Sorry if I am being stubborn but i want to understand why it won't work
40.8A is too much for a little battery like that i assume and probably will kill it fast?
how about keep the water warm (after it was boiled with a kettle)?
so if my battery is:
3.4 Ah, 3.7V - if i run it for 5min (300s) straight it would give out 40.8 A and that results in 45.3 kJ.
So if water requires 167.4kJ, i will need about 4 of these batteries to accomplish 80C temp increase? that sounds wrong to me
E = [500 grams * (80 C) * 4.186 J/g C ] = 167440.0 so that is for the energy to heat the water.
I just want to heat it up and leave it, like say for tea.
my main question is, what battery capacity and voltage would be best to use to heat up water by 10C and by 80C, how do you calculate that...
Hello everyone,
I'm stuck at a basic physics challenge that I randomly thought on the drive home from work.
The quick question is: how do i find out the amount of rechargeable batteries needed to boil water from room temp (20°C)?
My research:
A kettle uses a power cord connected to the...