No.
Your formula is quite off, and it starts with the nomenclature:
Watt
Watt is the unit of energy transfer which equals power.
The commonly used term "wattage" does not exist in science. It is a very despised shorthand only used in terms of electric power $P=UI$.
Both power $P$ (like work over time) and heat energy transfer $\Delta Q$ (which is one variant of power) use the unit $\text W=\frac {\text J} {\text s}$, which is confusing but a necessary distinction. Always remember that $P_\text{total}=\sum_{i=1}^nP_i$ - the total power in and out of an object is the sum of all partial powers!
Heat Energy transfer
The Heat energy transfer through an object is defined as the change of the heat energy $Q$ stored inside an object. $Q$ is given in $\text J$, so its change $\Delta Q$ is given in $\text J$ too. To get to the power, the energy change needs to be measured at several spots in time, so we make the derivate over time and get the power in $\text W$. We're looking at $\dot Q=\frac {\delta Q} {\delta t}$.
The absolute change of heat energy of an item is defined as $\Delta Q(t)=m(t) c \Delta T$: Increasing the temperature $T$ of an object with the mass $m$ and specific heat capacity $c$ by $\delta T$ (between times $t_0$ and $t$) results in a change of the stored energy by $\Delta Q(t-t_0)$.
So, we know $Q=c m \Delta T$ and $P=\dot Q=\frac \delta {\delta t} c m \Delta T$
Problem in question
We know that the drain (loss) of thermal energy from the system is via three ways:
- melting plastic (phase transition)
- extruding heated plastic
- convective heat loss to the air
- black body radiation of the heater block
We know that the total balance in equilibrium should be $P_\text{total}=P_\text{heating}+P_\text{melting}+P_\text{extrusion}+P_\text{convection}+P_\text{bb}=0$.
heat deposited into the system
Let's start at the simplest: we simply know the nominal heating power of the cartridge, it is usually written upon the cartridge itself, usually something in the area of 20 to 40 W. In praxis, it is not exactly that, but the ballpark fits. Otherwise, we'd plug in $P_\text{heating}=\epsilon \frac {U^2} {R}$ for our specific resistor, where $\epsilon$ is a coefficient between 0 and 1 telling us how good it is in converting electric to heat energy. Remember that since $U$ is technically a function of time (it is modulated to control heating behavior), our heating power also is, even though not explicit!
black body radiation loss $P_\text{bb}$
Black body radiation: $P_\text{bb}=A \sigma T^4$ where $A$ is the surface area of the object, $\sigma$ is a constant called Stefan-Boltzmann Constant. That much thermal energy is just lost due to radiation via photons, even if we don't see it glowing.
convection loss $P_\text{convection}$
The change of heat energy via heat convection is roughly defined as $H=\theta A (T-T_f)$ which brings us another coefficient $\theta$ about how good the block heats the air and the temperature of the medium (air) around $T_f$ - which we can replace as $(T-T_f)=\Delta T_a$.
And then we get to the biggest can of worms: the thermal heat transfer for melting the plastic and how much thermal energy is extruded from the system. For one of them, we can estimate some ballpark numbers, for the other, we will get into problems.
extrusion loss $P_\text{extrusion}$
The heat energy removed from the system by extruding plastic we can estimate from what we already established about thermal energy back in the Heat Energy transfer paragraph: $Q=mc(T_0+\Delta T)$ using the specific heat capacity $c [\frac {\text{J}}{\text kg K}] $ of the molten plastic as it is extruded (more about that later). But that's not the loss per time, but the heat energy stored in it in Joules. What factor is changing? In this case, it is the mass $m=r\times t$ where $r=\frac {\text kg} {\text s}$ is the extrusion rate. So $Q_\text{extrusion}=rtc\ \Delta T_\text{extrusion}$ and subsequently $P_\text{extrusion}=rc\ \Delta T_\text{extrusion}$
This leaves us with the big problem: as 0scar correctly pointed out by directing to the PolymerDatabase the specific heat capacity is not a constant and not linear but changes depending on the aggregate of the substance. We can make some estimate about it though from how we formulated the total power and adding a few absolutes for convenience:
$$P_\text{total}=P_\text{heating}+P_\text{melting}+P_\text{extrusion}+H_\text{convection}+P_\text{bb}=0$$
$$P_\text{heating}-H_\text{convection}-P_\text{bb}-P_\text{extrusion}=P_\text{melting}$$
$$\epsilon \frac {U^2}{R}-\theta A \Delta T_a-A \sigma T^4-rc\ \Delta T_\text{extrusion}=P_\text{melting}$$
Remember, that $U$ is a time-dependent factor (because of the control board activating it or disabling it), $T_f$ is also not a steady thing and changes depending on the airflow (though we can just pin it for our thought experiment) and thus $T$ itself might change over time as a result. $T$ is not equal to $\ \Delta T_\text{Extrusion}$ but is the temperature of the heater block system as a whole. $\ \Delta T_\text{Extrusion}$, on the contrary, is the temperature increase of the filament and not necessarily the same $\Delta T_{air}$, the differential between the heater block and the air. Why this differentiation is necessary becomes apparent if one realizes that the path of the filament might benefit from the heat that is lost from the heater block along that path, pre-heating the filament.
Phase Transition $P_\text{melting}$
$\propto$ is the proportionality sign and indicates that I might skip factors or constants.
What is that last part? That $P_\text{melting}$? It is the power of the Phase Transition. Matter does not shift between phases freely. There is energy stored in the state itself! So when transitioning from one phase to the other, that energy either has to be added (when going from solid to liquid or liquid to gas) or removed (when going the other way).
The "heat of fusion" is a material constant. For this look, I'll call it $\phi [\frac {\text{J}} {\text {g}}]$. We can make an estimation for the power that is put into melting the filament: there's an amount of filament that gets an amount of heat and undergoes the phase transition per time increment $$P_\text{melting}\propto \frac \delta {\delta t}\phi m_\text{melting}=\phi*\dot m$$ Now, we have the product of specific "latent heat" and melting mass derived over time... We had earlier the flow rate of material $m=r\times t$ and the "latent heat" is a constant. So, we pull out $\dot m=r$ again. So in the end we get that the power that is needed to melt our filament is proportional to the flow of the material and the material constant. $$P_\text{melting}\propto \phi r$$
Conclusion tl;dr
When eliminating the loss via convection and black body radiation and assuming them 0 or neglectable, we assume our heater is packed in perfect isolation - and call them losses. Assuming $\epsilon=1$ for a perfect heater, we are left with this equilibrium situation:
$\frac {U^2}{R}-P_\text{losses}=P_\text{melting}+rc\ \Delta T_\text{extrusion}$
The sum of the power of the phase transition (melting of the filament) and the energy stored in the extruded filament per time $(\frac{dQ}{dt})$ is equal to the energy deposited into the hotend over time [minus losses over time]
