First, it makes sense to determine the temperature of your tap water.
Suppose the temperature of tap water is 25℃ and it is heated to 90℃, then the temperature difference is 90-25 = 65℃
The specific heat capacity of water is 4.2x1000[J/Kg·℃)]
65 times 4.2x1000[J/Kg·℃)]=The amount of heat required to heat a kilogram of water from 25℃ to 90℃ (ignoring energy loss, i.e. under ideal conditions) = 273KJ/Kg
One kilowatt hour of electricity = 1KW·h = 3600KJ
3600/273=13.20Kg, that's the answer
As for the amount needed, the time depends on the amount of water you boil. Suppose you want to boil n kilograms of water, then
9KW=9KJ/s
273×n÷9=30.33×n seconds
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The calculations above are all theoretical. In real conditions, the energy loss must be taken into account, i.e. the boiling rateof water is slower than calculated. or water boiled by one degree of electricity is less than calculated.
Purely hand-made, I hope you can adopt it~~~~~