The heat absorbed by the water Q=cmt=4.2×10?×10?×50=2.1×10^8J
There are 60×60=3600 seconds in one hour
< p>Water heater power=2.1×10^8J/3600s=58333J/s=58kwThe estimated loss is 5KW
Choose a 65KW water heater.
If you are talking about raising the water to 90 degrees Celsius, then you need to use a 105 kilowatt electric heater, excluding losses. From W=Q, we obtain Pt=cm variable t, P ·. 3600s=4.2*10^3J/(kg·℃)*1m^3*1*10^3kg/m^3*90℃, so P=1.05*10^5W=105kW.