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.
You don't say what you use the batteries for, and if you are running your motor at full speed all the time. It's often better to have more voltage than you need and then modulate or regulate the current to run over a longer period of time. Then as the battery voltage decays you can go longer without recharging. A d.c. motor will rotate at a speed creating a counter voltage that equals the source voltage. Often motors are under some form of feedback, like a servo amplifier. I have controlled high inductance motors that put power back into the control amplifier. That extra energy needs to be put somewhere when slowing down the motor - it acts like a generator. Extra energy needs to be released as heat, or put back into the powering battery. You don't need any extra voltage as long as you can run at the speed you desire. Don't forget that voltage times current equals force, so can you usefully use extra torque? Do you mind accelerating faster than you have before? The insulation on a d.c. motor is at least twice the rated voltage, usually much more.
The constraints have more to do with force, top speed and control. Lighting a lamp with overvoltage will burn it out sooner, but that is not usually the main constraint in motors.
Best Answer
It depends on the battery climate temp. If sub 0’C use CCA , CA rating is sometimes 30% higher.
Since diesel has 22:1 compression ratio the start torque is much higher than a car at 8:1.
The proper way to choose a starting battery depends on the Battery/Motor ESR ratio under any condition as any drop in battery voltage is due to battery ESR and load current when starting to crank the pistons is due to the sum of both Battery ESR and starter ESR.
Battery ESR rises with Depth of Discharge rapidly above 80% and motor ESR rises due to the Reverse EMF of the motor rising with RPM. The motor rated current will stall at same voltage at about 8~10x rated current or ~ 1000A in your case.
simulate this circuit – Schematic created using CircuitLab
In Winterpeg, I had lots of CCA battery experience with cars and slow turning engines at -40'C and sometimes relied on battery and/or oil heaters or inline water heaters. Usually everyone had jump cables in the trunk for helping others that underestimated their battery needs and would willingly stop and help anyone. Occasionally to help themselves... For these climate conditions you always choose more CCA margin than you need in the summer.