Question

guys i relly need help on this question FAST
To compare the efficiency of a microwave with the efficiency of an electric kettle, Sara placed 500 mL of water into each. Sara’s teacher helped her calculate the energy required to boil the water, which was 168 kJ. The kettle, which had a power rating of 1.5 kW, took 132 s to boil the water. The microwave had a power rating of 1.2 kW and took 280 s to boil the water. Which is more efficient?

Answer 1

i know efficancy is found energy out divided energy in x 100 but i cant find the output energy

Answer 2

Power = (change in energy)/(change in time) = Energy/Time.
Hence Energy = Power x Time

So the kettle expended 1.5 kW x 132 s = 198 kJ
and the microwave 1.2 kW x 280 s = 336 kJ

See the answer now?

Answer 3

so is the kettle more efficant??

Answer 4

and how come the 168kj part is not needed

Answer 5

Yes, because it expended less energy than the microwave, both of which ended up doing the same effective work.

Answer 6

You can calculate the efficiency explicitly with the 168kJ if you want to.

Answer 7

but it is no revelent in this question right???

Answer 8

It's not required, no.

Answer 9

- im just in grade 9 new to this stuff

Answer 10

thx fir ur help i was really confused

Answer 11

...other than sense checking the answer. If either of the energy amounts we had found were less than 168kJ, then we would have known we had made an error.

Answer 12

can u explain more plz

Answer 13

so it had to be less than 168 because the input was 168??

Answer 14

For the record, the energy efficiency of the microwave is

Pragmatic work / Total work = 168 kJ / 336 kJ = 0.5 or 50%

======

I'm saying the total work done by the kettle or the microwave had to be at the very least equal to 168 kJ.

Answer 15

...because if it had been less, then there is no way the 168 kJ of pragmatic work would have been realized.

Answer 16

dont we divide output energy by input to find efficany

Answer 17

In other words, nothing has more than 100% efficiency.

Answer 18

What I'm calling "Pragmatic Work" is the output.

Answer 19

The output is 168 kJ.

Answer 20

oh so the output is 168 i was assuming that was the energy required being input

Answer 21

Hence for the microwave, output = 168 kJ, input = 336 kJ.

Therefore Efficiency = output/input = 168/336 = 0.5, or 50%

Answer 22

ok thank-you so much i apperichate ur help

if u want to see something cool and have google chrome tyoe tilt