# Current drawn from car battery to heat up water

1. Feb 13, 2006

### confusedbyphysics

A small immersion heater can be used in a car to heat a cup of water for coffee or tea. If the heater can heat 120 mL of water from 25 C to 95 C in 8 minutes, a) approximately how much current does it draw from the car's 12-V battery, and b) what is its reistance? Assume the manufacturer's claim of 60% efficiency.

My work:

T(f) - 95 C T(i)= 25 C
t = 8 min = 480 s
V= 12 V
volume = .12 L

First I converted .12 L to kg using density of water, comes out .12 kg.

Then I used the relationship between heat transfer and temperature change equation Q= mcT, specific heat of water is 4186 J/kg Cel

Q = .12kg (4186 J/kgC) (95 C-25 C) = 35162.4 J to heat the water.

Then I used the power equation, Power = work/time = 35162.4 J / 480 s = 73 Watts

Then I used I = P/V, Electric Current = Power / voltage = 73 W/12 V = 6.1 A

The answer in the back of the book says 10 A. I got 6.1 A, which if rounded up could be 10 A and it does say approximately, but I'm not sure if I did this right at all and if my work is wrong he will check it wrong. Am I doing this right or wrong? If wrong, what should I do?? And I'm not sure where the efficiency comes in? Thanks for any help

2. Feb 13, 2006

### andrevdh

Only 60% of the input power (or energy) is converted to heat by the resistor, that is
$$0.6P_{input}=P_{resistor}$$