac{dQ}{dt} = \rho V C \frac{dT}{dt}$ which brings me close to the desired equation, but now the factor $\rho CV$ ended up outside the derivative, which is not what I wanted. How can moving this factor inside the derivative be justified in a mathematically sound way? Or, if that is not immediately possible, what do I have to change in my derivation to obtain the desired equation?
 A: For any system, we can always write infinitesimal changes in the internal energy $U$ as
$$dU=\left(\frac{\partial U}{\partial T}\right)_{V,N_i\in N}dT+\left(\frac{\partial U}{\partial V}\right)_{T,N_i\in N}dV+\Sigma_i\left(\frac{\partial U}{\partial N_i}\right)_{T,V}dN_i,$$
which is just partial derivative expansion into three parameter types (because we're considering three ways to add energy: heat, work, and mass). I'll tell you why I picked $(T,V,N)$ as natural variables (rather than $(S,P,N)$, for example) in a moment.
Since $dU=T\,dS-P\,dV+\Sigma_i \mu_iN_i$, we recognize the first term as $T(\partial S/\partial T)_{V,N_i\in N}\equiv C_V$, where $C_V$ is the constant-volume heat capacity.
(This is why I chose to expand in $(T,V,N)$: I don't know of another expansion that gives the simple $C_V\,dT$ term that your expression makes me think we're looking for.)
With the assumption of constant volume that you mentioned, we have
$$dU=C_V\,dT+\Sigma_i\left(\frac{\partial U}{\partial N_i}\right)_{T,V}dN_i,$$
which we can integrate to give
$$\int dU=U-U_0=\int_0^T C_V(T^\prime)\,dT^\prime+\Sigma_i\int_0^{N_i}\left(\frac{\partial U}{\partial N_i^\prime}\right)_{T,V}dN_i^\prime,$$
where $U_0$ is a reference energy. We can differentiate with respect to time:
$$\frac{\partial U}{\partial t}=\frac{\partial}{\partial t}\left[\int_0^T C_V(T^\prime)\,dT^\prime+\Sigma_i\int_0^{N_i}\left(\frac{\partial U}{\partial N_i^\prime}\right)_{T,V}dN_i^\prime\right].$$
At this point, I feel like we have to make some simplifying assumptions. For a (1) closed system, (2) a photon gas, in which the particle number $N$ is not conserved, or (3) a scenario in which the partial molar internal energies cancel out, the last term disappears. If we also assume a temperature-independent heat capacity (which precludes case (2) of the photon gas), we have
$$\frac{d U}{d t}=\frac{d}{d t}\left(C_V T\right),$$
which we can also write as
$$\frac{d U}{d t}=\frac{d}{d t}\left(\rho V c_V  T\right)=V\frac{\partial}{\partial t}\left(\rho c_V\Delta T\right),$$ where $c_V$ is the specific constant-volume heat capacity. This matches your expression, with the additional simplification that the constant $V$ can be pulled out of the time derivative. Note that in the simple case of a single-phase closed system of mass $m$, $\rho V=m$, and so the constant $\rho$ can be pulled out as well. I'm also have a hard time thinking of a simple situation where $c_V$ is temperature independent but changes over time. This leads me to think that the authors wrote an expression with the appearance of generality but really is equivalent to simply $\frac{dU}{dt}=\rho Vc_V\frac{dT}{dt}$. Then again, I don't know the complete context, which is why I started with a quite general framework.
