Considering water of 20 litres 25 degree celsius of temperature needs to be raised. Using the heat equation q=mcpdt considering cp of water to be 4.2 kJ/kgk we get q=2100 kJ which equals 2100000 J. It means this amount of heat is required for heating the water under said conditions.
I considered 1500 W of heating coil made up of aluminium which equals 1500 J/s. So for this coil to generate 2100000 J of heat (2100000 J)/ (1500 J/s)=1400 secs =23.333 mins. Here to find the electric power consumption in kWh (1.5 kW)*(23.333/60)=0.583 kWh
Again if I consider 3000 W of heating coil made up of some other material say copper to generate the same amount of heat of 2100000 J the time required will be (2100000 J)/ (3000 J/s)= 700 secs= 11.66 mins. Electric power consumption in kWh (3 kW)*(11.66/60)= 0.583 kWh
The power consumption in both the coils of same dimension is same (0.583 kWh) as per calculations(only if I did it right). Does it specify that the neither the material of the coil nor the kW rating affect the power consumption to produce the said amount of heat ? Does it mean that the aluminium coil is as efficient as copper coil ? Where did the high and low resistances offered by the material for heating worked out in this calculation ? Does it mean resistance only affects the power rating and not the consumption ? Highly confused.