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 bat

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.

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Kiệt Gia 2 weeks 2021-08-29T00:04:13+00:00 1 Answers 0 views 0

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    2021-08-29T00:05:13+00:00

    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 \times 4.186 kJ/kg^{o}C \times 69^{o}C

                       = 5.77 kJ

    Now, we will calculate the power delivered as follows.

                        P = \frac{Q}{t}

                           = \frac{5.77 kJ}{13.9 \times 60}

                           = 69.1 W

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

              P_{input} = \frac{69.1}{75} \times 100  

                          = 92.13 W

    Now, we will calculate the current as follows.

                     P = V \times I

                     I = \frac{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.

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