Current drawn from car battery to heat up water

AI Thread Summary
A small immersion heater can effectively heat water in a car using a 12-V battery. The calculations indicate that to heat 120 mL of water from 25°C to 95°C in 8 minutes, approximately 6.1 A of current is drawn, based on a power requirement of 73 watts. However, considering the manufacturer's efficiency claim of 60%, the actual input power must be higher, suggesting the current could be closer to 10 A. The discussion highlights the importance of accounting for efficiency in electrical calculations. Clarification on how to incorporate efficiency into the calculations is sought by the poster.
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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
 
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