## a typical 12v batter can deliver about 7.5 x 10^5 C...

PROBLEM: A typical 12v battery can deliver about 7.5 x 10^5 C of charge before dying. This is not very much. To get a feel for this calculate the maximum number of kilograms of water (100 degrees celsius) that could be boiled into steam (100degrees celsius) using energy from this battery.

I think I'm supposed to use the formula, E= qv, but I dont know what to do after that.
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 Using E=qv, your energy will be in units of Joules. How many Joules does it take to raise 1 kilogram of water 1 degree Celcius? Edit: Oops. After re-reading the question, I see that the intent is to bring liquid water from 100C to vapor at 100C. For this, look up Latent heat of vaporization of water.

 Quote by Phrak Using E=qv, your energy will be in units of Joules. How many Joules does it take to raise 1 kilogram of water 1 degree Celcius? Edit: Oops. After re-reading the question, I see that the intent is to bring liquid water from 100C to vapor at 100C. For this, look up Latent heat of vaporization of water.
Thanks for responding. So all I have to do is look up that value for the latent heat of vaporization and that would be my final answer...I don't have to use any formula to calculate anything? I looked up the value and for 100 degrees C, the latent heat is 2260 kj/kg...would that be all?

## a typical 12v batter can deliver about 7.5 x 10^5 C...

Read the question carefully, then:-

So what you've got, is that you need 2260kJ of energy to change each kg of boiling water into steam.

How many joules of energy do you have available in the battery?

How does that help you answer the question that was asked.

 Tags car battery, couloumbs, energy, volts

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