Energy in the heat equation. Before getting to question, some background. Let $u(x,t)$ be the temperature in a laterally insulated rod of length $L$, at position $x$ and time $t$. The temperature satisfies the heat equation
$\partial_t u = \alpha \, \partial^2_x u$,
where $\alpha > 0$, thermal diffusivity of the rod, with Dirichlet (zero) boundary conditions say.
There are at least 2 math.stackexchange.com questions that involve the so-called Energy  integral
$E(t) = \int_0^L u(x,t)^2 dx$.
Here they are:
Energy for the 1D Heat Equation
and
Heat equation and energy
Now my question: Why does $E(t)$ represent energy? It does not have the units of energy! It's well-known if you look at any derivation of heat equation or if you know just basic thermodynamics that 
$c \rho u(x,t)$ 
does equal energy per unit length, where $c=$ specific heat, $\rho=$ density (so $c\rho =$ heat capacity of material). $\therefore$ the true energy of the bar equals (assuming $c$ and $\rho$ are constants)
$c \rho \int_0^L u(x,t)\, dx =$ true energy of rod.
Omitting constants, it would be OK to say
$\int_0^L u(x,t)\, dx=$ energy.
But, it is false to call $E(t) = \int_0^L u(x,t)^2 dx$ the energy of the bar. 
I know that people introduce $E(t)$ to prove uniqueness of heat equation (I've seen this proof many times) but why do teachers call $E(t)$ the energy when it is not? I do see that $E(t)$ does have resemblance to Kinetic energy (like $(1/2) m v^2$ with a squared term) but in the context of heat, $E(t)$ certainly does not equal energy, so why do teachers call it energy? Are they disingenuous? My final question then is, since $E(t)$ is not the true energy, what physical quantity does it represent?
 A: Nearly a year ago, I gave in the comments a non-answer that would justify calling the quantity
$$
\int u(x,t)^2~dx = E(t)
$$
the energy of the rod. Namely, one can think of minimizing the energy functional as satisfying a conservation law, and thus minimizing the energy leads you to solutions of the associated partial differential equations.
One year out, I think I have a more concrete, dimensional-analysis based answer. Let us examine the equation
$$
\partial_t u = -\alpha\partial_{xx}u
$$
for its dimensions. $u = u(x,t)$ represents temperature at the spacetime coordinate $(x,t)$, so it has units of temperature $T$. $\alpha$ here is the thermal diffusivity, which has units of length squared over time, $L^2/\tau$.  Finally, not in the equation but lurking in the background is the kinetic energy, which is related to temperature by the equation
$$
E_k = \frac{3}{2}kT,
$$
where $k$ is the Boltzmann constant and has units of energy over temperature, $E/T$.
Let me use $[unit]$ to represent the dimensions of any units in any expression. So for example, the units in the heat equation check out: $\partial_t u$ has units of temperature over time, while $\partial_xx u$ has units of temperature over length squared, which gives us
$$
[T\tau^{-1}] = [L^2\tau^{-1}][TL^{-2}].
$$
Now, in our scenario the total length of the rod is fixed, so without loss of generality we may treat it as a constant. Dimensionally, this is saying we will take $[L] = [1]$, so it drops out of any of our equations. This is all fine, as we see in the heat equation the units of length cancel each other anyway.
So let us turn to the energy functional. In terms of units,
$$
\int u(x,t)^2~dx = \int [T^2]~d[L] = [T^2L] = [T^2].
$$
So the energy functional has units $[T^2]$, which isn't where we want to be (we want units of energy). However, the primary thing we do with taking the energy functional is minimize it. Minimizing the energy functional $E(t)$ is equivalent to minimizing the square root of the energy functional, $\sqrt{E(t)}$, which has units of $[T]$.
Still not there. But aha! Energy and temperature, $[E]$ and $[T]$, are related by a proportionality law! That is,the equation
$$
E_k = \frac{3}{2}kT
$$
tells us precisely how to convert from temperature to energy, and the conversion preserves order. So if we can minimize the square root of the "energy functional" $\sqrt{E(t)}$, which has units of $T$, then we automatically know how to minimize the actual energy, which according to our dimensional analysis must be something like
$$
\sqrt{\int \frac{9}{4}k^2 u(x,t)^2~dx } = \frac{3k}{2}\sqrt{E(t)},
$$
where the right-hand side is actually integrating the physical notion of energy squared. And now, the units do indeed check out: these are quantities with units of energy. There are surely other ways to define a natural notion of a physical energy functional using an integral, but the units of this definition work and it has nice mathematical properties (pretty much exactly those of $\sqrt{E(t)}$, which we know has an excellent mathematical theory). This revolves around the observation that since $k$ is a constant, minimizing the "energy" functional defined using units of heat is equivalent to minimizing what you would expect to be the "physical energy" functional.
Morally, the fact that $k$ is a constant (despite having units of $[ET^{-1}]$) gives us a temperature-energy equivalence law, which tells us that for dimensional analysis purposes $[E]$ and $[T]$ are indistinguishable. This is what the physicist in me needs to say to satisfy himself. The mathematician in me chooses units for temperature and energy so that $k = 2/3$, and then I just have the simple equivalence $E_k = T$, and now I happily exchange temperature and energy at will. And presumably this is what mathematicians of the past have done.
Treating $L$ like a constant also needs some justification, but I think if we think of $L$ as very small and argue that computing the energy of large homogeneous bodies consists of computing the energy on small pieces and summing, then we can justify that assumption as well.
A: I have been thinking about this question for a bit and I think it might be helpful to consider a different context in which the heat equation might appear.
The heat equation is a special case of a reaction-diffusion equation, which describes the diffusion of a cloud of particles, for example. In this case $u$ describes the concentration of particles, i.e. the density of mass at a certain point. Then what you call "energy" corresponds to the total mass of particles, that is,
$$
M[u]=\int_{\Omega}u(x)\mathrm{d} x,
$$
and if you consider a closed system, which means we impose Neumann boundary conditions, then the conservation of "energy" translates to the conservation of total mass, that is, solutions of the heat equation satisfy $\frac{d}{dt}M[u(\,\cdot\,,t)]=0$. This way one can argue that it is not always natural to call $\int_{\Omega}u(x)\mathrm{d} x$ "energy".
In applications it is quite common to not impose Neumann conditions and often the total mass is not the important quantity to study and the functional
$$
E[u]=\int_\Omega u^2(x)\mathrm{d} x
$$
is more relevant.
The Partial Differential Equations community calls $E$ the energy functional for the heat equation, but it is known that it does not actually describe "energy" in the physical sense. My guess for why $E$ is called energy is that it measures the distance (in $L^2$-norm) from the trivial equilibrium solution $u=0$. The expression is reminiscent of the elastic potential energy of a spring or the gravitational potential energy of a rod suspended in the air.
The energy functional $E$ also has a clear relation with the Dirichlet energy $\int_\Omega(\partial_x u)^2$, for which solutions of the Laplace equation $\partial_{xx}u=0$ are minimizers.
For solutions of the heat equation it holds that
$$
\frac{d}{dt}E[u(\,\cdot\,,t)]=\frac{1}{2}\int_\Omega u(x,t)u_t(x,t)\mathrm{d}x=-\frac{1}{2}\int_\Omega (\partial_xu(x,t))^2\mathrm{d}x.
$$
Integrating over time tells us that $E[u(\,\cdot\,,t)]$ describes the accumulated loss of Dirichlet energy as time progresses.
