Ask Question
27 February, 12:54

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 = 0.08 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 10.0 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?

+2
Answers (1)
  1. 27 February, 12:57
    0
    t=17 min

    Explanation:

    Given that

    E = 10 V

    I=10 A

    Internal resistance Rin = 0.08 Ω

    E = I. R (net)

    E=I (R+r)

    10 = 10 (R+0.08)

    R = 1 - 0.0 8 = 0.02 Ω

    R = 0.98 Ω

    So the heat given as

    Q=I²Rt

    Q=10² x 0.98 x t

    Q = 98 t J

    Given that Q = 100 KJ

    98 t = 100 x 1000 J

    t = 100 x 1000/98 sec

    t=17 min
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers