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There is an electric kettle with a capacity of 2l. The name plate says "220 V 800 w 50 Hz". How long does it take to fill it with water from 20 ℃ to 100 ℃ When the voltage is 200V, the heating power of electric kettle
The rated power is 800W
The specific heat capacity of water is 4200j / (kg ·℃), and that of 2l water is 2kg
The total energy required from 20 ℃ to 100 ℃ is 2 × 80 × 4200 = 6.72 × 10 ^ 5 J
The time is t = w / P = 840S = 14min
This is the result of not considering the heat loss
The resistance of heating wire is r = 220 ^ 2 / 800 = 60.5 Ω
When the voltage is 200V, the power of electric kettle is 200 ^ 2 / R ≈ 661.2w
The time needed is t = w / P = 1016s ≈ 16.9min
The capacity of an electric kettle is 2L, the rated voltage is 220 V, the heating power is 800 W, and the frequency is 50 Hz
R=U^2/P
=(220V)^2/800W
=60.5 euro
Current of electric kettle with voltage of 220 V, power of 1360 W and frequency of 50 Hz during normal heating
Resistive load current = power / voltage = 1360 / 220 = 6.18 a
The capacity of an electric kettle is 0.8 L, the rated voltage is 220 V, the rated power is 800 W, and the frequency is 50 Hz
W=Pt
=0.8KW*2h
=1.6kWh
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