An electric kettle is required to heat 0.6 kg of water from 100C to boiling point in 5 minutes, the supply voltage being 230 V. The efficiency of the kettle is 80%. Calculate the resistance of the heating element.
Answers
Given : an electric kettle is required to heat 0.6 kg of water from 10° C to boiling point in 5 min
the supply voltage being 230 V
efficiency of the kettle is 80%.
specific heat capacity of water to be 4.2 kJ/kg K
To Find : calculate the resistance of the heating element
Solution:
change in thermal energy = mass × specific heat capacity × temperature change
mass = 0.6 kg
specific heat capacity = 4.2 kJ/kg K
Boling point = 100°C
temperature change = 100 - 10 = 90
=> change in thermal energy = 0.6 * 4.2 * 90 = 226.8 kJ
Electric Energy = V²t/R
t = 5 min = 300sec
V = 230V
= 230² * 300 / R
= 15870/R kJ
Efficiency of the kettle is 80%.
=> thermal energy = (80/100)15870/R kJ
= 12696/R kJ
12696/R = 226.8
=> R = 12696/ 226.8
=> R = 55.98 Ω
the resistance of the heating element = 55.98 Ω
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