It would indeed be better to use 12v heaters. 3d printer heating pads (presumably 120 v) are just one possibility. I like the RV water tank heater pads.
Personally my goal is not exactly to extend the range, but to prevent wear/damage on the cells when charging them under freezing temperatures, also possible wear/damage when using lots of power under freezing temperatures, and finally and mostly to prevent the sudden loss of SoC I've been experiencing on my last three battery packs while driving on "Hill of Despair". At some point it's covered by the warranty but I reckon there's a limit to what lithium batteries can deliver when they're cold. And two warranty claims is enough for me
From what I gather, if we could keep the pack above 5 C when charging and driving, we'd be golden.
Don't forget this will be a fairly inefficient process.
You are trying to heat a large mass, about 300kg of cells with some convected air.
The heat pads will heat the metal outer case which is un-insulated.
The heat will warm the metal case & start to heat the air inside which will convect up towards the large battery mass.
Some heat will escape from the outer case back to atmosphere, so perhaps 2/3 will be useful unless you insulate the outer battery case.
The heat inside the battery case will start to warm the outer skin of the cells by convection, the outer skin of the cells then will conduct heat towards the centre of the cells vvveeerrryyy slowly.
Convected air continues to rise to the outer metal skin of the battery case which will heat the air above & probably start to warm the cars metal floor which could slightly warm the cars interior.
I'm guessing that about 1/3 of the heat energy actually gets into the cells.
This of course assumes the car is stationary in a garage with few drafts.
According to Mr Google :-
The specific heat capacity of a material is the amount of energy required to raise the temperature of 1 kilogram of that material by 1°C. For example, water has a specific heat capacity of 4,200 joules per kilogram per degree Celsius.
I would assume lithium cells would be denser than water but use water for the calculation.
So 300kg would require 4200 x 300 = 1,260,000 Joules.
Mr Google also says that :-
How to calculate joules to kWh?
A watt is one joule per second; a kilowatt is 1000 joules per second; a kilowatt-hour (kWh) is 1000 joules per second times 1 hour times 3600 seconds per hour, which equals 3,600,000 joules. To convert joules to kWh, divide the number of joules by 3,600,000.
1260000/3600000 = 0.35 kwh which ties in nicely with the 300w heat pads, but as only 1/3 ( guess ) of the heat gets to the cells it would need the heater to be on for 3 hours to heat the cells by 1 degree C.
Could be that Mr Google has thrown me duff information, but thought i'd just add my 2 peneth ( cents ) of working out.
I expect there are other more knowledgeable folk out there who can add to or correct this.