Question

If the heater can heat 200 mL of water from 22 ∘C to 91 ∘C in 13.9 min , approximately how much current does it draw from the car's 12-V battery? Assume the manufacturer's claim of 75% efficiency. Express your answer using two significant figures.

Answer 1

Step-by-step explanation:

First, we will convert 200 ml into kg as follows.

200 ml = 0.2 kg

Now, we will take the density of water as 1 g/ml.

And, dT =
`(91 - 22)^(o)C`

=
`69^(o)C`

Also, we know that specific heat of water is 4.186
`kJ/kg^(o)C`. Therefore, we will calculate the heat energy as follows.

q =
`mC \Delta T`

=
`0.2 kg * 4.186 kJ/kg^(o)C * 69^(o)C`

= 5.77 kJ

Now, we will calculate the power delivered as follows.

P =
`(Q)/(t)`

=
`(5.77 kJ)/(13.9 * 60)`

= 69.1 W

It is given that efficiency is 75%. Therefore, input power will be calculated as follows.

= 92.13 W

Now, we will calculate the current as follows.

P =
`V * I`

I =
`(92.13 W)/(12 V)`

= 7.67 A

Therefore, we can conclude that current drawn from the car's 12-V battery is 7.67 A.