How long does it take for an electric kettle with a current of 4 A, voltage of 220 V to heat 2 kg of water from 20 degrees to a boil.
I = 4 A.
U = 220 V.
m = 2 kg.
C = 4200 J / kg * ° C.
t1 = 20 ° C.
t2 = 100 ° C.
To heat water of mass m from temperature t1 to boiling point t2, we express the formula: Qw = C * m * (t2 – t1), where C is the specific heat capacity of water.
The amount of heat Qh is released in the electric kettle, which we express by the formula: Qh = I * U * T, where I is the current strength, U is the current voltage, T is the operating time of the kettle.
The loss of thermal energy during the transfer from the kettle to water can be neglected: Qh = Qw.
I * U * T = C * m * (t2 – t1).
T = C * m * (t2 – t1) / I * U.
T = 4200 J / kg * ° C * 2 kg * (100 ° C – 20 ° C) / 4 A * 220 V = 763.6 s = 12 min 43.6 s.
Answer: to heat water in an electric kettle, T = 12 min 43.6 s.
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