It might be worth your effort to match the impedance of whatever load you come up with, to the impedance of the solar panels. This will maximize your power transfer.

 

Thank you!! That could work. As mentioned, I only need a small change of temprature to prove that something actually happens. I am so excited to try.

Another possibility is a small immersion heater, of the kind that are used to boil water in a mug such as for making tea. A unit rated for 1000 watts at 110 volts will have about 12 ohms of resistance, which is about right for getting 3.5 watts at 6 volts. Something like this.

 

Check your math. 6 volts divided by 0.5 ohms gives 12 amps of current. 12 amps times 6 volts is 72 watts, not 3.5 watts.

Oops ...
A 6 Volt 3.5 Watt Solar Panel could deliver 1/2 Amp, not a 1/2 Ohm resistor, as I previously wrote

Using the correct math ...
A 5 Watt, 12 ohm resistor / heater = 6 Volts / 1/2 amp ( from PV Panel )

 

1000W ÷ 110V = 9.1A
110V ÷ 9.1A = 12Ω

(@12V 20W PVP)
12V ÷ 12Ω = 1A
12V x 1A = 12W (watts)

(@6V 10W PVP)
6V ÷ 12Ω = 1/2A (0.5A or 500mA)
6V x 500mA = 3W (watts)

Is that enough to heat the water 1 to 5˚ C? I don't know how quickly (or long time) it will take. That's an equation I don't know. But if you ARE putting energy into a body of water then it is going to be warming up. But at the same time you'll be battling the loss of heat energy through the container or open air. In theory a candle can heat a warehouse "eventually". But in practicality, the warehouse will be losing heat through many different means. WILL a candle heat a warehouse? No. But in theory, and if you ignore all losses then the answer becomes yes.

So the questions I think need answering are "What kind of container?" "How much insulation?" "How long will it take to achieve the goal of one to five degrees celsius change (upwards)?"

 

Is that enough to heat the water 1 to 5˚ C? I don't know how quickly (or long time) it will take. That's an equation I don't know.

4.19 J energy is required to warm 1 g of water by 1 °C.
1 J = 1 W*s.
Then warming 100 g of water by 1 °C required 419 W*s energy.
So with 3.5 W power 100 g of water will warm by 1 °C every 419 / 3.5 = 119.7 s ≈ 2 min.

 

4.19 J energy is required to warm 1 g of water by 1 °C.
1 J = 1 W*s.
Then warming 100 g of water by 1 °C required 419 W*s energy.
So with 3.5 W power 100 g of water will warm by 1 °C every 419 / 3.5 = 119.7 s ≈ 2 min.

And you expect me to remember that ? ? ?

OK, so 2 to 10 minutes, barring any losses to ambient temperatures.

 

I would use a shot glass and put the resistor beneath it, inside some house insulation, then put some plastic over it with an electronic thermometer to test measure, time the experiment when it reaches full heat potential. Calculate how many ounces in a Gal to give a final Temp x Time. Make sure you test the beginning Temp.

kv