# 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

$$$$\frac{\partial U}{\partial t} = \frac{\partial}{\partial t} \left( \rho V C T \right),$$$$

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?

• Since $\rho V C$ is independent of time, we have that $\frac{\partial (\rho V C T)}{\partial t}=\rho V C \frac{\partial T}{\partial t}$. Commented Sep 2, 2020 at 12:40
• How do you know that $\rho$ is constant? I did not specify this.
– SSB
Commented Sep 2, 2020 at 12:42
• Is $\rho$ = constant a necessary assumption?
– SSB
Commented Sep 2, 2020 at 12:46
• since in partial derivative all other variables are assumed constant @RaduMoga comment is valid Commented Sep 2, 2020 at 12:59
• I am sorry to say I fail to see you point @HariRamakrishnanSudhakar As far as I can tell, $\frac{\partial (\rho VCT)}{\partial t} = (\rho VC) \frac{\partial T}{\partial t} + T \frac{\partial (\rho VC)}{\partial t}$. Do you claim this is not generally correct? If it is generally correct, why would the last term be zero in this case?
– SSB
Commented Sep 2, 2020 at 13:25

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.

• Thank you very much for your answer, @chemomechanics. I can tell you that the context of the equation is a derivation of (a variant of) the heat equation where $C_V$ is not assumed constant, as is usually done in all other derivations I have found. Having to assume $C_V$ to be independent of $T$ makes this equation appear not so useful though.
– SSB
Commented Sep 2, 2020 at 20:48
• Or only minimally dependent on $T$! For example, a context that would seem to work is small changes in temperature for a system undergoing a pressure-driven reaction or phase change. Commented Sep 2, 2020 at 21:27