watts
I'm trying to understand the difference between a power output in watts and a power capacity in watts, and I'm open to the possibility that I'm thinking about it all wrong.
If I want to charge a battery with a 2 watt-hour capacity, how long is it going to take me if I'm using a power source with a continuous output of 2 watts?
Your 75W fan required 220 Volts, it will not operate at typical battery voltages such as 12V.
A 75W fan operating (indirectly) from a 12V supply would draw 6.25 A not 0.34 A.
Even with a theoretical 100% efficient invertor, the current on the 12V side is greater than the current on the 220V side to deliver the same power across the system from battery to Fan.
Note that the reason that power is delivered through the grid at high voltage is to keep currents low. losses and cable sizes are proportional to (a function of) current. High V are used to keep currents low.
You convert 12V DC into 220 V AC using your inverter. Depending on capacity and utilisation the efficiency of an inverter might be the 80% you estimate or it may be worse - 50% Ref So your 6.25A may become 13A drawn from the 12V battery.
Therefore your 300 Ah battery-set might power a 75W fan for less than 24 hours.
Update: for 2 x nominal 12V SLA battery, the following diagram may clarify what is happening.
simulate this circuit – Schematic created using CircuitLab
On a very basic level you have it correct(or at least a reasonable estimate), but we can get more detailed and in depth.
First of all, 12v * 40 Ah gives 480 Wh NOT watts. Watt-hours, I.E. 480 watts for 1 hour or 240 watts for 2 hours etc. A technicality that doesn't change your calculations, but might make future calculations less confusing.
Your first assumption about lead acid batteries and not discharging more than 50-60% is, in general, a correct assumption. However, you call it a deep cycle battery, and a good quality deep cycle lead acid battery may tolerate up to 80% depth of discharge. Less discharge is always better, but the better deep cycle batteries do allow a lower discharge and still have a reasonable cycle life. The only way to know is to look at the data sheet from the manufacturer.
Your second assumption is that your battery will simply charge in a linear fashion. Although your calculations for how long it will take to charge your battery are an OK estimate, in reality a lead acid battery charges to 70% in 5-7 hours (if your solar panel can provide enough current) and the remaining 30% in the following 7-10 hours. Lead acid batteries follow a 3 stage charge algorithm (maybe 4 stages depending on your charge controller).
So in your battery if you discharged it 75% = 10a remaining. To get up to 70% need 18a (10+18=28a). 18A÷5h=3.6a/h which your panel is capable of and you said you have at least 6 hours of sunlight and it will take 5 hours to 70% so in the last hour the battery will continue to charge past 70%. So if you only have 6 hours of sunlight and your controller strictly follows the 5 hours to charge to 70% algorithm, then you will not fully charge the battery each day IF you discharge to 75%! In all likelihood your charge controller will take advantage of your panel's ability to provide more than the 3.6 amps and charge the battery a little quicker, but not much - there are limits.
Finally your router. Does it use AC power? If so, then you need an inverter which will convert the battery power to 120 volt AC which you will plug in your router. All inverters are somewhat inefficient and you will lose at least 15% of your solar power to the inverter. So add this to your calculation and you will find out how long your router will be powered.
Best Answer
"If I want to charge a battery with a 2 watt capacity, how long is it going to take me if I'm using a power source with a continuous output of 2 watts?"
There is no way to tell. The maximum instantaneous power output (2 W in your case) doesn't tell us how much total energy the battery can hold. For example, if this battery can only sustain this 2 W output for 15 minutes before it goes dead, then it can hold 1/2 Watt-hours, which is 1.8 kJ. If it can sustain 2 W output for 10 hours, then it can hold 20 Watt-hours, or 72 kJ.
You are charging the battery at 2 W, which is the same as saying you are transferring 2 J of energy into the battery every second. Obviously it will take a lot longer to transfer 72 kJ of energy at 2 J/s than it will take to transfer 1.8 kJ at 2 J/s.
On top of that, batteries have inefficiencies. If this battery is 75% efficient, for example, then you need to give it 1/75% = 1.33 times more energy during charging than it can ultimately hold and return to you during discharging.