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For example we have 1 electric kettle with 1 litter of water and we have another (the same type) electric kettle with 2 litters of water. How much time each kettle will need to boil and why?

How does the boiling time in electric kettle depend on the amount of water?

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I assume you mean that kettle 1 should have 1L water and kettle 2 should have 2L water? That is, I assume you are asking how the time to boil water scales with the amount of water, yes? – Flint72 May 17 '14 at 12:21
@Flint72 Yes, you are right. I edited and fixed the question. – webvitaly May 17 '14 at 22:10
up vote 2 down vote accepted

The amount of energy needed to heat 1kg of water by 1°C is given by the specific heat of water. This changes slightly with temperature, but it's around 4.2 kJ/kg/K.

So suppose we have a mass $m$ kg of water and we are heating it by $T$ degrees (e.g. from 20°C to boiling would be $T$ = 80°C). The amount of energy needed is:

$$ E = 4.2 m T \space \text{kJ} $$

If the power of the kettle is $W$ kW, i.e. $W$ kJ/s, and we assume no heat is lost then the time taken to heat the water is:

$$ t = \frac{4.2 m T}{W} \text{secs} $$

The time is proportional to the mass of water, so if you double the mass of water you double the time needed to boil it.

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The amount of energy needed to boil a certain amount of water from a certain initial temperature, scales linearly with the amount of water.

If your kettle would be ideal, than the boiling time will also be linear in the amount of water (double the amount of water, double the boiling time). This is assuming that your kettle will always use the same power.

Now, this is not the full story. Your kettle is not ideal. The water will not be uniformly heated, so some water will be already boiling, and may be converted into steam (which makes you to additionally add latent heat). Heat losses to the surrounding may also play different role, based on the water content, which should be taken into account. How this affects the boiling time is difficult to predict.

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I did actually tried to calculate this base on that equation with 0.33L of water and initial temperature of 24 so T=76, and 4200 joules, power of my kettle was 2000 W , based on the equation it gave 52 secs , yet experimental result was 120 secs.after measuring time with different quantity of water, It turns out my kettle only used 890 Watts initially and increased about 95+ watts each time as water was added.

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This appears to be a comment to John Rennie's post. – Kyle Kanos Dec 30 '15 at 11:42

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