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.
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Transcribed Image Text:Tutorial Exercise
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) = m„Cw(T – T)
692 V
0.692 kg)(41
4186 J/kg · °c)7
77
77
°C
= 2.23
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.

Transcribed Image Text:Step 3
Thus, the time At required to bring the water to a boil is
|x 105 )
At =
J/s
1 min
min.
60.0 s.
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