Question #7e831

1 Answer
Mar 3, 2017

Answer:

Here's how you can do that.

Explanation:

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 #65%#. This basically means that for every #"100 J"# of heat you supply to the kettle, only #"65 J"# will actually go into heating the kettle and its contents.

You've calculated that you need #7.40 * 10^4 color(white)(.)"J"# of energy in order to increase the temperature of the kettle and of the water by

#DeltaT = 96^@"C" - 12^@"C" = 84^@"C"#

However, this amount corresponds to a #100%# efficiency, which is something that you don't have here.

Your goal now is to figure out how much energy must be supplied to the kettle so that #65%# of that figure is equal to #7.40 * 10^4color(white)(.)"J"#.

#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 #7.40 * 10^4color(white)(.)"J"# of heat to a kettle that has an efficiency of #65%#, you need to deliver #1.14 * 10^5color(white)(.)"J"# of heat.

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 #"1 W"# of power, you need #"1 J"# of energy per second.

#color(blue)(ul(color(black)("1 W" = "1 J"/"1 s")))#