Respuesta :
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 = [tex](91 - 22)^{o}C[/tex]
= [tex]69^{o}C[/tex]
Also, we know that specific heat of water is 4.186 [tex]kJ/kg^{o}C[/tex]. Therefore, we will calculate the heat energy as follows.
q = [tex]mC \Delta T[/tex]
= [tex]0.2 kg \times 4.186 kJ/kg^{o}C \times 69^{o}C[/tex]
= 5.77 kJ
Now, we will calculate the power delivered as follows.
P = [tex]\frac{Q}{t}[/tex]
= [tex]\frac{5.77 kJ}{13.9 \times 60}[/tex]
= 69.1 W
It is given that efficiency is 75%. Therefore, input power will be calculated as follows.
[tex]P_{input} = \frac{69.1}{75} \times 100[/tex]
= 92.13 W
Now, we will calculate the current as follows.
P = [tex]V \times I[/tex]
I = [tex]\frac{92.13 W}{12 V}[/tex]
= 7.67 A
Therefore, we can conclude that current drawn from the car's 12-V battery is 7.67 A.
