Boiling a kettle?

[Gareth56]

Could someone settle an "difference of opinion" between myself and a friend? He says that it's cheaper to use a 1500W kettle to boil 1L of water than a 2kW. I think you use the same amount of electricity because the 1500kW kettle is heating the water for longer whilst the 2kW is heating for a shorter time thus canceling out the two effects. How can I show him this is the case if of course I'm correct?

 

Ta

[insane_alien]

They don't cancel.

 

the boiling water will be losing heat to the environment so, the longer it takes to boil it, the more heat it loses to the environment before it reaches boiling point. so, even with equal amounts of water and equal starting temperatures, the 1.5kW element will have to expend more energy to compensate.

 

if there were NO heat lost to the environment, both kettles would expend the same amount of energy on equal amounts of water.

[Gareth56]

So it's [marginally] cheaper to use the 2kW kettle. Presumably the difference would be greater if one were using a lower wattage kettle such as those little ones one takes on holiday.

[Klaynos]

So it's [marginally] cheaper to use the 2kW kettle. Presumably the difference would be greater if one were using a lower wattage kettle such as those little ones one takes on holiday.

 

Yep, it'll also make a difference how warm the room in in which your doing it, and other design aspects of the kettle.

[Gareth56]

So how do you work out the amount of electricity you use? Do you need the specific heat capacity of water or something like that?

[insane_alien]

yes. dH=C*M*dT

 

dH=enthalpy change (or energy used in this case) in joules

C=heat capacity of water in joules per kilogram-kelvin

M=mass of water in kilograms

dT=change in temperature in kelvin(or centigrade, for differences these are the same)

 

if you want to consider heat loss then we will need to know more about the construction of the kettle.

[Klaynos]

It's easier to measure it imo...

[insane_alien]

yeah, i suppose, an ammeter and a stopwatch would be much easier than considering the geometry and materials of the kettle. but the heat equation would give you a pretty close estimate.

[swansont]

Ideally, it's 4.18 J/g to get to 100 ºC, and then 2270 J/g to bring it to boil, assuming no energy loss. So starting at 20 ºC, it's 2604.4 J/g. Multiply by the amount of water to get total energy. Divide that by the power and you'll get the time.

 

You know that the 2 kW kettle will do it in 75% of the time of the other (assuming no energy losses) In real life this will be the minimum time, because you'll have to add in extra energy to make up for the losses due to imperfect insulation.

[CaptainPanic]

The 2 kW kettle uses more current (it has more power). More current runs through the cable that connects it. Therefore the cable will heat up a bit more than for the 1.5 kW kettle. This is only a very small effect though... but since you're discussing the heat loss of the kettle during its 1 minute of heating... I thought I should add this small effect too. Of course, the cable is hotter for a shorter period of time...

 

My point: modeling is cool, but measuring is better if it's easy to do!