Answer:
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
