Assume the power input to the bulb is 10 Watts.
Assume for now 100% efficiency from battery output to bulb input.
Efficiency of energy storage by the battery of energy supplied to it will vary with battery chemistry and how well the charger is designed. Best case using a Lithium battery of some sort, over 90% efficiency may be achievable. Lower or much lower efficiency is often achieved in practice.
Efficiency of energy provided at the battery terminals compared with energy out of the PV panel will depend on the interface design and will also vary with battery state of charge.
Power output from the panel at any moment (Wp) and compared to the maximum power the panel can make under ideal conditions (Wmpp) will vary with insolation level (sunshine level), panel conditions, atmospheric conditions and more.
SO overall, a say 100 Watt panel will produce 100 Watts in full sunshine when new and will produce the equivalent of 2 or 3 hours of equivalent sunshine in most continental US locations in winter and 5 to 6 hours of equivalent full sunshine.
ie you get 200 to 700 Watt-hours per day depending on season.
With the very best interface equipment (MPPT, intelligent battery sizing to minimise resistive losses, ... you may get 95% + of this energy at the battery terminals and, as above, 90%+ of this actually stored into the battery.
So PV Watts rating x 0.95 x 0.9 x hours_equivalent_per_day = Watt-hours available. Say 85%. Using 80% would be safer and still very optimistic in many cases.
At the start I assumed 100% battery out to bulb in power.
Regardless of load type (which is usually LED in this context), if you want constant brightness as battery varies or constant "bulb" input there will be some conversion losses. 90% from battery to bulb or LED would usually be excellent.
So overall PV "nameplate rating" watt-hours to 'bulb' input watt-hours is at best about 75%. Usually less.
When the sun is providing energy, some gains can be had by running the bulb from the panel without battery storage. This gain is useful but still a small part of the total energy needed via the battery. I'll ignore it in the following and it can be factored in later if needed.
From the above:
Watt hours available = (Panel Watts rated) x 75% x Sunshine hours.
Watt hours wanted = Load_Watts x 24.
Rearranging the above -
Panel Watts needed = Load Watts x 24 / (0.75 x Sunshine hours )
= Load_Watts x 32 / Sunshine_Hours
So eg 10 Watt load in winter with 2 hours/day sunshine hours /day (= equivalent full sunshine).
Panel Watts needed = 10 x 32 / 2 = 160 Watts !!!
10 Watt load in Summer with 6 sunshine hours/day.
Panel watts needed = 10 x 32 / 6 = 53 Watts.
In practice higher Watts will be needed.
Averge sunshine hours per day can be found at the wonderful Gaisma site here - this example is for Houston
Top line is insoltaion in kWh/m^2/day = sunshine hours/day = hours of equivalent full sunshine. I = January, II = February etc.
2.34 hours/day in January.
5.98 hours/day in July
These are means for many years and any year and any day in the montyh may vary widely from this. That's weather for you :-)
More later ...
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.
Best Answer
I didn't see if you are using a 120 V bulb for an AC lamp. I assume it's an incandescent bulb. If this is the case you will need to step up the voltage .
The filament they use glows white hot to produce light (and heat) which is wasteful. I agree with @Dampmaskin that LEDs are more efficient.