Physics help please?

Question

An electric filament can heat 200 cm^3 of water from 5 degrees celcius to 95 degrees celcius in 6 minutes. If the voltage source is a 12 V battery, how much current does it draw?

Answer 1

Ok first we want to know how much heat is involved, from the equation:
Energy(heat)= mass*specific heat*Change in Temperature
mass is in kilograms, specific heat of water is 4184 Joules per kG per degree Celsius
200cm^3 of water should be converted to kG. 1 gram of water is approximately 1 mL. 1 cm^3=1 mL, so here we have 200mL, or 200 grams. (which is 1/5 of a kG, or .2 kG)

Plug in:

.2*4184*change in Temp(90C)=7.53 e^4 (i.e. 75312)Joules Heat

If you know the equation for power, I(current)*V(voltage), and you realize it is unit of work/time, then if you multiply the Power by Time, you get Work, or here, Heat. Power is defined as Joules per Second if we use amps and volts as our units for current and voltage, respectively.

So we set up our equations:

[Current*12Volts]*360seconds (in 6 minutes)=75312J
4320*I(current)=75312K
Solve for I, current:
17.43 amps