Question

an electric kettle is required to heat 0.5 kg of water from 10 degree Celsius to boiling point in 5 min ,the supply voltage being 230 V. if the efficiency of the kettle is 80%. calculate the resistance of the heating element . assume the specific heat capacity of water to be 4.2 kJ/kg K​

Answer 1

Given : an electric kettle is required to heat 0.5 kg of water from 10 degree Celsius 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 = 0.5 kg

specific heat  capacity  = 4.2 kJ/kg K​

Boling point = 100°C

temperature change = 100 - 10 = 90

=> change in thermal energy = 0.5 * 4.2 * 90  = 189 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   = 189

=> R = 12696/189

=> R = 67.17 Ω

the resistance of the heating element   = 67.17 Ω

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