# Homework Help: [Help] battery power to boil water

1. Dec 2, 2013

### Inventor2014

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 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

2. Dec 3, 2013

### Staff: Mentor

Your estimates assume no loses of heat - so they are too low.

Power you draw from a battery is UI - potential in volts times current in amperes. Assume your battery works at 12 volts (or whatever) and is capable of delivering 1 A (or whatever), see for how long it has to work to deliver 260 kJ.

Do you want to boil the water out, or just to heat it to boiling (as in making coffee?) If it is the latter, you need only 33 kJ, plus some more to see it started to boil, but definitely not 260 kJ (after that you have no water left).

Using battery for boiling water is one of the most inefficient things that can be done with battery, any burner will be better. It is a matter of energy density.

Note: while technically not a homework, it is a homework like problem. I am moving it to a correct sub forum.

3. Dec 4, 2013

### Inventor2014

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?

edit:
i got the kj it takes to boil water
how do i find the kj each battery has if i have the voltage and the capacity (mAh) given?
this way it will answer my initial question of how many batteries would be needed depending on the battery i use

Last edited: Dec 5, 2013
4. Dec 5, 2013

V*A=W, W*s=J

5. Dec 5, 2013

### Inventor2014

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

6. Dec 5, 2013

### Staff: Mentor

You math is OK. I would go a slightly different route - Ah means 1A for 3600 seconds, so 3.4 Ah at 3.7 V means 3.4*3600*3.7=45288 J, which is 45.3 kJ that you got.

I told you heating water with batteries won't work.

7. Dec 5, 2013

### Inventor2014

Sorry if im 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)?

8. Dec 5, 2013

### Staff: Mentor

If you can store the boiled water in a well insulated container, you need very little energy to maintain it at boiling point. A near-perfect insulator such as a vacuum flask will keep water very hot for a long time without a need for constantly adding energy.

For camping you can buy an immersion resistance heater that operates off a car battery and will boil a cup of water without flattening the battery much.

9. Dec 5, 2013