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 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= mC,(AT), where the specific heat of water c= 4186 J/kg - °C. We have Q = mc(AT) = m c(T - T) 0.252 kg (4186 /kg · °c)(77, 77 °c) 0.252 - 0.81225 0.812 x 10 ) 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 = ( J/s.

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 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= mC,(AT), where the specific heat of water c= 4186 J/kg - °C. We have Q = mc(AT) = m c(T - T) 0.252 kg (4186 /kg · °c)(77, 77 °c) 0.252 - 0.81225 0.812 x 10 ) 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 = ( J/s.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

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 = mC„(AT),
where the specific heat of water c = 4186 J/kg · °C. We have
Q = m„Cw(AT) = m„Cw(T – T;)
0.252
0.252 kg )( 4186 J/kg · °C
77
77 °c)
0.81225
0.812 x 10° 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 =
A
J/s.
=

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