Electronic – Determining a solar panel, is this correct

Using a 120 VAC inverter and a light bulb to keep your pipes from freezing is just a dumb idea.

If you need 40W worth of heat in an enclosed space, just select a bunch of power resistors that add up to that amount, screw them to a heatsink and run them directly from your 12 V battery. For example, a 15-ohm resistor will dissipate about 10W @ 12V, use four of them.

You can also get CFL lamps that run directly from 12V.

Let's say you need a run time of 48 hours, and you have a minimum of 6 hours to recharge.

Your load is drawing a daily average of 3.33 (heater) + 5/24 × 1 (lights) = about 3.5 A. Multiply by 48 hours to get the battery capacity: 170 Ah

Batteries aren't perfectly efficient, so it'll take about 170 / 0.8 = 212.5 Ah to recharge it.

If you want to drive the load and recharge the battery in 6 hours, you need a panel capacity of about 40 A @ 12 V, or 480 W.

If you want NEVER run out of battery, you should come from ALWAYS extreme cloudy condition at shortest day, because weather statistics cannot guarantee cloudy time duration limits.

Note, solar battery manufacturer gives power output for very optimistic conditions, say clear day and battery oriented directly to sun.

Daylight wikipedia article gives you range of conditions you'd expect. According to it, in extreme cloudy day there's less than 0.002 of max. light intensity. Add here 40% daylight time. So you will need rated power of ~2000 times consumed to work forever (1 KW solar battery for 0.5 W device).

So you have to define less strict requirement to get reasonable values, say 10% per year failure rate. Then I expect about 10-20× solar power, say 5-10 W solar panel and larger accumulator (week of autonomous power). But accurate value require much statistical calculations or experiments.