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## / Boiling a kettle |

rubbercrumb - * * on 06 Nov 2012

She wants two cups of tea in fairly quick succession.

To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

Or boil a single cups' worth from cold each time?

Some important domestic point scoring at stake here...

To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

Or boil a single cups' worth from cold each time?

Some important domestic point scoring at stake here...

antdav - * * on 06 Nov 2012

separate

Flinticus - * * on 06 Nov 2012

In reply to rubbercrumb:

Just guessing but I would say single each time as surely it takes the same energy to heat (cup x 2) in one go as cup x 1 then cup x 1? Same volume of water.

If you heat both then leave one cup's worth of water to cool down is energy not dissipating out, energy which will have come from the kettle element.

Just guessing but I would say single each time as surely it takes the same energy to heat (cup x 2) in one go as cup x 1 then cup x 1? Same volume of water.

If you heat both then leave one cup's worth of water to cool down is energy not dissipating out, energy which will have come from the kettle element.

deepsoup - * * on 06 Nov 2012

In reply to rubbercrumb:

Depends - is the heating on? ;O)

If not, it takes more energy to boil the two cuppas' worth of water together because the second one is losing heat to the environment while its standing there and then has to be boiled again.

But the environment it's losing heat to is your house. If you're heating your house at the time, arguably the 'extra' energy is merely energy you'd be using to make your house warmer anyway.

Depends - is the heating on? ;O)

If not, it takes more energy to boil the two cuppas' worth of water together because the second one is losing heat to the environment while its standing there and then has to be boiled again.

But the environment it's losing heat to is your house. If you're heating your house at the time, arguably the 'extra' energy is merely energy you'd be using to make your house warmer anyway.

Monk - * * on 06 Nov 2012

rubbercrumb - * * on 06 Nov 2012

In reply to Monk:

teapot idea is a good one.

but in the original scenario she's claiming the water is already hot for the second boil so it's merely being 'topped up' to boiling point, as it were, therefore using less than if it were cold.

So it's one two-cup kettle from cold then one cup from warm/hot

versus

two x cold single cups

see the complexity?

teapot idea is a good one.

but in the original scenario she's claiming the water is already hot for the second boil so it's merely being 'topped up' to boiling point, as it were, therefore using less than if it were cold.

So it's one two-cup kettle from cold then one cup from warm/hot

versus

two x cold single cups

see the complexity?

Wonko The Sane on 06 Nov 2012 - *31.103.21.85 whois?*

In reply to rubbercrumb: In an IDEAL system, where all the energy is being transfered to the water, you would still have heat loss radiated from the surface area of the water. so, surface area would be less as a percentage of volume for TWO cups together.

So the answer is two cups, since the losses only get worse in a real kettle. I.E. not all the heat gets to the water.

So the answer is two cups, since the losses only get worse in a real kettle. I.E. not all the heat gets to the water.

tom_in_edinburgh - * * on 06 Nov 2012

In reply to rubbercrumb:

Measure it! Use a stopwatch and measure the amount of time the kettle is switched on for in both cases. Energy = Power x Time so whatever option takes less time has used less energy. (Assumes the electrical power is constant according to the rating of the element and that any dependence on water temperature can be ignored).

Measure it! Use a stopwatch and measure the amount of time the kettle is switched on for in both cases. Energy = Power x Time so whatever option takes less time has used less energy. (Assumes the electrical power is constant according to the rating of the element and that any dependence on water temperature can be ignored).

deepsoup - * * on 06 Nov 2012

In reply to rubbercrumb:

This is true - but it has already been boiled once. The "top up" is additional energy that wouldn't have been used otherwise.

No. There really isn't much complexity.

Boiling two cups requires twice as much energy as boiling one. Letting one cup cool down a bit and then boiling it again requires a bit more energy again.

> so it's merely being 'topped up' to boiling point, as it were, therefore using less than if it were cold.

This is true - but it has already been boiled once. The "top up" is additional energy that wouldn't have been used otherwise.

> see the complexity?

No. There really isn't much complexity.

Boiling two cups requires twice as much energy as boiling one. Letting one cup cool down a bit and then boiling it again requires a bit more energy again.

3 Names - * * on 06 Nov 2012

Wonko The Sane on 06 Nov 2012 - *31.103.21.85 whois?*

In reply to rubbercrumb:

Sorry, confused here.

Is the question that you boil exactly one cup, tip it out and boil another?

Or is the question that you have a partially filled kettle which you take a cup of water out of, top up and re boil?

> She wants two cups of tea in fairly quick succession.

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

>

> Or boil a single cups' worth from cold each time?

>

> Some important domestic point scoring at stake here...

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

>

> Or boil a single cups' worth from cold each time?

>

> Some important domestic point scoring at stake here...

Sorry, confused here.

Is the question that you boil exactly one cup, tip it out and boil another?

Or is the question that you have a partially filled kettle which you take a cup of water out of, top up and re boil?

Ben Sharp - * * on 06 Nov 2012

In reply to rubbercrumb: Older kettles sometimes have a minimum capacity which is greater than one cup as well. I don't think that would change the odds in the full fill's favour but worth considering.

As others have said it's more energy efficient to boil the two cups separately, it's even better to refill the kettle immediately with the extra cup so that the left over warmth from the kettle goes into the fresh water while the kettle is off. Obviously we all do this and then wrap the kettle up in dry tea towels and place it on the radiator.

Alternatively if you're just about to have a bath then put the kettle in a dry bag and take it in with you, that way some of the warmth from the bath water gets transferred to the water in the kettle, meaning less energy to bring it up to boiling.

I'm sure most people do this already but you can also decant water from the tap into hydration bladders worn against the skin and then fill your kettle from these. It's a bit chilly at first but we've all got to take a bit of punishment to save the planet.

Having said all that I don't like kettles myself and prefer to heat my water in situ (in the mug) with a blow torch, I'm not sure if this is more energy efficient or not but if you think about it, when you boil the kettle, the kettle's housing gets hot, which is just wasted energy because once you've poured the water out of it you don't need the kettle to be hot. Better to heat it in the mug because that'll get hot anyway.

As others have said it's more energy efficient to boil the two cups separately, it's even better to refill the kettle immediately with the extra cup so that the left over warmth from the kettle goes into the fresh water while the kettle is off. Obviously we all do this and then wrap the kettle up in dry tea towels and place it on the radiator.

Alternatively if you're just about to have a bath then put the kettle in a dry bag and take it in with you, that way some of the warmth from the bath water gets transferred to the water in the kettle, meaning less energy to bring it up to boiling.

I'm sure most people do this already but you can also decant water from the tap into hydration bladders worn against the skin and then fill your kettle from these. It's a bit chilly at first but we've all got to take a bit of punishment to save the planet.

Having said all that I don't like kettles myself and prefer to heat my water in situ (in the mug) with a blow torch, I'm not sure if this is more energy efficient or not but if you think about it, when you boil the kettle, the kettle's housing gets hot, which is just wasted energy because once you've poured the water out of it you don't need the kettle to be hot. Better to heat it in the mug because that'll get hot anyway.

deepsoup - * * on 06 Nov 2012

In reply to Ben Sharp:

If the radiator is on, you'd save more energy by turning the heating off and wrapping the tea towels around the kettle and yourself. ;O)

> Obviously we all do this and then wrap the kettle up in dry tea towels and place it on the radiator.

If the radiator is on, you'd save more energy by turning the heating off and wrapping the tea towels around the kettle and yourself. ;O)

bullybones - * * on 06 Nov 2012

In reply to rubbercrumb:

Make 2 cups at the same time and put a lid on one. If the second one's gone a bit cool by the time you remember to drink it, microwave for 20s.

Make 2 cups at the same time and put a lid on one. If the second one's gone a bit cool by the time you remember to drink it, microwave for 20s.

Wonko The Sane on 06 Nov 2012 - *31.103.21.85 whois?*

In reply to Ben Sharp:

It isn't quite that simple though.

I was taking it to mean a choice of boiling one cup then another vs two cups together with no reheating.

If we are saying 'is it more efficient to boil two cups seperately, or boil two cups, use one, wait, then re boil the other.......

You have to calculate the losses in the system for boiling each cup individually, then do the same excercise for both together and re boiling (AFTER A SPECIFIED TIME PERIOD!!) then subtract one from the other.

People appear to be treating this as an ideal system and it's not. A kettle is not 100% effieicent at turning electrical energy into heated water.

It isn't quite that simple though.

I was taking it to mean a choice of boiling one cup then another vs two cups together with no reheating.

If we are saying 'is it more efficient to boil two cups seperately, or boil two cups, use one, wait, then re boil the other.......

You have to calculate the losses in the system for boiling each cup individually, then do the same excercise for both together and re boiling (AFTER A SPECIFIED TIME PERIOD!!) then subtract one from the other.

People appear to be treating this as an ideal system and it's not. A kettle is not 100% effieicent at turning electrical energy into heated water.

Neil Williams - * * on 06 Nov 2012

Ava Adore - * * on 06 Nov 2012

In reply to rubbercrumb:

Someone's making me two cups of tea in fairly quick succession, I'm not going to worry about any delays

Someone's making me two cups of tea in fairly quick succession, I'm not going to worry about any delays

AWR on 06 Nov 2012

Me and my Mrs have a thermos type jug that hold 2 cups worth of water, we paid a fiver for it.

We boil the kettle once, put the left over water in the jug then use that for the next brew...therefore saving energy and money. No need for points scoring then, so all of our efforts can be put into the great "whos turn is it to make the brew" discussion instead

We boil the kettle once, put the left over water in the jug then use that for the next brew...therefore saving energy and money. No need for points scoring then, so all of our efforts can be put into the great "whos turn is it to make the brew" discussion instead

Milesy - * * on 06 Nov 2012

Tea shouldn't be made with reboiled water anyway. Looses too much oxygen in the process. Boil once and make a pot, remembering to remove the tea once brewed properly. Job done.

Philip on 06 Nov 2012

One big mug. My wife has a Pete's Eats pint mug and my bro-in-law has a 1.5 pint Darth Vader mug.

teflonpete - * * on 06 Nov 2012

In reply to Philip:

Darth Vader must have a terrible rust problem trying to pour 1.5 pints of tea through that mask.

> One big mug. My wife has a Pete's Eats pint mug and my bro-in-law has a 1.5 pint Darth Vader mug.

Darth Vader must have a terrible rust problem trying to pour 1.5 pints of tea through that mask.

Philip on 06 Nov 2012

In reply to teflonpete:

He likes it brewed strong, hence the comment "Your tea is weak old man" in Star Wars.

> (In reply to Philip)

> [...]

>

> Darth Vader must have a terrible rust problem trying to pour 1.5 pints of tea through that mask.

> [...]

>

> Darth Vader must have a terrible rust problem trying to pour 1.5 pints of tea through that mask.

He likes it brewed strong, hence the comment "Your tea is weak old man" in Star Wars.

DaveN - * * on 06 Nov 2012

In reply to deepsoup:

Do you need to factor in the effects of heating the mass of the kettle?

> (In reply to rubbercrumb)

>

> [...]

>

> This is true - but it has already been boiled once. The "top up" is additional energy that wouldn't have been used otherwise.

>

> [...]

>

> No. There really isn't much complexity.

>

> Boiling two cups requires twice as much energy as boiling one. Letting one cup cool down a bit and then boiling it again requires a bit more energy again.

>

> [...]

>

> This is true - but it has already been boiled once. The "top up" is additional energy that wouldn't have been used otherwise.

>

> [...]

>

> No. There really isn't much complexity.

>

> Boiling two cups requires twice as much energy as boiling one. Letting one cup cool down a bit and then boiling it again requires a bit more energy again.

Do you need to factor in the effects of heating the mass of the kettle?

Jim C - * * on 06 Nov 2012

In reply to rubbercrumb:

Put any excess water in a Thermos Flask, problem solved.

My gripe is she boils pots of food WITH lids then then removes the lids to bring it off the boil, or partially to put it on a simmer without altering the gas.

Why not turn down the gas until it is JUST off the boil, return the lid, and the water will simmer away on a boil , with no waste of energy.

"Too fiddly" is the response.

> She wants two cups of tea in fairly quick succession.

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.>

> Or boil a single cups' worth from cold each time?>

> Some important domestic point scoring at stake here...

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.>

> Or boil a single cups' worth from cold each time?>

> Some important domestic point scoring at stake here...

Put any excess water in a Thermos Flask, problem solved.

My gripe is she boils pots of food WITH lids then then removes the lids to bring it off the boil, or partially to put it on a simmer without altering the gas.

Why not turn down the gas until it is JUST off the boil, return the lid, and the water will simmer away on a boil , with no waste of energy.

"Too fiddly" is the response.

Flinticus - * * on 06 Nov 2012

In reply to Wonko The Sane:

Yeah, but is it not the same inefficient kettle being applied to both different boilng options?

Yeah, but is it not the same inefficient kettle being applied to both different boilng options?

deepsoup - * * on 06 Nov 2012

In reply to DaveN:

Don't think so, no.

Either way you heat it up, it cools down a bit and then you heat it up again.

Don't think so, no.

Either way you heat it up, it cools down a bit and then you heat it up again.

ablackett - * * on 06 Nov 2012

In reply to rubbercrumb:

It's easy. E=mc^2 so the Energy (E) is equal to the mass of the water x the speed of light squared.

Twice the mass is twice the energy. So just boil half the water and save twice the money.

It's easy. E=mc^2 so the Energy (E) is equal to the mass of the water x the speed of light squared.

Twice the mass is twice the energy. So just boil half the water and save twice the money.

spearing05 - * * on 06 Nov 2012

In reply to Flinticus: The same kettle is being used but in the first instance is heated once from cold then again from very hot (how hot depends on the time it is left for) but in the second instance it is heated twice from cold, how cold (and therefore how much of the energy stored in it from heating the first cup remains to be transfered to the second lot of water) will depend on the time between the two cups. However long this is though, a portion of it will be lost to the air and so need replacing on the second heat.

Boiling two cups together and using both takes less energy than two seperately. The kettle is only heated once and the surface area to volume ratio is smaller with a larger volume so less heat is lost to the surroundings during the heating cycle.

Boiling both together, using one then leaving to cool to room temperature before reboiling the other is less effecient than boiling seperately when required. Any gains from boiling together as above are lost by allowing it to cool.

There is a time therfore somewhere between these two where one becomes more effecient than the other - what that time is I have no idea.

Boiling two cups together and using both takes less energy than two seperately. The kettle is only heated once and the surface area to volume ratio is smaller with a larger volume so less heat is lost to the surroundings during the heating cycle.

Boiling both together, using one then leaving to cool to room temperature before reboiling the other is less effecient than boiling seperately when required. Any gains from boiling together as above are lost by allowing it to cool.

There is a time therfore somewhere between these two where one becomes more effecient than the other - what that time is I have no idea.

Pero - * * on 06 Nov 2012

In reply to rubbercrumb: You need to conduct an experiment! There are too many factors to be confident of a theoretical answer.

If you boil two cups together and you have to top up the kettle to get the second cup, then, as someone has said, top it up immediately, while the water is still boiling, for maximum efficiency!

It's brilliant to find out I'm not the only one who ponders questions like this. My girlfriend thinks I'm a bit mad when I talk about "critical path analysis" when making a meal!

If you boil two cups together and you have to top up the kettle to get the second cup, then, as someone has said, top it up immediately, while the water is still boiling, for maximum efficiency!

It's brilliant to find out I'm not the only one who ponders questions like this. My girlfriend thinks I'm a bit mad when I talk about "critical path analysis" when making a meal!

Wonko The Sane on 06 Nov 2012 - *31.103.21.85 whois?*

In reply to Flinticus:

Exactly.

So all the 'bottom line' inefficiencies, the energy lost before any water is even heated, are more for two operations than for one. Not exactly doubled, because it takes more time to heat twice the amount of water. And it will vary a little between appliances.

But it's anything but a simple asnswer.

The easiest way to calculate it would be on a meter of energy used. But obviously, this won't tell you where the losses are. I.E. element inefficiency, different insulating properties od different materials. Placing of the element (convection currents)

All sorts of things go into the sum. As does time between reheating,

> (In reply to Wonko The Sane)

> Yeah, but is it not the same inefficient kettle being applied to both different boilng options?

> Yeah, but is it not the same inefficient kettle being applied to both different boilng options?

Exactly.

So all the 'bottom line' inefficiencies, the energy lost before any water is even heated, are more for two operations than for one. Not exactly doubled, because it takes more time to heat twice the amount of water. And it will vary a little between appliances.

But it's anything but a simple asnswer.

The easiest way to calculate it would be on a meter of energy used. But obviously, this won't tell you where the losses are. I.E. element inefficiency, different insulating properties od different materials. Placing of the element (convection currents)

All sorts of things go into the sum. As does time between reheating,

cap'nChino - * * on 06 Nov 2012

In reply to rubbercrumb: Is the kettle on some sort of treadmill?

spearing05 - * * on 06 Nov 2012

In reply to Wonko The Sane: All sorts of things go into the sum. As does time between reheating,

Agreed - it is this that is the big variable there is a big difference between boiling two cups, pouring one then imediatly reboiling before pouring the second and boiling two cups, pouring one, waiting an hour then reboiling the second.

Agreed - it is this that is the big variable there is a big difference between boiling two cups, pouring one then imediatly reboiling before pouring the second and boiling two cups, pouring one, waiting an hour then reboiling the second.

crossdressingrodney - * * on 06 Nov 2012

In reply to ablackett:

You have a very advanced kettle.

> It's easy. E=mc^2 so the Energy (E) is equal to the mass of the water x the speed of light squared.

>

> Twice the mass is twice the energy. So just boil half the water and save twice the money.

>

> Twice the mass is twice the energy. So just boil half the water and save twice the money.

You have a very advanced kettle.

MonkeyPuzzle - * * on 06 Nov 2012

In reply to rubbercrumb:

Of course, the ideal temperature for the water depends on what type of tea you're brewing, so possibly the ideal situation would be to boil two cups of water together, make a perfect cup of strong tea - Assam, English breakfast, etc. - and then come back, don't re-boil the kettle but make a perfect cup of Earl Grey. Solved.

Of course, the ideal temperature for the water depends on what type of tea you're brewing, so possibly the ideal situation would be to boil two cups of water together, make a perfect cup of strong tea - Assam, English breakfast, etc. - and then come back, don't re-boil the kettle but make a perfect cup of Earl Grey. Solved.

malky_c - * * on 06 Nov 2012

In reply to rubbercrumb: Boiling two cups together is more efficient than boiling two single cups, as a smaller proportion of the water is either surface area or in contact with the side if the kettle.

However the question is - is the additional energy required to re-boil the water for the second cup more or less than the extra inefficiency of boiling two separate cups?

Obviously it depends a little on the efficiency of the kettle and its insulating properties, but mainly on how long you leave it between the two cups. As others have said, you would need to conduct an experiment to find out what this length of time is to get your answer - it may be so short that two separate boils is the only practical answer.

Also, how many of the more efficient cups of tea would you need to make to compensate for the energy used to conduct the experiment?

However the question is - is the additional energy required to re-boil the water for the second cup more or less than the extra inefficiency of boiling two separate cups?

Obviously it depends a little on the efficiency of the kettle and its insulating properties, but mainly on how long you leave it between the two cups. As others have said, you would need to conduct an experiment to find out what this length of time is to get your answer - it may be so short that two separate boils is the only practical answer.

Also, how many of the more efficient cups of tea would you need to make to compensate for the energy used to conduct the experiment?

Wonko The Sane on 06 Nov 2012 - *31.103.21.85 whois?*

In reply to rubbercrumb:

But to answer the OP, REAL efficiency lies in AGREEING with her and getting HER to make the tea

> She wants two cups of tea in fairly quick succession.

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

>

> Or boil a single cups' worth from cold each time?

>

> Some important domestic point scoring at stake here...

>

> To use the least energy should she boil two cups' worth initially so the remaining water is hot for the second boil.

>

> Or boil a single cups' worth from cold each time?

>

> Some important domestic point scoring at stake here...

But to answer the OP, REAL efficiency lies in AGREEING with her and getting HER to make the tea

deepsoup - * * on 06 Nov 2012

In reply to spearing05:

No, in the second instance as with the first its heated, allowed to cool down for a time and then heated again.

If its empty as it cools down it has less thermal mass than it does with the water in, and it'll cool to a lower temperature in the time between cuppas. As it'll have a lower average temperature during that time, it'll also lose less heat to the environment.

> but in the second instance it is heated twice from cold

No, in the second instance as with the first its heated, allowed to cool down for a time and then heated again.

If its empty as it cools down it has less thermal mass than it does with the water in, and it'll cool to a lower temperature in the time between cuppas. As it'll have a lower average temperature during that time, it'll also lose less heat to the environment.

LastBoyScout on 06 Nov 2012

elsewhere on 06 Nov 2012

In reply to rubbercrumb:

it depends if anybody is watching

it depends if anybody is watching

deepsoup - * * on 06 Nov 2012

In reply to Wonko The Sane:

The element's job is to convert electricity into heat. If it's less than 100% efficient, it'll waste a bit of energy by er.. converting it into heat.

> element inefficiency

The element's job is to convert electricity into heat. If it's less than 100% efficient, it'll waste a bit of energy by er.. converting it into heat.

deepsoup - * * on 06 Nov 2012

In reply to malky_c:

I don't think so. The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)

> Boiling two cups together is more efficient than boiling two single cups, as a smaller proportion of the water is either surface area or in contact with the side if the kettle.

I don't think so. The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)

malky_c - * * on 06 Nov 2012

In reply to deepsoup: Surely you're putting the same amount of energy into the water regardless of how much there is, so twice as much water will need twice as much energy.

However it isn't a perfect system so it is all about combatting the heat loss from the kettle. There will be marginally less heat loss per ml of water if you boil a larger amount.

However it isn't a perfect system so it is all about combatting the heat loss from the kettle. There will be marginally less heat loss per ml of water if you boil a larger amount.

Carless - * * on 06 Nov 2012

In reply to antdav:

I agree that it's a very important domestic discussion, but I think it's a bit extreme suggesting they need to split up over it

> separate

I agree that it's a very important domestic discussion, but I think it's a bit extreme suggesting they need to split up over it

Robert Durran - * * on 06 Nov 2012

In reply to deepsoup:

You have nailed the problem here. Anyone disagreeing is wrong.

Two good similar problems:

(1) Should a swimming pool be allowed to cool down over night after closing and then reheated before opening next morning or kept at the required temperature over night.

(2) If you make a cup of tea, but then find you have to wait 15 minutes before drinking it, should you put the milk in immediately or wait until you are ready to drink it (You want to drink the tea as hot as possible).

> Boiling two cups requires twice as much energy as boiling one. Letting one cup cool down a bit and then boiling it again requires a bit more energy again.

You have nailed the problem here. Anyone disagreeing is wrong.

Two good similar problems:

(1) Should a swimming pool be allowed to cool down over night after closing and then reheated before opening next morning or kept at the required temperature over night.

(2) If you make a cup of tea, but then find you have to wait 15 minutes before drinking it, should you put the milk in immediately or wait until you are ready to drink it (You want to drink the tea as hot as possible).

rmt - * * on 06 Nov 2012

In reply to Robert Durran: You're very confident Robert, but not necessarily correct in the real world. It's easy to be correct based on theoretical models. The only person who has 'nailed it' is Wonko at 13:17. Spearing does come very close though at 13:12.

Here's another 'real world' problem for you. If you put two buckets of water into the freezer, one filled with hot water and one filled with cold water which one will freeze first. Obvious right? Ideal model and all that? Hmmm. Not so much. Google it.

Here's another 'real world' problem for you. If you put two buckets of water into the freezer, one filled with hot water and one filled with cold water which one will freeze first. Obvious right? Ideal model and all that? Hmmm. Not so much. Google it.

spearing05 - * * on 06 Nov 2012

In reply to deepsoup:

'Fraid you are wronmg there. A 1cm cube has a surface area of 6cm2 so a ratio of 6:1 A cube 10cm x10cm x 10 cm has a volume of 1000cm3 and a surface area of just 600cm2 so a ratio of 0.6:1

This is why animals that huddle together, whilst keeping the same volume reduce the surface area through which to loose heat.

> (In reply to malky_c)

> [...]

>

> I don't think so. The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)

> [...]

>

> I don't think so. The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)

'Fraid you are wronmg there. A 1cm cube has a surface area of 6cm2 so a ratio of 6:1 A cube 10cm x10cm x 10 cm has a volume of 1000cm3 and a surface area of just 600cm2 so a ratio of 0.6:1

This is why animals that huddle together, whilst keeping the same volume reduce the surface area through which to loose heat.

deepsoup - * * on 06 Nov 2012

In reply to spearing05:

Not as far as my kettle is concerned, but it only works with liquid water really. What size cubes of water do you put in yours?

> 'Fraid you are wronmg there.

Not as far as my kettle is concerned, but it only works with liquid water really. What size cubes of water do you put in yours?

spearing05 - * * on 06 Nov 2012

In reply to deepsoup: The shape isn't important, cubes are just easier to demonstrate. Do the same with a sphere or any other shape you care to think of. The biger a volume of any given shape the smaller the surface area in relation to the volume. Granted in a kettle the water will have the same top and bottom surface but the amount touching the sides will be a lower ratio for a bigger volume of water.

Again to make things easier I'll assume a square kettle. Lets say the kettle is a square section tube of 10cm by 10cm. If the water is 1cm deep there is a surface area of 240cm2 for a volume of 100cm3 (2.4:1)

If the water is 10 cm deep there is a surface area of 600cm2 to a volume of 1000cm3 so 0.6:1

So as I said above, for a given shape the surface area will get smaller as a ratio to the volume as the volume increases. This will hold true if the kettle is cylindrical or oval or even (as is more realistic) a kind of oval shape that flares slightly as it gets deeper.

Again to make things easier I'll assume a square kettle. Lets say the kettle is a square section tube of 10cm by 10cm. If the water is 1cm deep there is a surface area of 240cm2 for a volume of 100cm3 (2.4:1)

If the water is 10 cm deep there is a surface area of 600cm2 to a volume of 1000cm3 so 0.6:1

So as I said above, for a given shape the surface area will get smaller as a ratio to the volume as the volume increases. This will hold true if the kettle is cylindrical or oval or even (as is more realistic) a kind of oval shape that flares slightly as it gets deeper.

deepsoup - * * on 06 Nov 2012

In reply to spearing05:

0.1 litres of water. Top and bottom both 10cm^2. 40cm^2 touching the sides

1 litre of water. Top and bottom both 10cm^2. 400cm^2 touching the sides.

Here's a cut/paste from the post you're telling me this proves wrong:

"The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)"

> Again to make things easier I'll assume a square kettle. Lets say the kettle is a square section tube of 10cm by 10cm. If the water is 1cm deep ...

0.1 litres of water. Top and bottom both 10cm^2. 40cm^2 touching the sides

> If the water is 10 cm deep ...

1 litre of water. Top and bottom both 10cm^2. 400cm^2 touching the sides.

Here's a cut/paste from the post you're telling me this proves wrong:

"The area of water in contact with the element is constant (in a modern kettle that's the bottom). In a cylindrical kettle, the surface area is also a constant, and the amount in contact with the sides is proportional to the amount of water. (So increases in proportion to the area in contact with the element as you add more water.)"

Robert Durran - * * on 06 Nov 2012

In reply to rmt:

Yes, I was overconfident. Spearing and Wonko make a good point. However, a back of the envelope calculation taking this into account and assuming that boiling both cups initially together is not used as a tactic for allowing the two cups to be drunk in quicker succession than is possible with separate boiling (ie the cooling time and reboiling time is at least as long as it takes to boil a single cup) strongly suggests that boiling the cups separately is more efficient. I would put money on it! I would certainly be very interested to see some experimental results.

Yes, this is a well known surprising effect, but I hardly think anything as weird is going on in our kettle.

> (In reply to Robert Durran) You're very confident Robert, but not necessarily correct in the real world. It's easy to be correct based on theoretical models. The only person who has 'nailed it' is Wonko at 13:17. Spearing does come very close though at 13:12.

Yes, I was overconfident. Spearing and Wonko make a good point. However, a back of the envelope calculation taking this into account and assuming that boiling both cups initially together is not used as a tactic for allowing the two cups to be drunk in quicker succession than is possible with separate boiling (ie the cooling time and reboiling time is at least as long as it takes to boil a single cup) strongly suggests that boiling the cups separately is more efficient. I would put money on it! I would certainly be very interested to see some experimental results.

> Here's another 'real world' problem for you. If you put two buckets of water into the freezer, one filled with hot water and one filled with cold water which one will freeze first. Obvious right? Ideal model and all that?

Yes, this is a well known surprising effect, but I hardly think anything as weird is going on in our kettle.

nufkin - * * on 06 Nov 2012

In reply to Robert Durran:

1) - The first choice, going by what my plumber told me. Though wouldn't it also depend how cold it gets without heating and how hot it needs to be (I think I read somewhere that it takes a similar amount of energy to heat water from 0 to 20 degrees as it does from 20 to boiling)?

2) - The latter works out fine, accidental experimentation has repeatedly shown. Didn't this come up in the Da Vinci Code as well?

> (In reply to deepsoup)

> [...]

>

> You have nailed the problem here. Anyone disagreeing is wrong.

>

> Two good similar problems:

>

> (1) Should a swimming pool be allowed to cool down over night after closing and then reheated before opening next morning or kept at the required temperature over night.

>

> (2) If you make a cup of tea, but then find you have to wait 15 minutes before drinking it, should you put the milk in immediately or wait until you are ready to drink it (You want to drink the tea as hot as possible).

> [...]

>

> You have nailed the problem here. Anyone disagreeing is wrong.

>

> Two good similar problems:

>

> (1) Should a swimming pool be allowed to cool down over night after closing and then reheated before opening next morning or kept at the required temperature over night.

>

> (2) If you make a cup of tea, but then find you have to wait 15 minutes before drinking it, should you put the milk in immediately or wait until you are ready to drink it (You want to drink the tea as hot as possible).

1) - The first choice, going by what my plumber told me. Though wouldn't it also depend how cold it gets without heating and how hot it needs to be (I think I read somewhere that it takes a similar amount of energy to heat water from 0 to 20 degrees as it does from 20 to boiling)?

2) - The latter works out fine, accidental experimentation has repeatedly shown. Didn't this come up in the Da Vinci Code as well?

Robert Durran - * * on 07 Nov 2012

In reply to spearing05:

I think this is wrong.

The numbers are easier with 8 cups versus 1 cup:

Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1, so heat loss ratio 8x4:1 = 32:1,so once all 8 cups are heated separately, the ratio is 32/8:1 = 4:1

ie 4 times as much heat lost boiling the cups together rather than separately (there will be some small second order affects).

For 2 cups the ratio will be 2^(2/3):1 = 1.6:1 (approx)

The surface ares to volume argument giving greater heat loss per volume in a given time for smaller volumes is not relevant here.

> Boiling two cups together and using both takes less energy than two seperately. The kettle is only heated once and the surface area to volume ratio is smaller with a larger volume so less heat is lost to the surroundings during the heating cycle.

I think this is wrong.

The numbers are easier with 8 cups versus 1 cup:

Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1, so heat loss ratio 8x4:1 = 32:1,so once all 8 cups are heated separately, the ratio is 32/8:1 = 4:1

ie 4 times as much heat lost boiling the cups together rather than separately (there will be some small second order affects).

For 2 cups the ratio will be 2^(2/3):1 = 1.6:1 (approx)

The surface ares to volume argument giving greater heat loss per volume in a given time for smaller volumes is not relevant here.

Robert Durran - * * on 07 Nov 2012

In reply to Robert Durran:

..............so it definitely more efficient to boil the two cups separately.

..............so it definitely more efficient to boil the two cups separately.

Robert Durran - * * on 07 Nov 2012

In reply to Robert Durran:

I have just done a more detailed analysis involving some fiddly differential equations and making the following assumptions:

(1) The rate of heat loss is proportional to the surface area of the volume of water and to the temperature difference between the water and room.

(2) The temperature of all the water is the same at any given time.

(3) The specifiic heat capacity of the water is the same at all temperatures.

(4) The rate of heat loss is small compared to the rate at which the kettle puts heat into the kettle (I think this is reasonable, otherwise the kettle would take absolutely ages to boil!)

It turns out that this gives a total heat loss proportional both to the volume of water and to the surface area of this volume. Heating two cups separately halfs the volume but we have to do the heating twice (cancels out), so the overall heat loss is proportional to the surface area. This will always be greater for two cups in the kettle than one, so it is confirmed that heating the two cups separately is more efficient.

> (In reply to spearing05)

> I think this is wrong.

> The numbers are easier with 8 cups versus 1 cup:

> Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1, so heat loss ratio 8x4:1 = 32:1,so once all 8 cups are heated separately, the ratio is 32/8:1 = 4:1

> ie 4 times as much heat lost boiling the cups together rather than separately (there will be some small second order affects).

> For 2 cups the ratio will be 2^(2/3):1 = 1.6:1 (approx)

> I think this is wrong.

> The numbers are easier with 8 cups versus 1 cup:

> Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1, so heat loss ratio 8x4:1 = 32:1,so once all 8 cups are heated separately, the ratio is 32/8:1 = 4:1

> ie 4 times as much heat lost boiling the cups together rather than separately (there will be some small second order affects).

> For 2 cups the ratio will be 2^(2/3):1 = 1.6:1 (approx)

I have just done a more detailed analysis involving some fiddly differential equations and making the following assumptions:

(1) The rate of heat loss is proportional to the surface area of the volume of water and to the temperature difference between the water and room.

(2) The temperature of all the water is the same at any given time.

(3) The specifiic heat capacity of the water is the same at all temperatures.

(4) The rate of heat loss is small compared to the rate at which the kettle puts heat into the kettle (I think this is reasonable, otherwise the kettle would take absolutely ages to boil!)

It turns out that this gives a total heat loss proportional both to the volume of water and to the surface area of this volume. Heating two cups separately halfs the volume but we have to do the heating twice (cancels out), so the overall heat loss is proportional to the surface area. This will always be greater for two cups in the kettle than one, so it is confirmed that heating the two cups separately is more efficient.

Robert Durran - * * on 07 Nov 2012

In reply to Robert Durran:

I have just realised that this assumption is not needed. All that is required is that the rate of heat loss is not great enough to stop the kettle boiling at all.

> (4) The rate of heat loss is small compared to the rate at which the kettle puts heat into the kettle (I think this is reasonable, otherwise the kettle would take absolutely ages to boil!)

I have just realised that this assumption is not needed. All that is required is that the rate of heat loss is not great enough to stop the kettle boiling at all.

tom_in_edinburgh - * * on 07 Nov 2012

In reply to Robert Durran:

I'm not sure I agree the relevant surface area ratio is 4:1. With a modern kettle with a plastic body the surface area of concern is the top face of the water where it is next to air, the water next to the plastic kettle is well insulated.

So, if you approximate the kettle as a perfect cylinder I'd argue the surface area ratio for 8 cups done separately compared to 8 cups done all at once is 8:1 because in the 8 cup case there are 8 'top surfaces' instead of one.

> (In reply to spearing05)

> [...]

>

> I think this is wrong.

> The numbers are easier with 8 cups versus 1 cup:

> Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1

> [...]

>

> I think this is wrong.

> The numbers are easier with 8 cups versus 1 cup:

> Volume ratio 8:1 so heating time ratio 8:1, surface area ratio 4:1

I'm not sure I agree the relevant surface area ratio is 4:1. With a modern kettle with a plastic body the surface area of concern is the top face of the water where it is next to air, the water next to the plastic kettle is well insulated.

So, if you approximate the kettle as a perfect cylinder I'd argue the surface area ratio for 8 cups done separately compared to 8 cups done all at once is 8:1 because in the 8 cup case there are 8 'top surfaces' instead of one.

spearing05 - * * on 07 Nov 2012

In reply to Robert Durran: Just a quicky coz I'm in work - not had time to follow all your argument through. Of couse the surface area will be greater for a larger volume BUT the ratio to volume is different.

Again imagine two cubes of water at the same (high) temperature, each has 6 faces through which to loose heat. Place these next to each other and now two of those faces are touching each other so there is no net heat loss through these faces. Therefore by taking two seperate volumes and making them (in effect) one single volume you have reduced the amount of heat lost by 1/6

When you heat water in the kettle the amount of heat that the element puts in has to be greater than the heat lost. Heat is only put in at the element. Heat is only lost at the surface area. The element has the same surface to transfer heat to the water however deep it is so the heat in will be a constant however deep the water is. A fixed volume of water requires a fixed amount of energy to raise its temperature a fixed amount whether it is done in one go or 10 goes. the only differences will be the energy that is wasted due to heat loss to the surroundings. As I showed above a large volume will loose heat slower than a small volume so boiling a larger volume will require less energy than two smaller ones.

Again imagine two cubes of water at the same (high) temperature, each has 6 faces through which to loose heat. Place these next to each other and now two of those faces are touching each other so there is no net heat loss through these faces. Therefore by taking two seperate volumes and making them (in effect) one single volume you have reduced the amount of heat lost by 1/6

When you heat water in the kettle the amount of heat that the element puts in has to be greater than the heat lost. Heat is only put in at the element. Heat is only lost at the surface area. The element has the same surface to transfer heat to the water however deep it is so the heat in will be a constant however deep the water is. A fixed volume of water requires a fixed amount of energy to raise its temperature a fixed amount whether it is done in one go or 10 goes. the only differences will be the energy that is wasted due to heat loss to the surroundings. As I showed above a large volume will loose heat slower than a small volume so boiling a larger volume will require less energy than two smaller ones.

Robert Durran - * * on 07 Nov 2012

In reply to tom_in_edinburgh:

Is plastic a much better insulator than air - my plastic kettle gets pretty hot!

I think you mean 1:8 (both cups together: separate cups is the way round I had it). Anyway, my ratio was comparing both cups together with ONE cup separately, so, for a cylindrical kettle with perfectly insulating walls the relevant ratio would be 1:1. This would give equal efficiency* for the two methods (same result for my more detailed analysis). Anyway, the plastic certainly isn't a perfect insulator, so heating both cups together will still be more efficient and once the cooling and reheating of the second cup is taken into account (as in the original problem) the more efficient method will certainly be to heat the cups separately.

*This situation is simple to understand - all relevant surface areas (top surfaces) are equal. With 8 separate cups the heating time for each is 1/8 that for all 8 cups together, so each loses 1/8 the heat, but this is cancelled out by the fact that there are eight of them (same argument for any number of cups).

> (In reply to Robert Durran)

> I'm not sure I agree the relevant surface area ratio is 4:1. With a modern kettle with a plastic body the surface area of concern is the top face of the water where it is next to air, the water next to the plastic kettle is well insulated.

> I'm not sure I agree the relevant surface area ratio is 4:1. With a modern kettle with a plastic body the surface area of concern is the top face of the water where it is next to air, the water next to the plastic kettle is well insulated.

Is plastic a much better insulator than air - my plastic kettle gets pretty hot!

>

> So, if you approximate the kettle as a perfect cylinder I'd argue the surface area ratio for 8 cups done separately compared to 8 cups done all at once is 8:1 because in the 8 cup case there are 8 'top surfaces' instead of one.

> So, if you approximate the kettle as a perfect cylinder I'd argue the surface area ratio for 8 cups done separately compared to 8 cups done all at once is 8:1 because in the 8 cup case there are 8 'top surfaces' instead of one.

I think you mean 1:8 (both cups together: separate cups is the way round I had it). Anyway, my ratio was comparing both cups together with ONE cup separately, so, for a cylindrical kettle with perfectly insulating walls the relevant ratio would be 1:1. This would give equal efficiency* for the two methods (same result for my more detailed analysis). Anyway, the plastic certainly isn't a perfect insulator, so heating both cups together will still be more efficient and once the cooling and reheating of the second cup is taken into account (as in the original problem) the more efficient method will certainly be to heat the cups separately.

*This situation is simple to understand - all relevant surface areas (top surfaces) are equal. With 8 separate cups the heating time for each is 1/8 that for all 8 cups together, so each loses 1/8 the heat, but this is cancelled out by the fact that there are eight of them (same argument for any number of cups).

Flinticus - * * on 07 Nov 2012

Robert Durran - * * on 07 Nov 2012

In reply to spearing05:

But it turns out this ratio is not important.

But what you (and others) are overlooking is that a bigger volume takes longer to heat, so there is more time for heat to escape (so more does escape!). This more than cancels out the reduction in heat loss due to reduction in total surface area (see my argument above).

> (In reply to Robert Durran) Just a quicky coz I'm in work - not had time to follow all your argument through. Of couse the surface area will be greater for a larger volume BUT the ratio to volume is different.

But it turns out this ratio is not important.

>

> Again imagine two cubes of water at the same (high) temperature, each has 6 faces through which to loose heat. Place these next to each other and now two of those faces are touching each other so there is no net heat loss through these faces. Therefore by taking two seperate volumes and making them (in effect) one single volume you have reduced the amount of heat lost by 1/6

> Again imagine two cubes of water at the same (high) temperature, each has 6 faces through which to loose heat. Place these next to each other and now two of those faces are touching each other so there is no net heat loss through these faces. Therefore by taking two seperate volumes and making them (in effect) one single volume you have reduced the amount of heat lost by 1/6

But what you (and others) are overlooking is that a bigger volume takes longer to heat, so there is more time for heat to escape (so more does escape!). This more than cancels out the reduction in heat loss due to reduction in total surface area (see my argument above).

Robert Durran - * * on 07 Nov 2012

In reply to Robert Durran:

Moeover, having established that the both cups together method is less efficient, it in fact takes even longer since the extra heat lost during heating has to be put back in - the time factor then becomes even more important and the relative inefficiency even worse (though the correction will be small for a powerful kettle).

> But what you (and others) are overlooking is that a bigger volume takes longer to heat, so there is more time for heat to escape (so more does escape!). This more than cancels out the reduction in heat loss due to reduction in total surface area (see my argument above).

Moeover, having established that the both cups together method is less efficient, it in fact takes even longer since the extra heat lost during heating has to be put back in - the time factor then becomes even more important and the relative inefficiency even worse (though the correction will be small for a powerful kettle).

deepsoup - * * on 07 Nov 2012

In reply to tom_in_edinburgh:

I'm not sure the relevant surface area is that of the water anyway. It's the outside of the kettle that's losing heat to the surroundings.

Even in the case of a plastic kettle (and obviously for a metal one) I don't think it actually is all that well insulated.

All this talk of volume/surface area ratios kinda sorta assumes the water is losing heat by radiation. The kettle might be, mostly, but that's certainly not going to be the case for the water.

I'd guess the water is mostly losing heat via conduction into the sides of the kettle, by conduction into the air in contact with the water's surface (both air and water will likely have convection currents flowing), and by the evaporation of steam (some of which will escape through the spout, some of which will condense out on the cooler areas of the inside of the kettle).

Also worth remembering, with all this talk of heat loss, that the purpose of a kettle is for the water to *gain* heat overall. If the water itself were perfectly insulated from its surroundings you wouldn't be able to heat it up. Whatever determines the rate that heat is lost, the amount of energy that's lost is also a function of the amount of time the kettle is being heated. Slower to boil = less 'efficient'.

> I'm not sure I agree the relevant surface area ratio is

I'm not sure the relevant surface area is that of the water anyway. It's the outside of the kettle that's losing heat to the surroundings.

Even in the case of a plastic kettle (and obviously for a metal one) I don't think it actually is all that well insulated.

All this talk of volume/surface area ratios kinda sorta assumes the water is losing heat by radiation. The kettle might be, mostly, but that's certainly not going to be the case for the water.

I'd guess the water is mostly losing heat via conduction into the sides of the kettle, by conduction into the air in contact with the water's surface (both air and water will likely have convection currents flowing), and by the evaporation of steam (some of which will escape through the spout, some of which will condense out on the cooler areas of the inside of the kettle).

Also worth remembering, with all this talk of heat loss, that the purpose of a kettle is for the water to *gain* heat overall. If the water itself were perfectly insulated from its surroundings you wouldn't be able to heat it up. Whatever determines the rate that heat is lost, the amount of energy that's lost is also a function of the amount of time the kettle is being heated. Slower to boil = less 'efficient'.

deepsoup - * * on 07 Nov 2012

Robert Durran - * * on 07 Nov 2012

In reply to deepsoup:

Only if the water was also insulated from the heating element - but that would be a really crap kettle!

Yes, this is the point that some people have been missing. A bigger volume in one go (2 cups) takes twice as long, so twice the heat loss (which more than cancels out gains in efficiency from surface area considerations).

> (In reply to tom_in_edinburgh)

> Also worth remembering, with all this talk of heat loss, that the purpose of a kettle is for the water to *gain* heat overall. If the water itself were perfectly insulated from its surroundings you wouldn't be able to heat it up.

Only if the water was also insulated from the heating element - but that would be a really crap kettle!

> Whatever determines the rate that heat is lost, the amount of energy that's lost is also a function of the amount of time the kettle is being heated. Slower to boil = less 'efficient'.

Yes, this is the point that some people have been missing. A bigger volume in one go (2 cups) takes twice as long, so twice the heat loss (which more than cancels out gains in efficiency from surface area considerations).

tom_in_edinburgh - * * on 07 Nov 2012

In reply to Robert Durran:

OK - I decided to do the experiment and nature has now spoken!

Experimental conditions: 1 large cup - about 450ml (had to be big enough to hit the minimum line on the kettle). Gap between making cups of tea 3min. Modern plastic jug kettle.

Option 1. Boil two cups at once. Kettle runs for 1min 48s to boil two cups then 3 mins later 11s to reheat the second cup. Total 1min 59s. Time is proportional to energy since E = (Power of kettle element) x Time.

Before running second experiment kettle was flushed with cold water and left until it was no longer warm to the touch.

Option 2. Boil first cup 1.08s. Empty kettle and wait three min. Boil second cup 58s. Total time 2 min 6s. Second cup is faster because the kettle itself is hot after the first boil.

So the measurements say it is more efficient to boil two cups at once.

I think the key factors which make the single boil option more efficient are the thermal mass of the kettle and the loss of heat from the surface of the water by vapour before the kettle actually boils. It was noticeable that the outside of the kettle was hottest at the top well above the surface of the water which suggests to me the mechanism is heat transfer from steam rising from the water surface.

OK - I decided to do the experiment and nature has now spoken!

Experimental conditions: 1 large cup - about 450ml (had to be big enough to hit the minimum line on the kettle). Gap between making cups of tea 3min. Modern plastic jug kettle.

Option 1. Boil two cups at once. Kettle runs for 1min 48s to boil two cups then 3 mins later 11s to reheat the second cup. Total 1min 59s. Time is proportional to energy since E = (Power of kettle element) x Time.

Before running second experiment kettle was flushed with cold water and left until it was no longer warm to the touch.

Option 2. Boil first cup 1.08s. Empty kettle and wait three min. Boil second cup 58s. Total time 2 min 6s. Second cup is faster because the kettle itself is hot after the first boil.

So the measurements say it is more efficient to boil two cups at once.

I think the key factors which make the single boil option more efficient are the thermal mass of the kettle and the loss of heat from the surface of the water by vapour before the kettle actually boils. It was noticeable that the outside of the kettle was hottest at the top well above the surface of the water which suggests to me the mechanism is heat transfer from steam rising from the water surface.

Robert Durran - * * on 07 Nov 2012

In reply to tom_in_edinburgh:

Yes, I suspect the steam is probably a big factor (and one which was not accounted for in my model - though I did have a lot of fun solving the differential equations!). It would be interesting to get some results for an enclosed kettle with no room for steam.

I shall repeat the experiment with a metal kettle at work before I go home (conduction might then be a bigger factor, favouring the separate cups as in my model) and then at home with my plastic kettle. I'll get back with the results later.

I suspect those who predicted that separate cups would be less efficient will now come back crowing that I was wrong, even though they were right for the wrong reasons (incorrect arguments about surface area)!

> (In reply to Robert Durran)

>

> OK - I decided to do the experiment and nature has now spoken!......

>

> OK - I decided to do the experiment and nature has now spoken!......

> ......... I think the key factors which make the single boil option more efficient are the thermal mass of the kettle and the loss of heat from the surface of the water by vapour before the kettle actually boils. It was noticeable that the outside of the kettle was hottest at the top well above the surface of the water which suggests to me the mechanism is heat transfer from steam rising from the water surface.

Yes, I suspect the steam is probably a big factor (and one which was not accounted for in my model - though I did have a lot of fun solving the differential equations!). It would be interesting to get some results for an enclosed kettle with no room for steam.

I shall repeat the experiment with a metal kettle at work before I go home (conduction might then be a bigger factor, favouring the separate cups as in my model) and then at home with my plastic kettle. I'll get back with the results later.

I suspect those who predicted that separate cups would be less efficient will now come back crowing that I was wrong, even though they were right for the wrong reasons (incorrect arguments about surface area)!

spearing05 - * * on 07 Nov 2012

In reply to Robert Durran:

I'm not sure where you are coming from with this time thing. Obviously it takes longer to boil 2 cups worth than 1 but that isnt what we are talking about. We are talking about the time to boil 2 cups either together or sepearately and as the experiment above showed the total time is shorter to boil them together.

As has been mentioned, by far the greatest heat loss is through evaporation at the surface. I work with vacuum, If we have a water leak in our vacuum chambers it creates ice as the pressure is way below the vapour pressure of water so it evaporates taking so much heat from the remaining water that it freezes. This is the reason that a saucepan without a lid takes much longer to boil than one with. With the lid on, the air above the water becomes saturated thus reducing the rate of evaporation and also the heat loss. For those talking about the lack of insulation on the side of the kettle and how warm it gets, try holding your hand over the spout and see how much heaat is lost there (actually don't unless you want serious burns and loss of flesh)

This loss of heat will be the same however much water is in the kettle as the top surface that is exposed to the air remains (roughly) the same. The difference being the amount of heat lost as a percentage of the total energy in the water will be lowr for a kettle with more water.

Glad someone else did the experiment as I was serriously considering having to do it to prove the point. Now we just need the experiment repeating with various time intervals and the results plotting on a graph till the time can be found where it is more economical to boil two individual cups rather than two together and reheating. As stated way back on the thread this time is the determining factor in which is the most economical.

> (In reply to deepsoup)

> [...]

>

> [...]

>

> Only if the water was also insulated from the heating element - but that would be a really crap kettle!

>

> [...]

>

> Yes, this is the point that some people have been missing. A bigger volume in one go (2 cups) takes twice as long, so twice the heat loss (which more than cancels out gains in efficiency from surface area considerations).

> [...]

>

> [...]

>

> Only if the water was also insulated from the heating element - but that would be a really crap kettle!

>

> [...]

>

> Yes, this is the point that some people have been missing. A bigger volume in one go (2 cups) takes twice as long, so twice the heat loss (which more than cancels out gains in efficiency from surface area considerations).

I'm not sure where you are coming from with this time thing. Obviously it takes longer to boil 2 cups worth than 1 but that isnt what we are talking about. We are talking about the time to boil 2 cups either together or sepearately and as the experiment above showed the total time is shorter to boil them together.

As has been mentioned, by far the greatest heat loss is through evaporation at the surface. I work with vacuum, If we have a water leak in our vacuum chambers it creates ice as the pressure is way below the vapour pressure of water so it evaporates taking so much heat from the remaining water that it freezes. This is the reason that a saucepan without a lid takes much longer to boil than one with. With the lid on, the air above the water becomes saturated thus reducing the rate of evaporation and also the heat loss. For those talking about the lack of insulation on the side of the kettle and how warm it gets, try holding your hand over the spout and see how much heaat is lost there (actually don't unless you want serious burns and loss of flesh)

This loss of heat will be the same however much water is in the kettle as the top surface that is exposed to the air remains (roughly) the same. The difference being the amount of heat lost as a percentage of the total energy in the water will be lowr for a kettle with more water.

Glad someone else did the experiment as I was serriously considering having to do it to prove the point. Now we just need the experiment repeating with various time intervals and the results plotting on a graph till the time can be found where it is more economical to boil two individual cups rather than two together and reheating. As stated way back on the thread this time is the determining factor in which is the most economical.

Robert Durran - * * on 07 Nov 2012

In reply to spearing05:

My point is that if you boil some water quickly, there is less time for heat to escape through conduction, so less will escape. If you follow my argument you will find that if all heat loss were through conduction, this would make boiling both cups together less efficient (DESPITE the smaller surface area).

Yes, the experimental result and your explanation suggests that this is the case.

If one could design a sealed kettle with no evaporation, I still contend that boiling the cups separately would always be nore efficient (assuming the ketle did not explode!).

The existence of this critical time was was arrived way back on the thread at using the erroneous argument about conduction and surface area which I fixed (see above). It will still exist, but for other reasons!

> (In reply to Robert Durran)

> I'm not sure where you are coming from with this time thing. Obviously it takes longer to boil 2 cups worth than 1 but that isnt what we are talking about.

> I'm not sure where you are coming from with this time thing. Obviously it takes longer to boil 2 cups worth than 1 but that isnt what we are talking about.

My point is that if you boil some water quickly, there is less time for heat to escape through conduction, so less will escape. If you follow my argument you will find that if all heat loss were through conduction, this would make boiling both cups together less efficient (DESPITE the smaller surface area).

> As has been mentioned, by far the greatest heat loss is through evaporation at the surface.

Yes, the experimental result and your explanation suggests that this is the case.

If one could design a sealed kettle with no evaporation, I still contend that boiling the cups separately would always be nore efficient (assuming the ketle did not explode!).

> Now we just need the experiment repeating with various time intervals and the results plotting on a graph till the time can be found where it is more economical to boil two individual cups rather than two together and reheating. As stated way back on the thread this time is the determining factor in which is the most economical.

The existence of this critical time was was arrived way back on the thread at using the erroneous argument about conduction and surface area which I fixed (see above). It will still exist, but for other reasons!

spearing05 - * * on 08 Nov 2012

In reply to Robert Durran:

Where do you get this boiling the water more quickly from?

You would be right if you did boil the water more quickly but you dont. Boiling both together or both seperately still requires the same amount of heat. It is quicker to boil one cup than two but you have to do this twice (which is what I think you may be missing out?) In a perfect system with no losses it would take exactly twice as long to boil two cups as it would one. In such a system boiling two cups together would take the same time as two individually. The small difference in time that we see in the experimntal data is due to the differences in the losses between the two systems - which are greater when boiling seperately than together hence the seperate time being slightly more than half the together time.

> (In reply to spearing05)

> [...]

>

> My point is that if you boil some water quickly, there is less time for heat to escape through conduction, so less will escape. If you follow my argument you will find that if all heat loss were through conduction, this would make boiling both cups together less efficient (DESPITE the smaller surface area).

>

> [...]

>

> My point is that if you boil some water quickly, there is less time for heat to escape through conduction, so less will escape. If you follow my argument you will find that if all heat loss were through conduction, this would make boiling both cups together less efficient (DESPITE the smaller surface area).

>

Where do you get this boiling the water more quickly from?

You would be right if you did boil the water more quickly but you dont. Boiling both together or both seperately still requires the same amount of heat. It is quicker to boil one cup than two but you have to do this twice (which is what I think you may be missing out?) In a perfect system with no losses it would take exactly twice as long to boil two cups as it would one. In such a system boiling two cups together would take the same time as two individually. The small difference in time that we see in the experimntal data is due to the differences in the losses between the two systems - which are greater when boiling seperately than together hence the seperate time being slightly more than half the together time.

rockchomper on 08 Nov 2012

In reply to rubbercrumb:

gggaawwwdddd....scientistsrus....puttit in a flippin' big flask and hush ya cribbin' y'all....sounds like 'you' need to put a bit more 'honey' with

her tea...

gggaawwwdddd....scientistsrus....puttit in a flippin' big flask and hush ya cribbin' y'all....sounds like 'you' need to put a bit more 'honey' with

her tea...

Robert Durran - * * on 09 Nov 2012

In reply to spearing05:

No, I am not missing this out.

I shall explain it again slightly differently:

Take your two cubes. Suppose each has a surface area A. If they are combined into a single lump, the surface area A' of this lump will be more than A (by a factor of 2^(2/3)=1.6 approx if they are moulded into a single larger cube).

Now, suppose the two cubes are heated separately, each taking time t. The heat loss from each will be proportional to A x t = At, and so the total heat loss from both will be propoptional to 2 x At = 2At (see, I have not missed this out!)

Now consider heating the combined lump. The time taken will be 2t (twice as much stuff to heat), so the heat loss will be proportional to A' x 2t = 2A't.

But, A' is greater than A, so the heat loss for the combined lump will be greater than the total for the two lumps separately.

In fact, the time taken to heat the combined lump will not be precisely twice that for a single cube, and an adjustment is needed taking into account the extra time needed to compensate for heat losses. My more detailed analysis using differential equations actually shows that this second order correction actually shows a greater relative inefficienccy using the combined lump than does the approximate argument above.

Yes.

Note that my analysis does not take account of losses via steam; as I have said before, I accept that this is the dominant factor, more important than conduction, meaning that, in practice, the combined method is more efficient.

> (In reply to Robert Durran)

> Boiling both together or both seperately still requires the same amount of heat. It is quicker to boil one cup than two but you have to do this twice (which is what I think you may be missing out?)

No, I am not missing this out.

I shall explain it again slightly differently:

Take your two cubes. Suppose each has a surface area A. If they are combined into a single lump, the surface area A' of this lump will be more than A (by a factor of 2^(2/3)=1.6 approx if they are moulded into a single larger cube).

Now, suppose the two cubes are heated separately, each taking time t. The heat loss from each will be proportional to A x t = At, and so the total heat loss from both will be propoptional to 2 x At = 2At (see, I have not missed this out!)

Now consider heating the combined lump. The time taken will be 2t (twice as much stuff to heat), so the heat loss will be proportional to A' x 2t = 2A't.

But, A' is greater than A, so the heat loss for the combined lump will be greater than the total for the two lumps separately.

In fact, the time taken to heat the combined lump will not be precisely twice that for a single cube, and an adjustment is needed taking into account the extra time needed to compensate for heat losses. My more detailed analysis using differential equations actually shows that this second order correction actually shows a greater relative inefficienccy using the combined lump than does the approximate argument above.

> In a perfect system with no losses it would take exactly twice as long to boil two cups as it would one.

Yes.

> The small difference in time that we see in the experimental data......

Note that my analysis does not take account of losses via steam; as I have said before, I accept that this is the dominant factor, more important than conduction, meaning that, in practice, the combined method is more efficient.

birdie num num - * * on 09 Nov 2012

In reply to rubbercrumb:

You can get two cups of tea out of one Yorkshire teabag

You can get two cups of tea out of one Yorkshire teabag

rockchomper on 09 Nov 2012

In reply to birdie num num:

...three cups if you hang it out on the line to dry.....

...three cups if you hang it out on the line to dry.....

Jim C - * * on 09 Nov 2012

In reply to rockchomper:

You. Are wasting your virtual breath, I suggested the thermos flask miles back and they ignored it. I would not mind, but the guy who did the practical stuff missed out the flask option!!

Even 20mins + gap and the flask would win in my experience.

I was recently travelling around Shetland from Bod to Bod making tea in the car from water boiled up from flask filled each morning. It was an efficient method both in time and gas saving.

> (In reply to rubbercrumb)

> gggaawwwdddd....scientistsrus....puttit in a flippin' big flask and hush ya cribbin' y'all....

> gggaawwwdddd....scientistsrus....puttit in a flippin' big flask and hush ya cribbin' y'all....

You. Are wasting your virtual breath, I suggested the thermos flask miles back and they ignored it. I would not mind, but the guy who did the practical stuff missed out the flask option!!

Even 20mins + gap and the flask would win in my experience.

I was recently travelling around Shetland from Bod to Bod making tea in the car from water boiled up from flask filled each morning. It was an efficient method both in time and gas saving.

rockchomper on 09 Nov 2012

In reply to Jim C:

hi y'all....just off to buy some 'more sugar' for me brew....one lump or two sweeties....hi ho, hi ho, its off to get some biccies i go...now then, how long would it take for a rich tea to go soggy if i dunked it in freshly made tea as opposed to tea from a flask....

hi y'all....just off to buy some 'more sugar' for me brew....one lump or two sweeties....hi ho, hi ho, its off to get some biccies i go...now then, how long would it take for a rich tea to go soggy if i dunked it in freshly made tea as opposed to tea from a flask....

Parrys_apprentice - * * on 09 Nov 2012

In reply to rockchomper:

has anyone just timed it yet, as suggested?

Surely this whole problem is totally dependant on how long you wait between cuppas.

has anyone just timed it yet, as suggested?

Surely this whole problem is totally dependant on how long you wait between cuppas.

deepsoup - * * on 09 Nov 2012

In reply to Parrys_apprentice:

Yep, tom_in_edinburgh did.

He found it was (just) more efficient to boil two cups worth together. (Well, gert big mugs worth.) In a plastic kettle, with three minutes between cuppas.

Seems like a short time between cups of tea to me, especially with almost a pint mug - but the mornings are cold and dark up there, so fair enough. ;o)

I've just done it too. (Clearly got way to much time on my hands.)

My results, under similar conditions but using a 350ml "cup" and a stainless steel kettle:

Boiled separately: 58s the first, and 52s the second - 1min 51s total.

Boiled together: 1:45s initially and 11s to re-boil - 1min 56s total.

So in my case*slightly* more efficient to boil the two cuppa's worth separately.

Yep.

And the time at which the two approaches are equally (in)efficient is dependant on how well insulated your kettle is.

Or, as some have suggested, whether or not you can be arsed to decant the other half of the water into a vacuum flask, or lag your kettle.

Or invent a vacuum flask kettle?

I don't like tea myself. Can anyone tell me the most energy efficient way to make my coffee in the morning? (*Not* instant coffee - ugh!)

> has anyone just timed it yet, as suggested?

Yep, tom_in_edinburgh did.

He found it was (just) more efficient to boil two cups worth together. (Well, gert big mugs worth.) In a plastic kettle, with three minutes between cuppas.

Seems like a short time between cups of tea to me, especially with almost a pint mug - but the mornings are cold and dark up there, so fair enough. ;o)

I've just done it too. (Clearly got way to much time on my hands.)

My results, under similar conditions but using a 350ml "cup" and a stainless steel kettle:

Boiled separately: 58s the first, and 52s the second - 1min 51s total.

Boiled together: 1:45s initially and 11s to re-boil - 1min 56s total.

So in my case

> Surely this whole problem is totally dependant on how long you wait between cuppas.

Yep.

And the time at which the two approaches are equally (in)efficient is dependant on how well insulated your kettle is.

Or, as some have suggested, whether or not you can be arsed to decant the other half of the water into a vacuum flask, or lag your kettle.

Or invent a vacuum flask kettle?

I don't like tea myself. Can anyone tell me the most energy efficient way to make my coffee in the morning? (*Not* instant coffee - ugh!)

Bulls Crack - * * on 09 Nov 2012

In reply to rubbercrumb:

Don't do what I did 2 nights ago with a hob-top espresso makers and forget to replace lower filter after repalcing a gasket.

Quite impressive

Don't do what I did 2 nights ago with a hob-top espresso makers and forget to replace lower filter after repalcing a gasket.

Quite impressive

deepsoup - * * on 09 Nov 2012

In reply to Bulls Crack:

Eek!

I have one of those but hardly ever use mine. Its quite a big-un and it doesn't seem to work well unless its full - if its just me drinking the coffee I end up shaking like a small nervous dog. ;o)

Eek!

I have one of those but hardly ever use mine. Its quite a big-un and it doesn't seem to work well unless its full - if its just me drinking the coffee I end up shaking like a small nervous dog. ;o)

Robert Durran - * * on 09 Nov 2012

In reply to deepsoup:

Now that is interesting!

My trial with a stainless steel kettle gave much greater efficiency boiling together (even not allowing for cooling time). As Spearing said this is probably due to heat losses through steam.

Not if steam is a significant factor (my argument above concerning conduction versus Tom's and my results shows that something other than conduction certainly can be important). Maybe your kettle is remarkably steam efficient!,

> My results, under similar conditions but using a 350ml "cup" and a stainless steel kettle:

> Boiled separately: 58s the first, and 52s the second - 1min 51s total.

> Boiled together: 1:45s initially and 11s to re-boil - 1min 56s total.

>

> So in my case slightly more efficient to boil the two cuppa's worth separately.

> Boiled separately: 58s the first, and 52s the second - 1min 51s total.

> Boiled together: 1:45s initially and 11s to re-boil - 1min 56s total.

>

> So in my case slightly more efficient to boil the two cuppa's worth separately.

Now that is interesting!

My trial with a stainless steel kettle gave much greater efficiency boiling together (even not allowing for cooling time). As Spearing said this is probably due to heat losses through steam.

> And the time at which the two approaches are equally (in)efficient is dependant on how well insulated your kettle is.

Not if steam is a significant factor (my argument above concerning conduction versus Tom's and my results shows that something other than conduction certainly can be important). Maybe your kettle is remarkably steam efficient!,

DancingOnRock - * * on 09 Nov 2012

In reply to rubbercrumb: Another issue here is that the 'thermostat' is no longer a bi-metallic strip. It's now a thermister that senses the steam temperature. So you have to have sufficient quantity of steam for a certain time to switch off. This will affect the times.

I suspect doing the experiment with an open pan and thermometer would give different results not just because of heat losses.

You need twice the amount of energy to boil twice as much water from the same starting temperature. For every degree warmer that the second cup is at you need 4.2 watts per gram less. So for 1000ml/1000g and a 2.4kw kettle it's 1.75s per degree.

I suspect doing the experiment with an open pan and thermometer would give different results not just because of heat losses.

You need twice the amount of energy to boil twice as much water from the same starting temperature. For every degree warmer that the second cup is at you need 4.2 watts per gram less. So for 1000ml/1000g and a 2.4kw kettle it's 1.75s per degree.

spearing05 - * * on 09 Nov 2012

In reply to Robert Durran:

You have missed something though, it is 2 x time and 2 x A so that should read 4At

I can't work out how to put this mathematically but if you say the time for both is 2t Which we agree on) and then say 2t = y then for the two cups seperate it is y x 2A and for 2 together it is y x A* and since 2A is bigger than A* it is less efficient.

Can't get my head round how to express this as following through your logic it seems to make sense. Ive made a similar mistake before when working out volume and converting ie 1" x 1" x 1" x 25.4 does not equal the volume in mm it is actually 1 x 25.4 x 1 x 25.4 x 1 x 25.4 which is a whole different number.

> (In reply to spearing05)

> [...]

>

> [...]

>

> No, I am not missing this out.

> I shall explain it again slightly differently:

>

> Take your two cubes. Suppose each has a surface area A. If they are combined into a single lump, the surface area A' of this lump will be more than A (by a factor of 2^(2/3)=1.6 approx if they are moulded into a single larger cube).

> Now, suppose the two cubes are heated separately, each taking time t. The heat loss from each will be proportional to A x t = At, and so the total heat loss from both will be propoptional to 2 x At = 2At (see, I have not missed this out!)

> [...]

>

> [...]

>

> No, I am not missing this out.

> I shall explain it again slightly differently:

>

> Take your two cubes. Suppose each has a surface area A. If they are combined into a single lump, the surface area A' of this lump will be more than A (by a factor of 2^(2/3)=1.6 approx if they are moulded into a single larger cube).

> Now, suppose the two cubes are heated separately, each taking time t. The heat loss from each will be proportional to A x t = At, and so the total heat loss from both will be propoptional to 2 x At = 2At (see, I have not missed this out!)

You have missed something though, it is 2 x time and 2 x A so that should read 4At

I can't work out how to put this mathematically but if you say the time for both is 2t Which we agree on) and then say 2t = y then for the two cups seperate it is y x 2A and for 2 together it is y x A* and since 2A is bigger than A* it is less efficient.

Can't get my head round how to express this as following through your logic it seems to make sense. Ive made a similar mistake before when working out volume and converting ie 1" x 1" x 1" x 25.4 does not equal the volume in mm it is actually 1 x 25.4 x 1 x 25.4 x 1 x 25.4 which is a whole different number.

spearing05 - * * on 09 Nov 2012

In reply to spearing05: Actually I think I may have got that wrong. Hmmmmm thinking here - and I should be working . . .

Robert Durran - * * on 09 Nov 2012

In reply to spearing05:

Indeed.

> (In reply to spearing05) Actually I think I may have got that wrong.

Indeed.

Robert Durran - * * on 09 Nov 2012

In reply to spearing05:

You might be better off working out the heat loss for a single cup and then for two cups together and establishing that the heat loss for both cups together is more than double that for a single cup.

> (In reply to spearing05) Actually I think I may have got that wrong.

You might be better off working out the heat loss for a single cup and then for two cups together and establishing that the heat loss for both cups together is more than double that for a single cup.

Robert Durran - * * on 09 Nov 2012

In reply to spearing05:

No you have to work it out for one cup and then double it (because there are two cups heated SEPARATELY). For each cup the loss is At. Doubling then gives 2At.

Yor 2A x 2t = 4At would be appropriate if we heated both cubes (cups) together at the same time in the same kettle without combining them into a single lump (ie A'=2A, so 2A't=4At) - ie without the efficiency advantage of losing some surface area by sticking them together. The heat loss would then be double that for heating them one at a time - a worst case scenario!

> (In reply to Robert Durran)

> You have missed something though, it is 2 x time and 2 x A so that should read 4At

> You have missed something though, it is 2 x time and 2 x A so that should read 4At

No you have to work it out for one cup and then double it (because there are two cups heated SEPARATELY). For each cup the loss is At. Doubling then gives 2At.

Yor 2A x 2t = 4At would be appropriate if we heated both cubes (cups) together at the same time in the same kettle without combining them into a single lump (ie A'=2A, so 2A't=4At) - ie without the efficiency advantage of losing some surface area by sticking them together. The heat loss would then be double that for heating them one at a time - a worst case scenario!

DancingOnRock - * * on 09 Nov 2012

In reply to Robert Durran:

A prism's volume is Al and its surface area is 2A+Pl where P is the perimeter of the area A. So for one double the volume, the volume is 2Al and area is 2A+2Pl. As A and P are constant in this case l is the only variable.

For 2 prisms you get surface area 4A+2Pl for volume 2Al.

It's not straight forward as A has a bigger influence for small l.

I think.

A prism's volume is Al and its surface area is 2A+Pl where P is the perimeter of the area A. So for one double the volume, the volume is 2Al and area is 2A+2Pl. As A and P are constant in this case l is the only variable.

For 2 prisms you get surface area 4A+2Pl for volume 2Al.

It's not straight forward as A has a bigger influence for small l.

I think.

Robert Durran - * * on 09 Nov 2012

In reply to DancingOnRock:

Yes. The A in my argument above is your 2A+Pl and my A' is your 2A+2Pl. So, as I said, A' is greater than A but less than 2A, which is your 4A+2Pl (my "worst case scenario" mentioned above).

> (In reply to Robert Durran)

>

> A prism's volume is Al and its surface area is 2A+Pl where P is the perimeter of the area A. So for one double the volume, the volume is 2Al and area is 2A+2Pl. As A and P are constant in this case l is the only variable.

> For 2 prisms you get surface area 4A+2Pl for volume 2Al.

>

> A prism's volume is Al and its surface area is 2A+Pl where P is the perimeter of the area A. So for one double the volume, the volume is 2Al and area is 2A+2Pl. As A and P are constant in this case l is the only variable.

> For 2 prisms you get surface area 4A+2Pl for volume 2Al.

Yes. The A in my argument above is your 2A+Pl and my A' is your 2A+2Pl. So, as I said, A' is greater than A but less than 2A, which is your 4A+2Pl (my "worst case scenario" mentioned above).

DancingOnRock - * * on 10 Nov 2012

Additionally the element is inductive and resistive.

You will get a large inrush current when you switch it on due to the inductive nature.

When the element is cold the resistance will be lower and the current higher. So more energy used.

My kettle is rated 1850-2200 watts.

I may do some experiments today with my son.

You will get a large inrush current when you switch it on due to the inductive nature.

When the element is cold the resistance will be lower and the current higher. So more energy used.

My kettle is rated 1850-2200 watts.

I may do some experiments today with my son.

crossdressingrodney - * * on 10 Nov 2012

richyfenn on 10 Nov 2012

In reply to DancingOnRock:

Sorry to be pedant But the inductive nature of the element will actually resist the rate of change of current and so reduce the inrush (slightly). Like a light bulb the resistance of the element will be very low when cold but will very quickly attain its normal working resistance when hot (light bulb resistance increases by about 10x when powered).

>

> You will get a large inrush current when you switch it on due to the inductive nature.

>

> When the element is cold the resistance will be lower and the current higher. So more energy used.

> You will get a large inrush current when you switch it on due to the inductive nature.

>

> When the element is cold the resistance will be lower and the current higher. So more energy used.

Sorry to be pedant But the inductive nature of the element will actually resist the rate of change of current and so reduce the inrush (slightly). Like a light bulb the resistance of the element will be very low when cold but will very quickly attain its normal working resistance when hot (light bulb resistance increases by about 10x when powered).

DancingOnRock - * * on 10 Nov 2012

In reply to richyfenn: Sorry you're right. I was confusing with the start up current of an induction motor which is something else entirely. Doh!

DancingOnRock - * * on 10 Nov 2012

In reply to richyfenn: Although won't it just reduce the 'apparent' current, the actual current will be the same due to power factor. I'm missing something.

richyfenn on 10 Nov 2012

In reply to DancingOnRock:

Now I bow to you as power factor is something I've never really got to grips with

> the actual current will be the same due to power factor.

Now I bow to you as power factor is something I've never really got to grips with

Robert Durran - * * on 10 Nov 2012

In reply to DancingOnRock:

Shouldn't that increase efficiency since heating will be faster and so less time for heat to be lost? So less energy used. Generally, the higher the power the the more efficient the kettle will be. To take an extreme example, a sufficiently low power kettle will never boil the water because above a certain temperature the rate of heat loss will exceed the power of the kettle.

> When the element is cold the resistance will be lower and the current higher. So more energy used.

Shouldn't that increase efficiency since heating will be faster and so less time for heat to be lost? So less energy used. Generally, the higher the power the the more efficient the kettle will be. To take an extreme example, a sufficiently low power kettle will never boil the water because above a certain temperature the rate of heat loss will exceed the power of the kettle.

Orgsm on 10 Nov 2012

In reply to rubbercrumb:

Get her a bigger t cup equivalent to 2 cups, then she won't need a second cup?

Get her a bigger t cup equivalent to 2 cups, then she won't need a second cup?

DancingOnRock - * * on 10 Nov 2012

In reply to Robert Durran: I don't know. The energy output will be proportional to the resistance so inversely proportional to the temperature. How this affects the rate of heating at increasing temperatures will depend on the temperature coefficient of resistance of the element.

All I'm saying is it's another factor in the time/energy in/heat loss/original temperature equation and definitely not linear.

All I'm saying is it's another factor in the time/energy in/heat loss/original temperature equation and definitely not linear.

Robert Durran - * * on 10 Nov 2012

In reply to DancingOnRock:

P=V^2/R, so power inversely proportional to resistance.

> (In reply to Robert Durran) The energy output will be proportional to the resistance....

P=V^2/R, so power inversely proportional to resistance.

deepsoup - * * on 10 Nov 2012

In reply to richyfenn:

Not so. A nichrome heating element isn't like a light bulb in this respect. Its resistance at working temperature will only be a few percent higher than it is at room temperature.

Even by the standards of this thread, talk of what happens as the element itself is warming up is silly. Compared to the time it takes to boil the kettle, it's instantaneous.

> Like a light bulb the resistance of the element will be very low when cold but will very quickly attain its normal working resistance when hot (light bulb resistance increases by about 10x when powered).

Not so. A nichrome heating element isn't like a light bulb in this respect. Its resistance at working temperature will only be a few percent higher than it is at room temperature.

Even by the standards of this thread, talk of what happens as the element itself is warming up is silly. Compared to the time it takes to boil the kettle, it's instantaneous.

yarbles - * * on 10 Nov 2012

In reply to rubbercrumb: Boil one cup then put the water for the second cup into the kettle. This can warm whilst waiting, absorbing the residual heat in the element. The water will then be at higher temp when kettle is turned on for second time and require less energy to boil.

tom_in_edinburgh - * * on 10 Nov 2012

In reply to yarbles:

This is definitely the best way to do the two boil cycle option. Probably better to swish the water round the kettle a bit as well because the top of the kettle body gets really hot from the steam during the first boil.

For the one boil option the best thing would be to fill the kettle long before you want to make the tea. That way the water will be at room temperature rather than the cold tap temperature which is at least 10 deg C higher at this time of year. That should reduce the time to boil by 10%.

> (In reply to rubbercrumb) Boil one cup then put the water for the second cup into the kettle. This can warm whilst waiting, absorbing the residual heat in the element. The water will then be at higher temp when kettle is turned on for second time and require less energy to boil.

This is definitely the best way to do the two boil cycle option. Probably better to swish the water round the kettle a bit as well because the top of the kettle body gets really hot from the steam during the first boil.

For the one boil option the best thing would be to fill the kettle long before you want to make the tea. That way the water will be at room temperature rather than the cold tap temperature which is at least 10 deg C higher at this time of year. That should reduce the time to boil by 10%.

spearing05 - * * on 11 Nov 2012

In reply to Robert Durran:

This thread is startignt to get a bit like the treadmill one but . . .

I am forced to bow to yoursuperior equations re the time/surface area thing and yet the results show differently. As above I think the greatest heat loss will be by far the loss through evaporation at the surface and 2 cup = twice the amount of this.

Also, I think there may be something in that heat los is proportional to the difference in temperature. Although it takes twice as long to boil two cups, how much of this time is spent at lower temps where the losses are less? This ties in with the whole evaporation thing. I have no idea of the figures but I suspect the kettle will be significantly more efficient at raising the water the first few degrees than it will be for the last few approaching boiling.

At these higher temps the losses through the sides will have increased and more importantly the rate of evaporation will have too. As there is twice the surface area to evaporate from when boiling two cups seperately does this mean that a larger amount of time is spent during the maximum losses part of the cycle when compared to boiling two together?

It is crazy really how many things come into play with what at first seems a simple problem.

> (In reply to spearing05)

> [...]

>

> You might be better off working out the heat loss for a single cup and then for two cups together and establishing that the heat loss for both cups together is more than double that for a single cup.

> [...]

>

> You might be better off working out the heat loss for a single cup and then for two cups together and establishing that the heat loss for both cups together is more than double that for a single cup.

This thread is startignt to get a bit like the treadmill one but . . .

I am forced to bow to yoursuperior equations re the time/surface area thing and yet the results show differently. As above I think the greatest heat loss will be by far the loss through evaporation at the surface and 2 cup = twice the amount of this.

Also, I think there may be something in that heat los is proportional to the difference in temperature. Although it takes twice as long to boil two cups, how much of this time is spent at lower temps where the losses are less? This ties in with the whole evaporation thing. I have no idea of the figures but I suspect the kettle will be significantly more efficient at raising the water the first few degrees than it will be for the last few approaching boiling.

At these higher temps the losses through the sides will have increased and more importantly the rate of evaporation will have too. As there is twice the surface area to evaporate from when boiling two cups seperately does this mean that a larger amount of time is spent during the maximum losses part of the cycle when compared to boiling two together?

It is crazy really how many things come into play with what at first seems a simple problem.

DancingOnRock - * * on 11 Nov 2012

In reply to rubbercrumb: I think the heat loss due to the evaporative cooling at the top surface will be reduced due to the lid holding the steam close to the surface. Instead of the surface area being double, only the surface area in contact with the sides of the kettle will lose significant heat. The heat loss won't be double.

spearing05 - * * on 11 Nov 2012

In reply to DancingOnRock: Not all the steam is kept in by the lid. On our plastic kettle I'd be quite prepared to dab my hand against the side - you wouldn't catch me passing it over the spout.

Parrys_apprentice - * * on 12 Nov 2012

In reply to spearing05:

Should we start using pressure cookers to boil water? would that be better?

> (In reply to DancingOnRock) Not all the steam is kept in by the lid.

Should we start using pressure cookers to boil water? would that be better?

yarbles - * * on 12 Nov 2012

In reply to Parrys_apprentice: Don't think so, as when you release the pressure - ie pour the water the temp will drop. PV=nRT

Jim C - * * on 12 Nov 2012

In reply to spearing05:

Hot tip, when making tea in a cup (usually camping) I always put my teabag ON the kettle spout of the kettle when it is boiling, the steam/water vapor softens up the leaves and as soon as I pour the teabag falls into the cup and the tea flavour seems to flow out better than just pouring hot water onto a cold bag.

I would be interested if others have tried this think the same, or am I kidding myself.

> (In reply to DancingOnRock) Not all the steam is kept in by the lid. On our plastic kettle I'd be quite prepared to dab my hand against the side - you wouldn't catch me passing it over the spout.

Hot tip, when making tea in a cup (usually camping) I always put my teabag ON the kettle spout of the kettle when it is boiling, the steam/water vapor softens up the leaves and as soon as I pour the teabag falls into the cup and the tea flavour seems to flow out better than just pouring hot water onto a cold bag.

I would be interested if others have tried this think the same, or am I kidding myself.

Toerag - * * on 12 Nov 2012

In reply to yarbles:

Nail.Head. Everyone's neglected the fact that you're not just heating the water, you're heating the element too. Someone needs to do the experiment with a microwave which only heats the water and cup(s).

> (In reply to rubbercrumb) Boil one cup then put the water for the second cup into the kettle. This can warm whilst waiting, absorbing the residual heat in the element. The water will then be at higher temp when kettle is turned on for second time and require less energy to boil.

Nail.Head. Everyone's neglected the fact that you're not just heating the water, you're heating the element too. Someone needs to do the experiment with a microwave which only heats the water and cup(s).

Wonko The Sane on 12 Nov 2012 - *customer18941.107.wv.cust.t-mobile.co.uk*

In reply to Toerag:

Tell that to it's magnatron.

> (In reply to yarbles)

> [...]

>

> Nail.Head. Everyone's neglected the fact that you're not just heating the water, you're heating the element too. Someone needs to do the experiment with a microwave which only heats the water and cup(s).

> [...]

>

> Nail.Head. Everyone's neglected the fact that you're not just heating the water, you're heating the element too. Someone needs to do the experiment with a microwave which only heats the water and cup(s).

Tell that to it's magnatron.

DancingOnRock - * * on 12 Nov 2012

In reply to spearing05:

No but the steam can't all escape at the same time can it. Without a lid it will escape a lot quicker than with a lid. Try timing the kettle boiling with and without the lid. You'll definitely see a big difference.

> (In reply to DancingOnRock) Not all the steam is kept in by the lid. On our plastic kettle I'd be quite prepared to dab my hand against the side - you wouldn't catch me passing it over the spout.

No but the steam can't all escape at the same time can it. Without a lid it will escape a lot quicker than with a lid. Try timing the kettle boiling with and without the lid. You'll definitely see a big difference.

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