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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 base and receives 120V-220V (depending on the kettle).

To boil water from room temp (20°C) to boiling point (100°C), I calculated the power to be:

M=100ml ~ 100g

C= 4.186 J/g C

T = 80°C

Q=MCT

= 100 grams * (100°C - 20°C) * 4.186 J/g C = 33488 J = 33.488 kJ

I've been told that once the water is at 100°C, still more energy must be added.

This is given by the heat of vaporization, which for water is 40.7 kJ / mol or 2261 Joules per gram.

once the 100 grams of water is at 100 °C, this amount of energy must be added to boil it:

100 g * 2261 Joules/gram = 226100 Joules = 226.1 kJ

so in total 226 + 33.5 = 259.5 kj (260kJ for simplicity) to boil 100mL of room temp water.

Is this correct? if yes, how do i calculate the amount of batteries from knowing that i need to generate 260kJ and boil water?

i am trying to make it so i don't use the power outlet but fully portable and just use rechargeable batteries to boil water

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# [Help] battery power to boil water

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