The mass of water in the kettle is 0.50 kg. The specific heat capacity of water is 4200 J/kg °C. The initial temperature of the water is 100 °C. Calculate the average power output from the water in the kettle to the surroundings in 2 hours

1.27 Mega watt approx

Explanation:

Given data

Mass= 0.5kg

c= 4200 J/kg °C

T1= 100°C

Time = 2 hours >>>> 7200 seconds

From the given data the water is already at the boiling point

Hence if continuous heat is being added the latent heat of vapourization will result, the latent heat of vaporization is about 2,260 kJ/kg

Q= mL

Q= 0.5*2,260

Q = 1130kJ

Power= Energy * Time

Power=  1130* 1130

Power= 1.27mega watt approx

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