Boat Auxiliary battery – this simple ok

The amount of "loudness" you need is uncertain. but you can almost certainly get more than enough power from say 6 x AA cells and a Piezo sounder.

Ignoring your battery spec for the moment.

6 x 2500mAh AA Alkaline have about N x V x C = 6 x 1.2 x 2.5 = 16.5 Watt hours capacity. For rapid discharge lets allow 5 to 10 Watt hours.

10 minutes operation at even 5 Watt hours is 20 Watts input. That is vastly more power than you need and gives a sensibility check on the battery size.
6 x AA cells
= 6 x 14500
= 6 x 14mm dia x 50mm long
= about 30mmm x 45mm x 50mm envelope
= easily inside size spec.

6 x AAA give around half that
3 x AAA give around 1/4 that or say 1 to 4 Watt hours energy or maybe 5 Watts siren power.

Piezo sounders are relatively flat and available in as powerful as required sizes. Good acoustic coupling to the housing used is probably your main area of effort.

Electronics volume need is minimal

Fewer cells - lower voltage = harder to get power out. 6 x AAA = 9V start = 6V finish, when driving a piezo push pull give ~18V down to ~12V peak to peak output voltage.

A switch mode voltage booster MAY be needed. Component cost under $1 in manufacturing volumes.

**All the following are available from Digikey:

Piezo elements to investigate

Mallory Sonaalert - 105 dBA at one metre.Just apply DC and go. And similar

Probably too low powered. Mybe not

80 dBA at 5V RMS

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.