Battery heating can be roughly defined using its internal resistance. We can get a rough estimate of a battery's internal resistance using it's cold cranking amp rating. This is rated at -18C, which is close to -20C.
For a nominal 12V battery:
$$
R = \frac{12V - 7.2V}{CCA}
$$
Let's assume this is constant vs. current draw.
Let's assume at time t=0, the battery is completely insulated from the world and has a heat capacity similar to pure water.
The amount of energy required to heat 17 kg of water from say even -20C to -18C is:
$$
E = 4.2 \frac{J}{g K} \cdot 17 kg \cdot (20^oC - 18^oC) = 142.8 kJ
$$
For a 9A draw, this would take:
$$
t = \frac{E}{(9A)^2 \cdot R}
$$
Suppose you have a 200 CCA battery. Then we can find t = 73 seconds.
Problems with this analysis: Obviously the full weight of the battery is not just the liquid inside. Thus the time required to heat the battery 2 degrees C would be less. However, a larger problem is that you probably don't have a perfectly insulated battery. The 9A draw only produces ~2W, which could easily be lost from the battery (the cables are extremely good conductors away from the battery).
At temperatures much colder you're going to have to start worrying about fluids freezing, thus many things could just stop working. You're much better off following proper car winterization guide.
I think batteries usually follow a CC/CV or constant current/constant voltage system to charge up batteries. The LiPoly batteries need such a mechanism. So, this can be achieved by a system that sends out PWMs that control the current and voltage in a more controlled manner. Ofcourse lead acid charging will be different. But please read up on that as I believe you are bypassing that.
And yes over-charging is definetily an issue in your case. You can ofcourse have a dedicated IC or even a combination of op-amp/comparators to do the job for you. When the voltage in the battery reaches a specific value then have the comparator/opamp circuitry turn off or isolate it via a MOSFET(again check the current rating and heat dissipation here).
What you are doing is blindly charging the battery which can very well end up in a pile of ash.
Best Answer
Charging Mode:
Lead Acid current charge efficiency is high - say 90% +
BUT Wattage charge efficiency is lower.
Data sheet says max charge current is 60 A. That would be at from about 12V to say 15V.
At that high rate the power would be about 15V x 60A = 900 Watts.
You can charge at substantially lower than that. eg 20A is the nominal 10 hour rate and power would be 12 to 15V x 20A = 240 - 300 Watt range.
Slower again is possible, but very low rates may not fully charge battery. A higher rate may be needed to achieve a topping or boost charge 'at the end' needed to equalise all cells.
Thermal power dissipation would be expected to to in the 10% - 30% range of input power and vary across the cycle. The energy input to replace output would probably be 110%-130% depending on various factors.
Boost charge serves mainly to equalise cells - so current should be low but efficiency may not be good. BUT ...
Float "Charging"
is not really "charging" at all - it is maintaining the battery in charged state and only needs to make up self discharge losses so should be very low.
For a self discharge of say 10% of capacity in a month (high) you'd need a makeup charge of
200 Ah x 10% / (30 days x 24 hours) ~+ 30 mA
and 360 mW for a 12V battery.
So "a few Watts" seems a good real world answer.
Easiest is to measure float current and see.
Cyclical & Standby use:
Note that two float voltage ranges are specified by the manufacturer.
14.4 - 14.7V for deepish discharge use. and
13.6 - 13.8V for standby or "float" use.
The former is for where the battery frequently does substantial work and is then recharged after each use. The latter is for when the battery is seldom used to work but "lurks" in full charge mode waiting the time when it will be needed.
Discharge formulae:
I take the two formulae given to mean as below.
1.65V is "rather low" and should only be used at high rates where IR voltage drops account for quite a lot of the drop and actual cell voltage is higher.
Where did you get those formulae.
The battery will not last 12 years if discharged in that manner frequently.
Can be discharged from fully charged at a constant 337 amps for 15 mins to an end point voltage of 1.65V per cell.
Can be discharged from fully charged at a constant 597 Watts for 15 mins to an end point voltage of 1.65V per cell.
For interest:
337A x 15/60 h x (2.25 + 1.65)/2 Vavg x 6 cells = 985 Wh.
597 W x 15/60 h x 6 cells = 895 Wh
For 337A discharge to yield 895 Wh the average voltage per cell = 1.77V (!).