If the heater can heat 130 mL of water from 25 ∘C to 93 ∘C in 8.6 min , approximately how much current does it draw from the car's 12-V battery? Assume the manufacturer's claim of 75% efficiency.

The current draws from the car's 12-V battery is 8 A.

Explanation:

heat capacity of water: 4186 J / kg C

mass water = 0.13 kg

To heat water:

               Q = 0.13 4186 (93-25)

                  = 37004.2 J

       Power = Q / t

                   = 37004.2 / (8.6 60)

                   = 71.713 J/s

            Pbat =  71.713 / 0.75

                    = 95.617 J/s

                   r = 144 / 95.617  

                     = 1.50 ohms

                    I = v/r

                     = 12/1.5

                  I = 8 A.

im guessing the west of station b