To practice problem-solving strategy 25.1 power and energy in circuits. a device for heating a cup of water in a car connects to the car's battery, which has an emf e = 10.0 v and an internal resistance rint = 0.04 ω . the heating element that is immersed in the cup of water is a resistive coil with resistance r. david wants to experiment with the device, so he connects an ammeter into the circuit and measures 12.0 a when the device is connected to the car's battery. from this, he calculates the time to boil a cup of water using the device. if the energy required is 100 kj , how long does it take to boil a cup of water?

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

    V = i (r + R)

We replace we calculate

     r + R = V / i

     R = v / i - r

     R = 10/12 -0.04

     R = 0.793 Ω

We calculate the power supplied

     P = V i = I² R

     P = 12² 0.793

     P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

     P = E / t

     t = E / P

     t = 100 10³/114

     t = 877 s

The time to boil the water is 877 s

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we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer