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Tealight

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #750 on: January 29, 2017, 01:44:08 AM »
How come all the models and papers talk about a shutdown in the thermohaline circulation when the data we are watching this winter is the North Atlantic Drift flowing into the Arctic Ocean?

I guess I just know the right question to ask Google.

The thermohaline circulation in the North Atlantic is driven by cold salty water sinking to the bottom off the coast of Greenland. If the Atlantic warms then the water isn't dense enough to sink to the sea floor. Someone or his model concluded that this would completly stop the thermohaline circulation.

I think the circulation just shifts further north into the Barents Sea and beyond. The Arctic has the right topographical features for it. All cold salty water can sink into the Arctic Basin and exit through the deep Fram straight where it rejoins the current thermohaline circulation.

CognitiveBias

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #751 on: January 30, 2017, 12:22:36 AM »
Jai,
  The nullschool view is certainly dramatic...  I've been tracking it as well for a while now.

I'm unclear how you get ~4C though.  I see light blue as 0 to -10C.  Anything over 0 I see as green.  Am I missing something?

https://earth.nullschool.net/#2017/02/03/1800Z/wind/surface/level/overlay=temp/orthographic=-92.96,92.34,1339/loc=-15.747,75.726


Also interesting is the heat and wind on the Fram side.  Export should be picking up.

Thanks!  CB
 

logicmanPatrick

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #752 on: January 30, 2017, 06:47:44 AM »
How come all the models and papers talk about a shutdown in the thermohaline circulation when the data we are watching this winter is the North Atlantic Drift flowing into the Arctic Ocean?

I guess I just know the right question to ask Google.

The thermohaline circulation in the North Atlantic is driven by cold salty water sinking to the bottom off the coast of Greenland. If the Atlantic warms then the water isn't dense enough to sink to the sea floor. Someone or his model concluded that this would completly stop the thermohaline circulation.

I think the circulation just shifts further north into the Barents Sea and beyond. The Arctic has the right topographical features for it. All cold salty water can sink into the Arctic Basin and exit through the deep Fram straight where it rejoins the current thermohaline circulation.

Slowdown is highly likely.  Previously, the warm water would start to be cooled as soon as it passed under the marginal ice near Iceland.  Also, thick sea ice in process of rejecting salt would make the current saltier.  Year by year the trend is for the warm Atlantic water to penetrate further north, all the time being cooled less and mixing with less rejected saline. 

Ultimately the long-term trend to lower specific gravity must lead to a slower rate of descent, hence a slower thermohaline circulation.  This will carry heat from the tropics more slowly, hence more heat to travel south.  After a few years the polar see-saw kicks in and Antarctic glaciers begin to accelerate.

That's just my humble opinion and as such is open to critical assessment.
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jdallen

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #753 on: January 30, 2017, 09:50:17 AM »
Stupid question follow on...

What is the estimate for the total flow into and out of the Arctic basin, in KM3 of water?

The reason I ask is, salt purge out of freezing ice is only going to be a small fraction of the volume of ice melted (which is going to be less than 6000 or so KM3 a season).  If we presume 10% of that - or 600KM3/year, what fraction would that be of the total circulation?

Point I'm making here is, is the refreeze really what's driving the circulation?
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Pmt111500

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #754 on: January 30, 2017, 11:41:49 AM »
Stupid question follow on...

What is the estimate for the total flow into and out of the Arctic basin, in KM3 of water?

<clip>

Point I'm making here is, is the refreeze really what's driving the circulation?

No numbers from me, but the latter point such a good one I thought to add in the expected snowfalls on open ocean during future polar nights.

DavidR

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #755 on: January 30, 2017, 12:27:56 PM »
Stupid question follow on...

What is the estimate for the total flow into and out of the Arctic basin, in KM3 of water?

The reason I ask is, salt purge out of freezing ice is only going to be a small fraction of the volume of ice melted (which is going to be less than 6000 or so KM3 a season).  If we presume 10% of that - or 600KM3/year, what fraction would that be of the total circulation?

Point I'm making here is, is the refreeze really what's driving the circulation?

According to wikipedia total flow out of the Arctic is about 11 Sverdrups.

( A sverdrup) is equivalent to 1 million cubic metres per second (264,000,000 USgal/s). The entire global input of fresh water from rivers to the ocean is equal to about 1.2 sverdrup.

Therefore a Sverdrup is equivalent to 1/1000th of a Km^3. So  in a day the outflow is approx 8640 * 11 / 1000 Km^3. per day so about 95 Km^3 per day. The inflow from rivers would be at best 100th of this so the rest must come from the Atlantic and Pacific.

600Km^3 therfore amounts to  about 2% of the flow.
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shmengie

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #756 on: February 02, 2017, 03:49:32 AM »
Is this right, or am I just making this stuff up?

Quote
Since 2015, 80°North hasn't been below 245K(elvin)
In 1960, 80°North was rarely above 245K in Arctic winter

Over the last decade, the average temperature 80°North has shifted almost~10°K warmer during winter.

The extreme North climate has changed more drastically in the past decade than lower latitudes.

Artic multi-year sea ice is a thing of the past leaving only Greenland's snow/ice & pack surviving in the northern hemisphere.  After Greenland melts, Antarctica will be the last multi-year ice (MYI) on the planet.

After Antarctica, it may be centuries before MYI returns to the surface of the planet, barring a global volcanic, nuclear or astrological event cause a cooling period sooner.

Been eyeballing graphs, thats what the trends spell out to me.
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bairgon

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #757 on: February 02, 2017, 02:16:13 PM »
I have a heat budget question. This may be all complete rubbish but comes out of thinking about the after-effects of the GAC with many gaps in the ice freezing over, and therefore increasing the area graph. 

When ice melts, heat is absorbed to make the transition from solid to liquid (latent heat). Conversely, when ice is made, heat must be released. That's a strange way of looking at it, but the enthalpy of the system is constant.

So, assuming that the heat is released into the air as it seems to me that the air drives ice formation, what is the impact on air temperature? Some very rough calculations:

The heat of fusion for water is 334 joules per gram. There are 10^6 g in 1 m^3 of water so the heat of fusion for 1 m^3 of water is 334 x 10^6 joules or 0.334 x 10^6 kj.

Assume an open sea area of 10km x 10km in which of 0.25m thick forms. The volume of ice is 0.25 x 10^8 m^3. Therefore the total heat of fusion is 0.0835 x 10^14 kj, or 8.35 x 10^12 kj.

Assuming cold temps and taking some approximations, the density of air is 1.39 kg / m^3. The specific heat of air is defined in kj/kg K and is close enough to 1. That means that 1 kj will raise the temperature of 1 kg of air by 1 degrees K.

If we take a volume of air 100km x 100km x 100m thick (assuming it is blowing over the ice) then that is 10^12 m^3. The weight is 1.39 x 10^12 kg. Dividing the kj released from freezing the ice by that weight gives around 6 degrees K temperature rise.

Therefore it seems to me that the freezing of the open areas opened up by the GAC will have had some not insignificant effect on the air temperatures. This will reduce the FDDs by some factor.

I guess the effect on air temperature could be determined by correlating rises with polynyas refreezing. Probably going to be difficult to pick out the values.

So there appears to be a triple impact of the GAC:
- air temperature increased due to new ice formation, decreasing FDDs for other ice
- ice export to warmer ocean areas
- thin ice formed is less rigid and more prone to eventual melting, breakup and warming via insolation in the summer


Tigertown

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #758 on: February 02, 2017, 03:04:34 PM »
@bairgon
No problem bairgon, the heat can just escape out into space and not bother anything at all.
OOPS! It can't escape that easy anymore, can it? Now it and every other source, like heat from upwelling warm water, is a problem. Upwelling hurt ice before, but now it can't escape even in open water areas  as easily. It's a heat trap extraordinaire.
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oren

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #759 on: February 03, 2017, 02:27:02 PM »
Is this right, or am I just making this stuff up?

Quote
Since 2015, 80°North hasn't been below 245K(elvin)
In 1960, 80°North was rarely above 245K in Arctic winter

Over the last decade, the average temperature 80°North has shifted almost~10°K warmer during winter.

The extreme North climate has changed more drastically in the past decade than lower latitudes.

Artic multi-year sea ice is a thing of the past leaving only Greenland's snow/ice & pack surviving in the northern hemisphere.  After Greenland melts, Antarctica will be the last multi-year ice (MYI) on the planet.

After Antarctica, it may be centuries before MYI returns to the surface of the planet, barring a global volcanic, nuclear or astrological event cause a cooling period sooner.

Been eyeballing graphs, thats what the trends spell out to me.
Shmengie, referring to some of your statements: "Arctic multi-year sea ice is a thing of the past" - not true. There is still 3, 4, and 5-year sea ice in the Arctic, just much less of it than used to be. Even after a virtually ice-free September there will still be multi-year sea ice, unless there was a complete melt-down of every last ice floe. This will happen at some point, but not necessarily in the very near future.
In addition, The ice on Greenland and Antarctica is not referred to as multi-year ice / MYI (though linguistically it might be correct). Not sure the correct terminology - Ice Sheets and Glaciers? Glaciated ice?
Finally, the Antarctic ice sheet is expect to survive in some form at least for centuries to millenia (and more).

Graham P Davis

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #760 on: February 04, 2017, 11:09:10 AM »
How come all the models and papers talk about a shutdown in the thermohaline circulation when the data we are watching this winter is the North Atlantic Drift flowing into the Arctic Ocean?

I guess I just know the right question to ask Google.

According to something I read almost fifty years ago, the sudden shutdowns of the NAD were associated with the current circulation of the North Atlantic being bimodal. What happened in the alternate mode to the present situation was that the NAD no longer spun off from the Gulf Stream but was replaced by an extension of the Labrador Current. I assume that this change meant that the volume of water in the NAD remained unchanged but was somewhat colder and less saline.

According to what I read then, the cause of the sudden switched in the circulation was unknown. The best guess was that it was due to a slowing of the Gulf Stream as a response to a weakening of the subtropical high. When I heard about the shutdown in the thermohaline circulation, I assumed this was the answer but just recently I have wondered whether it has more to do with salinity changes in the surface waters with a lowering of salinity of the Labrador current causing it to stop sinking below the Gulf Stream and flow over the NAD instead. The warm NAD would then sink to replace the submarine extension of the Labrador Current.
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Neven

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #761 on: February 04, 2017, 02:30:01 PM »
Welcome, Graham. Your profile has been released.
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shmengie

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #762 on: February 04, 2017, 04:07:11 PM »
Shmengie, referring to some of your statements: "Arctic multi-year sea ice is a thing of the past" - not true.

Thanks oren.  I was scratching my head about how old some of the MYI in the Artic was.  Thought there was some remaining, but for the most part, its waning toward being gone.

Not been to the Arctic. A set of before & after pictures of glacial landscapes, formed the impression of MYI sea ice gone, in my head.  Been reluctant to post a link to these pictures here, because they're not entirely relevant, but stunning. Couldn't find them on weather.com, where I first saw them.  Don't know why... This site presents them better, tho.

Stunning before & after photos:
http://www.snowaddiction.org/2014/06/photos-from-alaska-then-and-now-this-is-a-get-ready-to-be-shocked-when-you-see-what-it-looks-like-now.html
« Last Edit: February 05, 2017, 04:09:28 PM by shmengie »
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Tigertown

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #763 on: February 04, 2017, 04:24:12 PM »
Some of the individuals at the weather channel have strong opinions about changes in the climate and the Arctic, but it is kind of like someone is holding them back from talking about it very much. This is reflected in their website at weather.com .

It is a shame that they don't cover the Arctic more, on both media outlets, as they have such a large following.
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Pmt111500

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #764 on: February 05, 2017, 05:38:13 AM »
Where are the dmi +80 numbers for 2016? I downloaded the two pieces and the other one ends at 2015/12/31 and the current starts from 2017/1/1. This is no biggie though, doing historical stats ( if i get to it) won't depend on one recent year, only asking for the sake of completeness. Maybe i hit the dataset before the yearly update, so should check it myself too.

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #765 on: February 10, 2017, 06:20:29 PM »
is this really true?  I'm having a bit of trouble believing that the distance between freezing and boiling in pure water at standard pressure has any relationship to the percentage change in energy at arbitrary temperatures.

What ktnonine said is correct, but it has nothing to do with the freezing/boiling temperature of water. Converted to Celsius, he just said "-172.15°C is 1% more energy than -173.15°C". It's a bit neater in Kelvin  ;)

The only bit that's incorrect is that it should be 101K, not 101°K, as Kelvin is an absolute scale and therefore not in degrees (c.f. degrees in an angle vs radians)

I don't get it, and if this is true then I need a very careful explanation of the units of measurement and physics such that a 1 degree K (or C) equals ONE PERCENT.

Seumas

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #766 on: February 10, 2017, 06:26:38 PM »
I don't get it, and if this is true then I need a very careful explanation of the units of measurement and physics such that a 1 degree K (or C) equals ONE PERCENT.
Ah, right, I think I see what you don't get.

It doesn't make any difference how large a unit Kelvin is. Let's imagine I create a new absolute temperature scale. We can call it Seulcius! And 1S == 2°C (== 2K).

Now, the difference between freezing and boiling water is 50S. But I can still say, "101S is 1% more energy than 100S". Which, in Kelvin, is like saying, "202K is 1% more energy than 200K".

There is in fact an absolute scale that uses Farenheit spaced units, called the Rankine scale. And 101R is 1% more energy than 100R.

If that hasn't made it clear, think of it as distance. If you have a standard length Stride, then 101 Strides is 1% more than 100 Strides, whether the Stride is 10cm or 20 miles.

ktonine

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #767 on: February 10, 2017, 06:31:58 PM »
The only bit that's incorrect is that it should be 101K, not 101°K, as Kelvin is an absolute scale and therefore not in degrees (c.f. degrees in an angle vs radians)

Aaaaack!  Yes, I am chagrined at the mistake.

Physicist Matt Strassler had a nice post on energy and temperature back in 2014  - Happy (Chilly) New Year

Jim, go read the Matt Strassler link above.  It's a nice physical explanation that a layperson can understand by a physicist.


Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #768 on: February 10, 2017, 07:05:50 PM »
If that hasn't made it clear, think of it as distance. If you have a standard length Stride, then 101 Strides is 1% more than 100 Strides, whether the Stride is 10cm or 20 miles.
No....one stride (two steps, step one left and step one right) is infinitely more than 0 strides, and and it is an indeterminate amount more than one step -- most people step more with one foot than another.  101 strides is merely one stride more than 100 strides -- it is no percents more.  (Figures don't lie, but liars figure.)

Seumas

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #769 on: February 10, 2017, 07:28:13 PM »
101 strides is merely one stride more than 100 strides -- it is no percents more.
You haven't put a smiley in there, so I can't tell if you're joking or genuinely confused.

If the standard Stride is 10 cm, then 100 Strides is 1000 cm and 101 Strides is 1010 cm, which is 1% greater.

If the standard Stride is 20 miles, then 100 Strides is 2000 miles and 101 Strides is 2020 miles, which is 1% greater.

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #770 on: February 10, 2017, 07:33:37 PM »
101 strides is merely one stride more than 100 strides -- it is no percents more.
You haven't put a smiley in there, so I can't tell if you're joking or genuinely confused.

If the standard Stride is 10 cm, then 100 Strides is 1000 cm and 101 Strides is 1010 cm, which is 1% greater.

If the standard Stride is 20 miles, then 100 Strides is 2000 miles and 101 Strides is 2020 miles, which is 1% greater.
Then this is purely a case of liars figure.  1% of which measure?  The previous stride or the current stride?

crandles

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #771 on: February 10, 2017, 07:51:43 PM »

Then this is purely a case of liars figure.  1% of which measure?  The previous stride or the current stride?

I'm lost. Do you accept that 21K is 5% more than 20K? Or are you complaining that this is 5% and not the same 1%?

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #772 on: February 10, 2017, 09:13:25 PM »

Then this is purely a case of liars figure.  1% of which measure?  The previous stride or the current stride?

I'm lost. Do you accept that 21K is 5% more than 20K? Or are you complaining that this is 5% and not the same 1%?
No....I do not accept that.  If you take the 21K as the basis than it is one thing, but if you take the 20K as the basis it is another.  Liars figure.  You have to specify what you are talking about extremely precisely or percent is meaningless.  Actually....percent is basically meaningless to begin with unless you are talking about dice odds.


Seumas

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #773 on: February 10, 2017, 09:19:10 PM »
No....I do not accept that.
Then I'm afraid the problem is simply that you don't understand percentages. I suggest you look at the Khan Academy videos which explain how percentages work.

Quote
Actually....percent is basically meaningless to begin with unless you are talking about dice odds.
It's really not. It means "per hundred". An increase of 1% means you multiply the figure you are talking about by 1.01 because you are increasing it by one part in a hundred. That's it. There is literally no-one here confused about what the percentage increase refers to, except you.

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #774 on: February 10, 2017, 09:23:49 PM »
No....I do not accept that.
Then I'm afraid the problem is simply that you don't understand percentages. I suggest you look at the Khan Academy videos which explain how percentages work.

Quote
Actually....percent is basically meaningless to begin with unless you are talking about dice odds.
It's really not. It means "per hundred". An increase of 1% means you multiply the figure you are talking about by 1.01 because you are increasing it by one part in a hundred. That's it. There is literally no-one here confused about what the percentage increase refers to, except you.

You must not have bothered to read what I wrote.  Go away.

ktonine

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #775 on: February 10, 2017, 09:29:31 PM »
You must not have bothered to read what I wrote.  Go away.

???

Percent and percent error have long been used both colloquially and scientifically.  Obviously correctly defining the dividend and divisor makes a difference -- but that's true in *any* similar mathematical statement. 

Like Seumas, I at first assumed you were joking.  Now you just seem a bit off.

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #776 on: February 10, 2017, 09:33:28 PM »
Wouldn't the Boltzmann constant be applicable here?  1.380655·10-23 J/K.
Since it is linear in regards to energy and temperature the percent increase of energy is dependent upon the starting temperature (a measure of the energy in the system).

The change in energy is proportional to the change in Kelvin, but the percent change in energy per degree Kelvin is not.  I think this is where the disagreement arises. 

In regards to it being tied to something, it is - the triple point of water, which can vary depending upon the isotopic composition of the water.  There are attempts to minimize the uncertainty in the measurement of the Boltzmann constant.  Once that is achieved it essentially will not be tied to the properties of water.  Not sure that has been achieved as of yet.

The kelvin: The Boltzmann Project
https://www.ptb.de/cms/en/research-development/research-on-the-new-si/ptb-experiment/the-kelvin-the-boltzmann-project.html
« Last Edit: February 10, 2017, 10:23:01 PM by dj »

Seumas

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #777 on: February 10, 2017, 09:53:24 PM »
You must not have bothered to read what I wrote.  Go away.

I think I'll assume you're posting drunk and will make more sense tomorrow. Have a good evening.

Neven

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #778 on: February 10, 2017, 10:19:00 PM »
Wouldn't the Boltzmann constant be applicable here?  1.380655·10-23 J/K.
Since it is linear in regards to energy and temperature the percent increase of energy is dependent upon the starting temperature.

In regards to it being tied to something, it is - the triple point of water, which can vary depending upon the isotopic composition of the water.  There are attempts to minimize the uncertainty in the measurement of the Boltzmann constant.  Once that is achieved it essentially will not be tied to the properties of water.  Not sure that has been achieved as of yet.

The kelvin: The Boltzmann Project
https://www.ptb.de/cms/en/research-development/research-on-the-new-si/ptb-experiment/the-kelvin-the-boltzmann-project.html

Welcome, dj, your profile has been released.
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dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #779 on: February 10, 2017, 10:24:13 PM »
Wouldn't the Boltzmann constant be applicable here?  1.380655·10-23 J/K.
Since it is linear in regards to energy and temperature the percent increase of energy is dependent upon the starting temperature.

In regards to it being tied to something, it is - the triple point of water, which can vary depending upon the isotopic composition of the water.  There are attempts to minimize the uncertainty in the measurement of the Boltzmann constant.  Once that is achieved it essentially will not be tied to the properties of water.  Not sure that has been achieved as of yet.

The kelvin: The Boltzmann Project
https://www.ptb.de/cms/en/research-development/research-on-the-new-si/ptb-experiment/the-kelvin-the-boltzmann-project.html

Welcome, dj, your profile has been released.

Thank you.

sidd

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #780 on: February 10, 2017, 10:45:56 PM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #781 on: February 10, 2017, 10:51:05 PM »
The only bit that's incorrect is that it should be 101K, not 101°K, as Kelvin is an absolute scale and therefore not in degrees (c.f. degrees in an angle vs radians)

Aaaaack!  Yes, I am chagrined at the mistake.

Physicist Matt Strassler had a nice post on energy and temperature back in 2014  - Happy (Chilly) New Year

Jim, go read the Matt Strassler link above.  It's a nice physical explanation that a layperson can understand by a physicist.

Now I'm going to add confusion to the issue by mentioning that a slightly negative absolute temperature is in fact possible...

viz: https://phys.org/news/2013-01-atoms-negative-absolute-temperature-hottest.html

... it spoils the fun somewhat though to add that this is an artifact of the definition of temperature; a negative value exists if the pressure goes down as you add energy - Paradoxically this can be made to happen happen by creating a system in which an atomic gas is very hot and still being heated but the individual atoms have an upper limit to the amount of kinetic energy they can possess (as in the above experiment, which trapped them in a lattice of laser interference patterns).

For those wondering how under the above circumstances adding further energy reduces the pressure, I'm kind of fuzzy on that myself - but my hand-waving understanding of it is that there's a normally insignificant amount of quantum tunneling going on, which suddenly becomes more significant if kinetic energy is maxed-out.


Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #782 on: February 10, 2017, 10:55:04 PM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

Thank you.  I think I will state the original question over again even though you you have answered it...just so the rest will understand.

The statement I was questioning was that the change from 100 degrees K to 101 degrees K entailed 1% change in energy.  I could not understand how that could be true...and as you point out, it was not.  In fact, you have pointed out that the statement was meaningless and irrelevant.


dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #783 on: February 10, 2017, 11:11:58 PM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

Because you are undergoing a phase change.  That is a poor example as it relies upon the heat capacity of water (for example A) and the heat of fusion of water (example B).  See the Boltzmann constant above.

Edit: Corrected spelling

dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #784 on: February 10, 2017, 11:17:52 PM »
Quote
Thank you.  I think I will state the original question over again even though you you have answered it...just so the rest will understand.

The statement I was questioning was that the change from 100 degrees K to 101 degrees K entailed 1% change in energy.  I could not understand how that could be true...and as you point out, it was not.  In fact, you have pointed out that the statement was meaningless and irrelevant.

The Boltzmann constant is 1.38064852 × 10-23 J/K. 
The percent change in temperature is equal to the percent change in energy.
The percent change in energy per degree Kelvin would depend upon the starting temperature.

I think this is the source of your confusion.

dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #785 on: February 10, 2017, 11:35:54 PM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

Thank you.  I think I will state the original question over again even though you you have answered it...just so the rest will understand.

The statement I was questioning was that the change from 100 degrees K to 101 degrees K entailed 1% change in energy.  I could not understand how that could be true...and as you point out, it was not.  In fact, you have pointed out that the statement was meaningless and irrelevant.

To further elaborate. 
Temperature is simply a measure of the amount of energy present in the molecules of whatever you are measuring.

Temperature as measured in Kelvin is set at the triple point of water.  And is defined such that 0 K is where there is no energy in the molecule, it is not moving or vibrating because it has no energy associated with it.

The Boltzmann constant describes the linear relationship between the amount of energy represented by Kelvin units, which is 1.38064852 × 10-23 J/K.

Because this is a linear relationship, the percent change in Kelvin equals the percent change in energy.  The percent change of energy per degree Kelvin does change.

I hope this helps.

DrTskoul

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #786 on: February 11, 2017, 12:04:34 AM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

Thank you.  I think I will state the original question over again even though you you have answered it...just so the rest will understand.

The statement I was questioning was that the change from 100 degrees K to 101 degrees K entailed 1% change in energy.  I could not understand how that could be true...and as you point out, it was not.  In fact, you have pointed out that the statement was meaningless and irrelevant.

To further elaborate. 
Temperature is simply a measure of the amount of energy present in the molecules of whatever you are measuring.

Temperature as measured in Kelvin is set at the triple point of water.  And is defined such that 0 K is where there is no energy in the molecule, it is not moving or vibrating because it has no energy associated with it.

The Boltzmann constant describes the linear relationship between the amount of energy represented by Kelvin units, which is 1.38064852 × 10-23 J/K.

Because this is a linear relationship, the percent change in Kelvin equals the percent change in energy.  The percent change of energy per degree Kelvin does change.

I hope this helps.

Except when we are talking about the internal energy of a substance with a phase change as given with examples above.

dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #787 on: February 11, 2017, 12:11:12 AM »
True, but that was not the source of confusion that Jim was having previously. He was having trouble understanding why the percent change of energy is equal to the percent change in temperature as measured in Kelvins.

As I pointed out previously, the example given is poor one because it does include a phase change and has nothing to do with the relationship between Kelvin and Energy.

DrTskoul

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #788 on: February 11, 2017, 12:14:09 AM »
True, but that was not the source of confusion that Jim was having previously. He was having trouble understanding why the percent change of energy is equal to the percent change in temperature as measured in Kelvins.

As I pointed out previously, the example given is poor one because it does include a phase change and has nothing to do with the relationship between Kelvin and Energy.

Late in the discussion :)

Thanks!

Hyperion

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #789 on: February 11, 2017, 12:04:31 PM »
Stupid question follow on...

What is the estimate for the total flow into and out of the Arctic basin, in KM3 of water?

The reason I ask is, salt purge out of freezing ice is only going to be a small fraction of the volume of ice melted (which is going to be less than 6000 or so KM3 a season).  If we presume 10% of that - or 600KM3/year, what fraction would that be of the total circulation?

Point I'm making here is, is the refreeze really what's driving the circulation?

According to wikipedia total flow out of the Arctic is about 11 Sverdrups.

( A sverdrup) is equivalent to 1 million cubic metres per second (264,000,000 USgal/s). The entire global input of fresh water from rivers to the ocean is equal to about 1.2 sverdrup.

Therefore a Sverdrup is equivalent to 1/1000th of a Km^3. So  in a day the outflow is approx 8640 * 11 / 1000 Km^3. per day so about 95 Km^3 per day. The inflow from rivers would be at best 100th of this so the rest must come from the Atlantic and Pacific.

600Km^3 therfore amounts to  about 2% of the flow.

Does anyone get any devious notions from numbers like this or is it just me.  ;)

specific heat capacity of water = 4.2 kj/kg/K
~42 000 000 000 000 kJ per day for 1% of flow out of arctic ocean (~100km^3) at 10k difference
= 486 000 000 kw = 486 000 megawatts = 486 gigawatts
world = 5 terawatts = 5million megawatts
therefore 10% of flow is equal to world electricity consumption
Irminger Current 11- 27 sverdrup = 11m-27m cubic m per sec.

Wouldn't it be nice to heat the greenland low salinity meltpool with gulfstream waters so it can radiate heat away, and not cause these cyclone canons? And thereby cool the Salty hot stuff to preserve overturning, but keep it south of the faroes- iceland ridge so it dinae get up north where its truble?.  :o  8)  ???
Anyone heard of heatpipes?
Policy: The diversion of NZ aluminum production to build giant space-mirrors to melt the icecaps and destroy the foolish greed-worshiping cities of man. Thereby returning man to the sea, which he should never have left in the first place.
https://en.wikipedia.org/wiki/McGillicuddy_Serious_Party

Neven

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #790 on: February 11, 2017, 12:06:49 PM »
Welcome, Hyperion, your profile has been released.
Il faut comparer, comparer, comparer, et cultiver notre jardin

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #791 on: February 11, 2017, 03:19:10 PM »
There is a relation between Temperature and Energy, but it is not necessarily linear ...

For a simple example, consider the change in the internal energy U of 1 gm of water as it is cooled successively from a) 1.5C to 0.5C b) 0.5C to -0.5C

Step a) is a change in U of 4.2 calorie.
Step b) is a change of 84.2 calorie
 
both steps involve a 1C change in temperature, but quite different amounts of energy

sidd

Thank you.  I think I will state the original question over again even though you you have answered it...just so the rest will understand.

The statement I was questioning was that the change from 100 degrees K to 101 degrees K entailed 1% change in energy.  I could not understand how that could be true...and as you point out, it was not.  In fact, you have pointed out that the statement was meaningless and irrelevant.

To further elaborate. 
Temperature is simply a measure of the amount of energy present in the molecules of whatever you are measuring.

Temperature as measured in Kelvin is set at the triple point of water.  And is defined such that 0 K is where there is no energy in the molecule, it is not moving or vibrating because it has no energy associated with it.

The Boltzmann constant describes the linear relationship between the amount of energy represented by Kelvin units, which is 1.38064852 × 10-23 J/K.

Because this is a linear relationship, the percent change in Kelvin equals the percent change in energy.  The percent change of energy per degree Kelvin does change.

I hope this helps.
I think I follow this, though I would like the see the dimensions spelled out -- especially the dimensions for temperature.  Am I correct that 100K to 101KM is an extremely special case, and that a 101K to 102K change would not yield a 1% change in energy?  Or is it the case that a 100K to 101K and a 101K to 102K change would cause the same percent, but not 1%?  (I don't think it is the second...)

oren

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #792 on: February 11, 2017, 03:42:32 PM »
For anyone who has read the Matt Strassler link mentioned above, I think the whole thing is quite clear:
The ratio of temperatures (measured in Kelvin) is the ratio of kinetic energy of the molecules. From 300k to 330k means 10% more kinetic energy.
So: 1K means 1% only if you happen to be at 100k.
And: a phase change means the molecules gain or lose chemical bond energy, as long as temperature doesn't change that means their kinetic energy (and therefore motion speed) is the same.

It seems there have been some inaccurate statements in some of the posts above, but as far as I can tell this gives you the correct bottom line. (Until someone corrects me...)

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #793 on: February 11, 2017, 04:01:18 PM »
The internal energy of an ideal gas is proportional to its mass (number of moles) N and to its temperature T

 U = c N T,. C = heat capacity. Similar laws apply for diatomic, polyatomic gases etc.

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #794 on: February 11, 2017, 04:45:48 PM »
Lots of interesting stuff.  I did not post  in freezing because it was way off topic.  Working a lot of hours.  But going back to the catalyst of this post, quote unquote, technically minded writers adding or subtracting 32 when they should only be multiplying for temperature differentials. 

I have no idea how to stop that all too common error. I have seen it all too many times, and I do not recall retractions. I would guess that it is singular to U.S. citizens trying to convert from the original Celsius.  That might also explain the lack of retractions for that mistake from that subset.  If Rep. Charles Grassley (later Senator and Finance Chair) had not basically singularly stopped the metric system in the US, we would not be having this discussion. 

I was going to do a silly question of at what point does Kelvin and Fahrenheit intersect because I was too lazy to do the math but
  http://www.answers.com/Q/At_what_temperature_are_kelvin_and_fahrenheit_the_same   does.  574.5875


dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #795 on: February 11, 2017, 04:54:18 PM »
The Boltzmann constant is 1.38064852 × 10-23 J/K

In dimensions: K = Joules = W*s = N*m = (kg*m2)/s2

Example
1 K = 1.3806485 ×10-23 Joules.
2 K = 2.7612970 x10-23 Joules.

change in K: 2 K - 1 K = 1 K
change in Joules: 1.3806485 x 10-23 Joules
% change K = 100%
% change in Joules = 100%

Example
100 K = 138.06485 x10-23 Joules.
101 K= 139.4454985 x10-23 Joules.

change in K: 1 K
change in Joules: 1.3806485 x 10-23 Joules
% change in K: 1%
% change in Joules: 1%

The % change of energy per 1 Kelvin changes.  The % change of energy = % change in Kelvin.
« Last Edit: February 11, 2017, 05:04:29 PM by dj »

Tor Bejnar

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #796 on: February 11, 2017, 05:35:52 PM »
1% of 100 is, "magically" 1, so a 1% increase yields 101.  In a sense, this is a "special case".
1% of 101 is 1.01.  [Simply divide 101 by 100 - that is what the "0/0" symbol means - "percent" = "per-cent" = "for each 100" or "divided by 100"]  So a 1% increase from 101 is 102.01.*

"At what temperature is there a 1% energy increase from 0oF [~-17.77oC]?" is a dangerous question.  "In the real world", the temperature must be converted to an 'absolute' scale like Kelvin, so 0oF ~ 255.37 K.  1% of 255.37 is ~2.55; a 1% increase yields 257.92K or, converted, -15.23oC or ~4.59oF.

From the MathNotations blog:
Quote
"Your arithmetic is correct but your result is arbitrary. When expressing a temperature change as a percentage, one must use a temperature scale whose zero point is the temperature of absolute zero, and then use that selected scale consistently. The Kelvin scale is such a temperature scale. All other temperature systems, like Fahrenheit and Celsius, have arbitrary "zero points," and calculations of percent temperature change using those scales will give arbitrary results.

In the example you provided [edit by Tor: not related to my question above], on the Fahrenheit scale the temperature rise is 100 percent (from 50(degrees) to 100(degrees) F.), but that same change on the Celsius scale is 280 percent (from 10 to 38 C.). Using the Kelvin scale, whose zero point is absolute zero (-460(degrees) F., -273(degrees) C.), the rise is actually 10 percent, from 283(degrees) to 311(degrees) K."
----------
Tom Skilling is chief meteorologist at WGN-TV.

___
* - At the end of such sentences, I appreciate the European practice of using commas for decimal points!
Arctic ice is healthy for children and other living things.

Jim Williams

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #797 on: February 11, 2017, 05:43:53 PM »
The Boltzmann constant is 1.38064852 × 10-23 J/K

In dimensions: K = Joules = W*s = N*m = (kg*m2)/s2

Example
1 K = 1.3806485 ×10-23 Joules.
2 K = 2.7612970 x10-23 Joules.

change in K: 2 K - 1 K = 1 K
change in Joules: 1.3806485 x 10-23 Joules
% change K = 100%
% change in Joules = 100%

Example
100 K = 138.06485 x10-23 Joules.
101 K= 139.4454985 x10-23 Joules.

change in K: 1 K
change in Joules: 1.3806485 x 10-23 Joules
% change in K: 1%
% change in Joules: 1%

The % change of energy per 1 Kelvin changes.  The % change of energy = % change in Kelvin.

The lightbulb dawns!

Since they are all in the same dimensions we don't want to talk in either Kelvin or Celsius.  We want to talk in Joules -- or rather Peta-joules (or something).

dj

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #798 on: February 11, 2017, 05:55:18 PM »
Well that is what temperature scales, like Kelvin, does. 
However, it provides easier units to be able to do the math.

1 K = 0.00000000000000000000000138 Joules. = 1.38 x 10-23 Joules

The math is a bit easier using the Kelvin units ;)

sidd

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Re: "Smart" and "Stupid" Questions - Feel Free To Ask
« Reply #799 on: February 11, 2017, 05:56:06 PM »
The Boltzmann constant is more appropriate in discussions of ideal gases. In other cases, even those involving no phase change, say the internal energy of a crystal at low temperatures which goes as T^4 , temperature changes correspond to very nonlinear energy changes. Each increment of temperature gives nonlinear change in internal energy of a crystal. If the internal energy of a crystal at T=1K is U1, and that at 2K is U2, then the ratio U2/U1 is 2^4=16  even absent any phase change.

A deeper context for Boltzmann's constant than the ideal gas laws appears in the relationship of entropy S to number of microstates W in the formula S=k*ln(W)

Coupled with the thermodynamic definition of entropy, dS = integral(dQ/T) , one immediately sees the fundamental relation between entropy, internal states and information theory. I much prefer this as the defining relation for Boltzmann's constant, but others may differ.
 
sidd