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.