Question: I'm trying to design a liquid fueled rocket engine, with copper walls cooled by kerosene.

While designing it, the following formula comes out:

$$Q=qA=wc(T-T(0))$$ 
Where:

* Q = total heat transferred
* q = average heat transfer rate
* A = heat transfer area
* w = coolant flow rate
* c = specific heat of coolant
* T = temperature of coolant leaving the engine
* T(i) = temperature of coolant entering the engine

This formula gives the total heat transferred from the combustion chamber to the cooling liquid.

To proceed with my calculations, I need to know the value of **"q"**, known as "**overall heat transfer coefficient**" or "**average heat transfer rate**", between a copper chamber and the coolant (kerosene).

I need this number to find w:

$$w =\frac{Q}{c-(T-T(i))}$$

"A", from the first formula, is known: "q" is needed to find the value of Q and finally find the actual cooling liquid flow rate

I'm currently working with Btu, lb, sec, in as units

Thank you in advance

Answer: 
This is essentially the *definition* of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically.

If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is defined to be the eigenvalue of the Hamiltonian. And we have
$$H \Psi(x, t) = i \hbar \frac{\partial}{\partial t} \Psi(x, t) = i \frac{\partial}{\partial t} (\psi(x) \phi(t)) = i \hbar \left(- \frac{iC}{\hbar}\right) \Psi(x, t) 
 = C \Psi(x, t).$$
So $C$ is the energy. 

Of course, this just reduces the question to "why is the time-dependent Schrodinger equation true", and the answer is, again, that in quantum mechanics energy is $\hbar$ times the rate of change in phase.
The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities.

In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E\_K+V$ in standard systems) you know that
$$\dot{A} = \{ A, H \} +\frac{ \partial A }{\partial t}$$
with $\{ a,b\}$ the Poisson brackets. Following quantization rules we replace the Poisson brackets with commutation relations (divided by $i\hbar$), and all quantities in operators, to get
$$\frac{d}{dt} \langle A \rangle = -\frac{i}{\hbar}\langle \left[ A, H \right] \rangle + \frac{ \partial }{\partial t}\langle A \rangle$$
and you can see how $H$ here plays the role of the energy. This is just the Heisenberg picture, which is equivalent to the Schroedinger picture, so we can identify the term as a Hamiltonian, which measures the energy.

Another (closely related) way to see this: start from the fundamental relations of QM $[x,p]=i\hbar$, from which you can derive $[x, f(p)]=i\hbar \partial\_p f$ and $[f(x),p]=i\hbar \partial\_x f$. Then you take Hamilton equations from classical mechanics and quantize them

$$ \frac{d}{dt}\langle x \rangle = \langle \frac{\partial H}{\partial p} \rangle = -\frac{i}{\hbar}\langle \left[x, H\right]\rangle$$
and again you got Heisenberg picture, from which you can derive Schroedinger equation, with the $H$ the energy as it comes from classical mechanics.