The heating element of a coffeemaker operates at 120 V and carries a current of 7.70 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.252 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 = mFw(AT) = m„„(T – T) 0.252 0.252 kg) (4186 J/kg - °c)(7 77 °C 0.81225 0.812 x 105 J. Step 2 The rate Pat which the heating element converts electrical potential energy into the internal energy of the water is P (AV)I = (120V 120 v7.7 7.7 A- 924 924 J/s. Step 3 Thus, the time At required to bring the water to a boil is |× 105 ) At | J/s) I min 60.0 s min.
The heating element of a coffeemaker operates at 120 V and carries a current of 7.70 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.252 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 = mFw(AT) = m„„(T – T) 0.252 0.252 kg) (4186 J/kg - °c)(7 77 °C 0.81225 0.812 x 105 J. Step 2 The rate Pat which the heating element converts electrical potential energy into the internal energy of the water is P (AV)I = (120V 120 v7.7 7.7 A- 924 924 J/s. Step 3 Thus, the time At required to bring the water to a boil is |× 105 ) At | J/s) I min 60.0 s min.
Physics for Scientists and Engineers, Technology Update (No access codes included)
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Author:Raymond A. Serway, John W. Jewett
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Chapter27: Current And Resistance
Section: Chapter Questions
Problem 27.55P: Review. The healing element of an electric coffee maker operates at 120 V and carries a current of...
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Transcribed Image Text:The heating element of a coffeemaker operates at 120 V and carries a current of 7.70 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.252 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 = mwCw(AT) = mwCw(T – T;)
0.252
0.252 kg
4186 J/kg · °C
77
7기 °C
= 0.81225
x 105 J.
0.812
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
v7.7
7.7 A
= 924
924 J/s.
Step 3
Thus, the time At required to bring the water to a boil is
105 J)
At =
J/s
1 min
min.
60.0 s.
Expert Solution
Step 1
Given information:
The voltage at which the coffee maker works (V) = 120 V
The current it carries (I) = 7.70 A
The mass of the water (m) = 0.252 kg
The initial temperature of the water (Ti) = 230 C
The final temperature of the water required (Tf) = Boiling point = 1000 C
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