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It depends on how many liters of water. An average kettle holds up to 1.7 liters and takes 3KW. If it takes 3 minutes( 1/20 of an hour) to boil that's 3/20 of a KWh, = 0.15 KWh.

Doing the thermal energy sum, you have say 1.5 liter = 1.5 kg of water raising from say 10 C to 100 C, that's 90 C raise and 1500 grams of water so that makes 135,000 calories. 1 calorie = 4.2 Joules, so we have 567,000 Joules.

1 watt = 1 Joule/sec , 1 KW = 1000 Joules/sec, so 1 KWh = 1000 x 3600 Joules (3600 secs in 1 hour), 1 KWh = 3,600,000 Joules. So 567,000 Joules = 0.1575 KWh.

So the two sums give similar results. If you do a more exact comparison, you could find the efficiency of the kettle from the difference in the two figures, there is bound to be some loss of electrical power, but it should not be much.

Alternative Answers.

For a more accurate answer, use the following formula -

W - Q = [m c (T1 - T2)]water + [m c (T1 - T2)]kettle

Where

W = work done in joules

Q = heat transfer away from the kettle

m = mass of water (and mass of kettle)

c = specific heat capacity of water (and of the material from which kettle is made)

T1 = initial temperature of water/kettle

T2 = final temperature of water/kettle

Then convert your answer from joules to kilowatt hours.

However, the REALLY simple way is to time how long it takes for your kettle to boil, then multiply its nameplate power rating (in kilowatts) by that time (expressed in hours)!

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โˆ™ 2014-03-24 11:34:08
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Q: How many kWh does boiling a kettle use?
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