# In deriving the heat transfer equation, why do we use heat capacity at constant pressure?

I have seen many derivations of the heat transfer equation. It always has a form something like the following:

$$\rho C_{P} \frac{\partial T}{\partial t}-\nabla\cdot(k\nabla T)=\dot{q}_{V}$$ No matter how you write it, there is always a $C_{P}$ term, for specific heat capacity at constant pressure even though you are not necessarily considering an example that is at constant pressure. None of the derivations explain why they choose $C_{P}$ specifically either (they just say that one should use heat capacity, but why not $C_{V}$, for instance). Where does the constant pressure part come into play?

The underlying equation is the entropy equation $$\frac{\partial s}{\partial t } = \frac{1}{T}\nabla \cdot(\kappa\nabla T) + \ldots$$ where $s$ is the entropy density. In general, $s$ is a function of two variables, e.g. $T$ and $P$, and we have to make some assumption to reduce this result to an equation for $T$. It does not make sense to assume $n$ (this implies $V$) to be constant, because materials expand when heated. It makes sense to assume that $P$ is approximately constant, as long as the system is in mechanical equilibrium and fluid velocities are small.
When first law is written in terms of enthalpy you get \begin{align} \frac{dh}{dt}-v\frac{dp}{dt} & =\dot{q} \\ \rho C_p \frac{dT}{dt}-v\frac{dp}{dt} & =\dot{q} \\ \rho C_p \frac{dT}{dt}-v\frac{dp}{dt} & =\nabla \cdot(k~\nabla T)+\dot{q}_v \end{align} This form of the equation is preferred because mostly we deal with constant pressure processes occurring in open atmosphere, in which case you get your equation \begin{align} \rho C_p \frac{dT}{dt}-\nabla \cdot(k~\nabla T)=\dot{q}_v \end{align} However if pressure is not constant, a work term, $\dot{w}_v\equiv -v\frac{dp}{dt}$, must appear in the equation: \begin{align} \rho C_p \frac{dT}{dt}-\nabla \cdot(k~\nabla T)=\dot{q}_v-\dot{w}_v \end{align}
• Surely I can have an incompressible fluid that has different value for $C_{P}$ and $C_{V}$, right? So, shouldn't it matter? – Argon Jan 7 '17 at 3:41
• @Argon Be careful what you mean by "incompressible" here. "Incompressible flow" and "an incompressible material" are two different concepts. Does an "incompressible material" also have a zero coefficient of thermal expansion $\alpha$? If $\alpha = 0$, then $C_P$ and $C_V$ are necessarily equal. If not, they can be different. – alephzero Jan 7 '17 at 7:27