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?

Question

Physics

Lillian Compton

10 February, 20:18

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?

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Answers (1)

Know the Answer?

Ean Sanford · Answer 1

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