The heating element of a coffeemaker operates at 120 V and carries a current of 3.70 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.692 kg of water from room temperature (23.0°C) to the boiling point. Step 1 The energy required to raise the temperature of an amount of water of mass m, from T, = 23.0°C to the boiling point, T = 100°C, is Q = m„C„(AT), where the specific heat of water c, = 4186 J/kg · °C. We have Q = mCw(AT) = mC(T - T) .692 0.692 kg)(4186 J/kg · °C)(77v 7기 0C 3D 2.23 V 2.23 x 105 J. Step 2 The rate P at which the heating element converts electrical potential energy into the internal energy of the water is P = (AV)I = 120 120 V 3.7 3.7 A) = 444 V 444 J/s.

The heating element of a coffeemaker operates at 120 V and carries a current of 3.70 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.692 kg of water from room temperature (23.0°C) to the boiling point. Step 1 The energy required to raise the temperature of an amount of water of mass m, from T, = 23.0°C to the boiling point, T = 100°C, is Q = m„C„(AT), where the specific heat of water c, = 4186 J/kg · °C. We have Q = mCw(AT) = mC(T - T) .692 0.692 kg)(4186 J/kg · °C)(77v 7기 0C 3D 2.23 V 2.23 x 105 J. Step 2 The rate P at which the heating element converts electrical potential energy into the internal energy of the water is P = (AV)I = 120 120 V 3.7 3.7 A) = 444 V 444 J/s.

Physics for Scientists and Engineers

10th Edition

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter27: Direct-current Circuits

Section27.4: Rc Circuits

Problem 27.5QQ: Consider the circuit in Figure 27.17 and assume the battery has no internal resistance. (i) Just...

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