Question: A typical 12-V car 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 deg celsius) that could be boiled into steam (100 deg celsius) using energy from this battery.
My Thoughts:
I know V = EPE/q is supposed to utilized in this problem. i tried to substitute some equations into the problem, but they don't seem to fit!! i want to get a mass amount...i just don't know how to do that. someone help me out by showing me how to do the problem! appreciate all the help i can get.
Book Answer: 4 kg (just for checking purposes!)
(i will award a best answer - i promise)
2007-01-27 15:41:39 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You need a statement of energy relating voltage and charge.
E = QV
Energy = charge * voltage
Then you need to find out the heat of vaporization needed per Kg to boil water and since heat is energy multiply it out, keeping units straight.
(battery joules)* (calories/joule)*(Kg/calorie) = Kg
Hope that makes sense.
2007-01-27 16:26:15 · answer #1 · answered by themountainviewguy 4 · 0⤊ 0⤋
The total energy is E=VQ=12V*7.5E5C = 9.0E6 J
The latent heat of water is L= 2272 J/g for vaporization at 100 C.
The mass of water vaporized is E/L = 9.0E6J/2272 J/g = 4.0E3g=4.0 kg.
Note that you have only two sig figs so the final answer must be to that precision.
2007-01-28 00:02:13 · answer #2 · answered by d/dx+d/dy+d/dz 6 · 0⤊ 0⤋