Power consumption of filament extrusion I'd like to calculate the power lost through the filament being extruded (or in other words, at how many Watts I'd have to run an ideal heater that loses heat ONLY through filament so that it stays at constant temperature).

Power is defined in Watts as $\text W =\frac{\text J}{\text s}$
Specific heat capacity of a material is defined as $C =\frac{\text J}{ \text K \times\text{kg}}$

When extrusion happens, the filament of higher temperature leaves the hotend while the same weight of filament of lower temperature enters the hotend.
Let's say the specific heat capacity of the filament is $C$ and the extrusion rate $r$ is given with units kg/s. The temperatures are $T$.
Is it correct then to say that the power consumption of filament extrusion is $W = (c\times T_\text{Nozzle} - c\times T_\text{Environment}) \times r$
meaning that if I would run a heater cartridge at exactly &quot;W&quot; watts and extrude filament with rate &quot;r&quot; and the block would not loose heat through any other means than through the filament extrusion, then the nozzle temperature would stay constant?
This is very simply stated, in fact the specific heat is a function of temperature and state of the material (liquid or solid). Also you need to consider which type of specific heat you use, e.g. the one for constant volume $C_V$ or for constant pressure $C_P$. Constant pressure is probably preferred considering the mechanics of the printer (pressing filament into the nozzle-heatbreak assembly).

A very interesting source of information is the PolymerDatabase.com.

This source confirms that:


  In the case of polymers, we have to distinguish between the heat capacity of liquid, rubbery and glassy polymers. The heat capacity increases with increasing temperature, therefore, a liquid or rubbery polymer can hold more energy than a solid polymer. All materials show this increase in heat capacity with temperature.


also:


  Specific heat capacities as a function of temperature have been published for only a limited number of polymers. In many cases, the heat capacity (at constant pressure) as a function of temperature can be approximated by straight lines.


In such cases you can use the value of the specific heat at a predefined temperature (in thermodynamics that is frequently 298 K) to get approximations for your thermoplastic material. These formulae can then be used to integrate over the temperature rise.

Please remember that a cartridge is of a certain value of Watt; to have a lower power, the cartridge modulates voltage to keep the heating block within a predefined temperature range.