Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?Do my noodles cook quicker when the water is boiling or when it is just about to boil?Why is it faster (as in proportion to volume) to boil 4 cups of water than to boil 2 cups?Does coffee cool faster than tap water?Why does hot water clean better than cold water?Hot water freezing faster than cold waterDoes ice made with hot water melt faster than ice made with cold water?Why does splitting hot tea from one glass in two glasses makes it cool faster?How does the hot cup of tea cool down?Why does hot water freezes faster than normal water?Boiling water in metal pot

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Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?


Do my noodles cook quicker when the water is boiling or when it is just about to boil?Why is it faster (as in proportion to volume) to boil 4 cups of water than to boil 2 cups?Does coffee cool faster than tap water?Why does hot water clean better than cold water?Hot water freezing faster than cold waterDoes ice made with hot water melt faster than ice made with cold water?Why does splitting hot tea from one glass in two glasses makes it cool faster?How does the hot cup of tea cool down?Why does hot water freezes faster than normal water?Boiling water in metal pot













9












$begingroup$


I always make tea with boiling water - not water that is almost boiling, not water that has boiled and stopped boiling a few seconds earlier - in my opinion, for the best tea flavour the water must be boiling as it comes out of the kettle into the pot (or cup).



My wife is not so obsessive, and will often leave the kettle to switch itself off, then, after as long as a minute, pour the water onto the tea (she does admit that tea made by me is far superior to the tea she makes).
Anyway, I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water). But that seems scientifically impossible - or could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water? Just clutching at physics straws here!



Maybe the whole thing is my imagination.










share|cite|improve this question









New contributor




Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
    $endgroup$
    – Vectorjohn
    4 hours ago






  • 1




    $begingroup$
    Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
    $endgroup$
    – R.M.
    3 hours ago















9












$begingroup$


I always make tea with boiling water - not water that is almost boiling, not water that has boiled and stopped boiling a few seconds earlier - in my opinion, for the best tea flavour the water must be boiling as it comes out of the kettle into the pot (or cup).



My wife is not so obsessive, and will often leave the kettle to switch itself off, then, after as long as a minute, pour the water onto the tea (she does admit that tea made by me is far superior to the tea she makes).
Anyway, I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water). But that seems scientifically impossible - or could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water? Just clutching at physics straws here!



Maybe the whole thing is my imagination.










share|cite|improve this question









New contributor




Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
    $endgroup$
    – Vectorjohn
    4 hours ago






  • 1




    $begingroup$
    Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
    $endgroup$
    – R.M.
    3 hours ago













9












9








9





$begingroup$


I always make tea with boiling water - not water that is almost boiling, not water that has boiled and stopped boiling a few seconds earlier - in my opinion, for the best tea flavour the water must be boiling as it comes out of the kettle into the pot (or cup).



My wife is not so obsessive, and will often leave the kettle to switch itself off, then, after as long as a minute, pour the water onto the tea (she does admit that tea made by me is far superior to the tea she makes).
Anyway, I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water). But that seems scientifically impossible - or could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water? Just clutching at physics straws here!



Maybe the whole thing is my imagination.










share|cite|improve this question









New contributor




Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I always make tea with boiling water - not water that is almost boiling, not water that has boiled and stopped boiling a few seconds earlier - in my opinion, for the best tea flavour the water must be boiling as it comes out of the kettle into the pot (or cup).



My wife is not so obsessive, and will often leave the kettle to switch itself off, then, after as long as a minute, pour the water onto the tea (she does admit that tea made by me is far superior to the tea she makes).
Anyway, I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water). But that seems scientifically impossible - or could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water? Just clutching at physics straws here!



Maybe the whole thing is my imagination.







thermodynamics temperature everyday-life water cooling






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New contributor




Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Qmechanic

108k122021255




108k122021255






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asked 11 hours ago









PaulPaul

461




461




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Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
    $endgroup$
    – Vectorjohn
    4 hours ago






  • 1




    $begingroup$
    Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
    $endgroup$
    – R.M.
    3 hours ago
















  • $begingroup$
    I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
    $endgroup$
    – Vectorjohn
    4 hours ago






  • 1




    $begingroup$
    Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
    $endgroup$
    – R.M.
    3 hours ago















$begingroup$
I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
$endgroup$
– Vectorjohn
4 hours ago




$begingroup$
I feel like you should just get a couple thermometers, identical cups, and test it. It sounds like it's your imagination.
$endgroup$
– Vectorjohn
4 hours ago




1




1




$begingroup$
Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
$endgroup$
– R.M.
3 hours ago




$begingroup$
Related, but probably not directly applicable: the Mpemba effect (which is somewhat disputed).
$endgroup$
– R.M.
3 hours ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

There is a reasonable chance that your tea is cooling faster.



This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.



Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.



The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.



If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.



Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
    $endgroup$
    – Peter Cordes
    5 hours ago











  • $begingroup$
    @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
    $endgroup$
    – JMac
    4 hours ago


















2












$begingroup$


Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?




In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.




I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)




Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.




could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?




The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.



Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.



Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @Paul In other words, it all depends on multiple factors as well as your bias. :)
    $endgroup$
    – Bill N
    10 hours ago


















1












$begingroup$

The answer is no. It does not cool faster to a drinkable temperature.



The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.



Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.



If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
    $endgroup$
    – JMac
    4 hours ago










  • $begingroup$
    I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
    $endgroup$
    – Vectorjohn
    4 hours ago










  • $begingroup$
    There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
    $endgroup$
    – JMac
    3 hours ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

There is a reasonable chance that your tea is cooling faster.



This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.



Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.



The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.



If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.



Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
    $endgroup$
    – Peter Cordes
    5 hours ago











  • $begingroup$
    @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
    $endgroup$
    – JMac
    4 hours ago















5












$begingroup$

There is a reasonable chance that your tea is cooling faster.



This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.



Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.



The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.



If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.



Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
    $endgroup$
    – Peter Cordes
    5 hours ago











  • $begingroup$
    @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
    $endgroup$
    – JMac
    4 hours ago













5












5








5





$begingroup$

There is a reasonable chance that your tea is cooling faster.



This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.



Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.



The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.



If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.



Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.






share|cite|improve this answer











$endgroup$



There is a reasonable chance that your tea is cooling faster.



This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.



Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.



The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.



If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.



Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 4 hours ago

























answered 11 hours ago









JMacJMac

9,17321936




9,17321936







  • 1




    $begingroup$
    So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
    $endgroup$
    – Peter Cordes
    5 hours ago











  • $begingroup$
    @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
    $endgroup$
    – JMac
    4 hours ago












  • 1




    $begingroup$
    So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
    $endgroup$
    – Peter Cordes
    5 hours ago











  • $begingroup$
    @PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
    $endgroup$
    – JMac
    4 hours ago







1




1




$begingroup$
So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
$endgroup$
– Peter Cordes
5 hours ago





$begingroup$
So to be specific, you're saying the extra variable is the mass of water: if more is lost to evaporation because of starting hotter, there's less thermal mass to cool off. (And the ratio of area to volume changes). So the smaller amount of non-evaporated water that started hotter can overtake the water that started not quite boiling.
$endgroup$
– Peter Cordes
5 hours ago













$begingroup$
@PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
$endgroup$
– JMac
4 hours ago




$begingroup$
@PeterCordes Ignore my other comment if you saw it, this would be one of the major factors here.
$endgroup$
– JMac
4 hours ago











2












$begingroup$


Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?




In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.




I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)




Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.




could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?




The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.



Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.



Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @Paul In other words, it all depends on multiple factors as well as your bias. :)
    $endgroup$
    – Bill N
    10 hours ago















2












$begingroup$


Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?




In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.




I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)




Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.




could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?




The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.



Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.



Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @Paul In other words, it all depends on multiple factors as well as your bias. :)
    $endgroup$
    – Bill N
    10 hours ago













2












2








2





$begingroup$


Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?




In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.




I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)




Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.




could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?




The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.



Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.



Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.






share|cite|improve this answer











$endgroup$




Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?




In short, yes. The rate of cooling (temperature drop) will be higher when the liquid is boiling than when it has already reached a temperature closer to ambient temperature. However, as it cools, its rate of cooling will slow down, and it will reach the ambient temperature of the room more slowly than another cup in the same environment with a starting temperature closer to ambient.




I've found that tea made by be (boiling water) cools to a drinkable temperature faster than tea made by her (boiled, then very slightly cooled water)




Your classification for drinkability probably includes how well steeped the tea is and how much of the detectable compounds have transferred into the water. That process happens faster with the hotter starting temperature. There may also be some psychological bias in wanting to be right.




could it be explained in the same way that warm water freezes (when put in a domestic freezer) quicker than cold water?




The notion about warm water freezing more quickly than cold water is probably because water passing through the hot water heater has picked up more minerals/sediment/seeds for crystallization than water which has not.



Think about it this way: You bring a pot of water to a boil. Once it's well boiling, you pour equal amounts into identical mugs, one with tea and the other without. You start the timer for tea made your style. You let the other cool down for a minute, then add tea to the second and start the timer for tea made your wife's way. You stop the timers when the cups reach the temperature you classify as "drinkable." You are going to stop both of those timers at pretty much the same time, but the timer for tea made your way is going to display more time on it, because you started the other one after the cooling process was already underway.



Note: If you leave the second cup in a closed, insulated tea kettle not exposed to open air above, and put the first cup in a poorly insulated mug with an open top where steam and heat can easily leave, that first cup might indeed reach "drinkable" temperature more quickly than the second.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 11 hours ago









WBTWBT

3311311




3311311











  • $begingroup$
    @Paul In other words, it all depends on multiple factors as well as your bias. :)
    $endgroup$
    – Bill N
    10 hours ago
















  • $begingroup$
    @Paul In other words, it all depends on multiple factors as well as your bias. :)
    $endgroup$
    – Bill N
    10 hours ago















$begingroup$
@Paul In other words, it all depends on multiple factors as well as your bias. :)
$endgroup$
– Bill N
10 hours ago




$begingroup$
@Paul In other words, it all depends on multiple factors as well as your bias. :)
$endgroup$
– Bill N
10 hours ago











1












$begingroup$

The answer is no. It does not cool faster to a drinkable temperature.



The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.



Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.



If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
    $endgroup$
    – JMac
    4 hours ago










  • $begingroup$
    I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
    $endgroup$
    – Vectorjohn
    4 hours ago










  • $begingroup$
    There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
    $endgroup$
    – JMac
    3 hours ago















1












$begingroup$

The answer is no. It does not cool faster to a drinkable temperature.



The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.



Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.



If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
    $endgroup$
    – JMac
    4 hours ago










  • $begingroup$
    I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
    $endgroup$
    – Vectorjohn
    4 hours ago










  • $begingroup$
    There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
    $endgroup$
    – JMac
    3 hours ago













1












1








1





$begingroup$

The answer is no. It does not cool faster to a drinkable temperature.



The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.



Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.



If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.






share|cite|improve this answer









$endgroup$



The answer is no. It does not cool faster to a drinkable temperature.



The key thing other answers seem to be ignoring is that you're asking about the time between pouring the water and the tea becoming a drinkable temperature (emphasized because someone else interpreted your words as "drinkability"). It doesn't matter what's happening in the kettle, weather the air is saturated in it or not. Because you are asking about tea that is in a cup not a kettle.



Your tea certainly cools faster at first, because it's still evaporating quickly and heats up the cup more. The temperature differences and steam all lead to faster cooling. But at some point in time T+X, the temperature of your tea is the same temperature as your wife's was when she poured it. At that point, there is no magical process that will keep it cooling faster than hers, and in the mean time hers has already had X seconds of cooling. It will always be cooler than your tea.



If you really want to answer the question, it's a trivial experiment to do and might be fun. Make sure you use identical room temperature cups.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









VectorjohnVectorjohn

1134




1134











  • $begingroup$
    This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
    $endgroup$
    – JMac
    4 hours ago










  • $begingroup$
    I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
    $endgroup$
    – Vectorjohn
    4 hours ago










  • $begingroup$
    There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
    $endgroup$
    – JMac
    3 hours ago
















  • $begingroup$
    This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
    $endgroup$
    – JMac
    4 hours ago










  • $begingroup$
    I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
    $endgroup$
    – Vectorjohn
    4 hours ago










  • $begingroup$
    There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
    $endgroup$
    – JMac
    3 hours ago















$begingroup$
This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
$endgroup$
– JMac
4 hours ago




$begingroup$
This is a good point, and my answer was inaccurate. I lost track of why I brought up evaporation in the first place. The extra heat has the benefit of increasing the evaporation potential, which reduces the overall mass (if poured to the same level); which has the potential to overtake the temperature difference.
$endgroup$
– JMac
4 hours ago












$begingroup$
I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
$endgroup$
– Vectorjohn
4 hours ago




$begingroup$
I think it's still incorrect to say the answer is "yes" his cup cools faster. He specifically is asking about how long it takes to go from pouring the tea to the same temperature. His wife's tea has to have a delta-T of X, and his tea needs a delta-T of (X+Y). There is no way, once the tea is poured, that it could do that faster. Consider, that once his tea has cooled slightly, it stops evaporating as much and the rate of cooling slows down.
$endgroup$
– Vectorjohn
4 hours ago












$begingroup$
There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
$endgroup$
– JMac
3 hours ago




$begingroup$
There's also a delta M though, that his warmer cup gets extra of because it is allowed to evaporate more during the cooling. It is possible that the effects of the reduced mass could lead to a bigger heat difference than the heat difference caused by his wife's water cooling
$endgroup$
– JMac
3 hours ago










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