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 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 11.0 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?

Answer 1

Answer:

It takes 951 seconds to boil a cup of water.

Explanation:

Given:

EMF of the battery (E) = 10.0 V

Internal resistance of the battery (r) = 0.04 Ω

Resistance of the circuit = 'R'

Current measured in the circuit (I) = 11.0 A

Energy required to boil water (U) = 100 kJ = 100 × 10³ J [1 kJ = 10³ J]

Time taken for boiling (t) = ?

We know that, the emf of the battery is given as:

$E=I(R+r)$

Plug in the given values and solve for 'R'. This gives,

$10=11(R+0.04)\\\\R+0.04=\frac{10}{11}\\\\R=\frac{10}{11}-0.04=0.909-0.04=0.869\ \Omega$

Now, energy required to boil the water is given as:

$U=I^2Rt$

Plug in the given values and solve for 't'. This gives,

$100\times 10^3\ J=(11.0\ A)^2(0.869\ \Omega)t\\\\t=\frac{100000\ J}{121\times 0.869\ A^2\Omega}\\\\t=\frac{100000\ J}{105.149\ A^2\Omega}\\\\t=951\ s$

So, it takes 951 seconds to boil a cup of water.