Here's how you can do that.
The idea here is that some of the energy that could be transferred as heat to increase the temperature of the kettle and of the water is actually lost on account of the efficiency of the kettle.
The kettle is said to have an efficiency of
You've calculated that you need
#DeltaT = 96^@"C" - 12^@"C" = 84^@"C"#
However, this amount corresponds to a
Your goal now is to figure out how much energy must be supplied to the kettle so that
#100 color(red)(cancel(color(black)("J supplied"))) * (7.40 * 10^4color(white)(.)"J needed")/(65color(red)(cancel(color(black)("J supplied")))) = color(darkgreen)(ul(color(black)(1.14 * 10^5color(white)(.)"J needed")))#
Therefore, you can say that in order to supply
If you want to convert this to power, you need to have a measure of time, since
#color(blue)(ul(color(black)("1 J = 1 W s")))#
In other words, power is expressed as energy delivered per unit of time. In order to have
#color(blue)(ul(color(black)("1 W" = "1 J"/"1 s")))#