MrSoftware
- Joined Oct 29, 2013
- 2,015
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.
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.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.
Oops ...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.
4.19 J energy is required to warm 1 g of water by 1 °C.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.
And you expect me to remember that ? ? ?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.