Guest Opinion by Kip Hansen – 27 July 2022
A great deal of controversy arose in response to the first three parts of this series, which came as no surprise. One well-respected reader implied that the Global Warming hypothesis and all the fuss for 30 years wasn’t about the Earth Climate gaining heat energy (despite all the emphasis on the Global Energy Budget) but only about temperature rise (as measured by thermometers). Some seemed to be appalled at the suggestion that temperatures from disparate places and times could not be averaged to produce meaningful numeric results.
This brief addendum is to give an example, which if extended to the broader question, will help to explain the chief problem inherent in averaging 2-meter air temperatures (in degrees by any method) from around the world and claiming the result as evidence of Global Warming.
Take a perfectly insulated box 2 meters by 1 meter by 1 meter. This box will have a volume of two cubic meters and, since it is perfectly insulated, will not allow any heat/energy to flow into or out of the box.
At the first instant, the box contains one cubic meter of water at 20 °C and one cubic meter of air with a temperature of 30 °C.
Allow enough time for the temperatures to become equalized – for the energy in the water and air in the box to reach equilibrium (we will ignore any phase changes such as evaporation).
What will be the resultant equilibrium temperature of the whole system (air and water)?
And why will it NOT be the mathematical average (the mean)calculated as 20+30/2 = 25 °C?
Bonus Question: (My high school science teacher always included Bonus Questions – allowing some of us to score over 100% on tests.)
If we raised the initial air temperature by 2°C, to 32°C, how much would it change the final answer of the system temperature at equilibrium?
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Author’s Comment:
I’ll ask readers to provide the correct answer, at least to good back-of-envelope estimates. Most important is the second question, why won’t result be the “average” (the mean) between 20 and 30 °C?
Those familiar with the Earth climate as a system can then explain to others why this also answers the question as to why, if the global warming hypothesis is based on the imbalance of the Global Energy Budget, at least in part due to the increase in the Greenhouse Effect due to rising atmospheric CO2 concentrations (and other anthropogenic GHG emissions), then this example also explains one of the reasons why any Global Average Surface Air Temperature metric (in degrees, or as anomalies, or as land and sea skin surface temperature combinations) is not a proper metric to use as evidence for Global Warming by CO2 .
Thanks for reading.
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Density of water 997 kg/m3
Density of air 1.16 kg/m3
Specific heat capacity of water 4180 j/kg.C
Specific heat capacity of air 1000 j/kg.C (isobaric).
Tw = 20 C
Ta = 30 C
Mw = 997 kg/m3 * 1 m3 = 997 kg
Ma = 1.16 kg/m3 * 1 m3 = 1.16 kg
Qa = 1.16 kg * 1000 j/kg.C * (30 C -20 C) = 11600 j
ΔTw = 11600 j / (997 kg * 4180 j/kg.C) = 0.003 C
The equilibrium temperature will be approximately 20.003 C.
Yes. Its the amount of heat the different bodies contain that matters. Obvious as soon as it occurs to you. Back to another cup of coffee!
You forgot the significant digits rules.
Do you see now why averaging temperatures (even if it was allowed under proper scientific practice) provides little knowledge towards the allocation of energy distribution in the atmosphere? Do you think the standard radiation diagrams that are in use take this into account properly?
I don’t know the answer but one big elephant in the room is what is the relative humidity of the air? Granted air between water the water is the 600lb gorilla regardless of humidity of the air!
You have to make a few assumptions here because the problem is underdefined but they are reasonable ones: the pressure is one atmosphere, heat capacities are constant (very close to the truth), and as stated above, ignore heat involved in phase changes. The mass of water is 998.21 kg (NIST) and the mass of air is 1.165 kg. The heat capacity of water is 4184.1 J/kg-K and the heat capacity of dry air is 1006 J/kg-K. The approach is to determine what temperature between 20°C and 30°C balances the amount of heat transferred from the air to the water.
deltaH(water) = mass(water)*Cp(water)*(20+x)
deltaH(air) = mass(air)*Cp(air)*(30-x)
Set these two equations equal to each other and solve for x.
I calculate that the final temperature of the system at equilibrium is 20.0028°C. If we include the heat required to humidify the dry air, the water evaporating from the liquid will actually cool the system so the final temperature will be 19.993°C.
But should we use the heat capacity of air at constant volume, because the experimental setup is a closed box ?
Here you have used the heat capacity at constant pressure (Cp).
The heat capacity at constant volume is approx. 717 J/K/kg and not 1006 J/K/kg as used in your calculation.
Probably so.
Kip,
one further question that is more relevant. Suppose you take measure the air temperature regularly and over the course of a year you notice that it has risen by a degree can you say that the box has gotten warmer or not?
If over the year it has gotten colder can you say that the box has gotten cooler or not 😉
That is not exactly the point of the articles. Even attempting to average temps at a single location doesn’t include the varying water vapor in the air at each measurement. At the same temp, you may have more or less energy in a volume air. A rising average just doesn’t let you determine scientifically what is going on.
Now to your question. What does “warmer” mean to you? Higher temps, higher enthalpy, higher water vapor? You must answer these in your own mind before making a decision. “Warmer” has a different meaning colloquially than scientifically.
Yes of course you can. “Warmer” is a description of temperature. It is the same as stating that the temperature of the box has increased.
However you did say the box is “perfectly insulated” so the temperature would not change.
Because water at 50’C holds a lot more energy than air at the same temperature this winter Britons will be using hot water bottles to keep warm.
Yes indeed. There are actually people in the UK who will be hesitating to turn on the kettle for a hot water bottle this winter because of the cost of electricity. Many will be miserable, and some will die. All of whom could have been saved by a sensible use of the money wasted on renewable subsidies.
Good job there was no CO2 in that box.
According the well established and settled science of the green house gas effect – you’d have a supernova on your hands
:-O
Well, always ready to have a rough guess at these things! I don’t understand the math and theory of specific heat, so I tried a shortcut.
Air I think (from Wikipedia, which I may have misunderstood) will contain about one quarter of the heat of water, and if this is right it will take four times as much heat to warm the water through one degree as it does the air.
Suppose we cool the air by one degree and transfer the heat lost to the water. This will raise the water temp by 0.25. Lets cool it by 8 degrees, and transfer the heat. This will raise the water temp by 2 degrees and lower the water temp by 8.
The water will now be 22 degrees, as will the water, which should be equilibrium.
I’m a bit surprised by this, if its right – I would have expected water to contain much more heat than air, by much more than a ratio of 4:1. Maybe I didn’t understand Wikipedia on specific heat?
[checking what others have said before posting, I see Kip reports the difference is 6:1. So lower air by 9 degrees and get enough heat to raise water by 1.5. The answer must be a bit over 21 degrees in this case.]
And this was completely wrong, because it doesn’t take account of mass. Oh dear, not enough coffee this morning! There’s too little heat in that mass of air to make any but the tiniest of differences. Obvious as soon as it occurs to you.
“I’ll ask readers to provide the correct answer, at least to good back-of-envelope estimates. “
Is bugger all good enough?
The heat capacity (not specific which is heat capacity per unit of mass) of the air will be something like 4000 less than of the water, so the 2°C extra heat of the air is enough energy to raise the temperature of the system by a couple of 10 000th of a degree, so bugger all – unless you measure the temperature a million times.
Robert ==> In the U.S. correct but not be to said in mixed company. In the U.K., you could print it in the daily newspaper with no objections.
Time change. The village in Under Milk Wood was called Llareggub, but in the 50s the BBC objected and it had to be changed to Llaregyb.
I was a bit slow to spot that.
It’s like “fanny” raises a few eyebrows in England.
It’s not about temperature, it’s about energy — something we all should have learned in high school.
I learned early on that some people seem to have a natural affinity and interest in science and engineering while others prefer humanities. Is it nature or nurture?
Next question — why do humans sweat?
Because they can. 🙂
It requires 3.4 watt hrs of energy to change temp of 1 cu M dry air at 20 C and 1 bar 10 deg C
It requires 11,400 watt hrs of energy to change the temp of 1 cu M of 20 C water 10 deg C
There would be quite a few zeros after 20 and the decimal to state the actual temp rise of water, maybe 20.000298 Deg C
Now consider the heat sink capacity of the ocean compared to the temp rise of the atmosphere do to green house gasses.
Dip ==> Yes, quite correct. This is why all the worry about slightly changes in temperature at a Global level (which I do not believe we can or should calculate) is not evidence of Global Warming by CO2.
Kip are you familiar with the Nenana Ice classic; where the spring ice breakup time is turned into a lottery pool? I have been following it for more than twenty years. https://www.nenanaakiceclassic.com/
My data shows a linear trend over the past 106 years of a nine day earlier breakup. Actual breakup ranges from April 14 to May20th.
I had difficulty finding the specific heats of air and water and probably* got it wrong ; but here is my take on this:
specifics heats are 83.64 kj/kg for water and 0.0718 kj/kg for air. at their respective temperatures. Total combined specific heat would be circa 83.70 kj/kg.
Thus the average combined temperature would be circa 25.a wee smidge deg C.
As for the bonus question the result would be 25. a wee smidge plus a minuscule smidge deg C.
Indeed NOT something to get hysterical about and * yes undoubtedly got it wrong but can’t be bothered to put it right.
Alasdair ==> I hope you meant 20 plus a wee smidge…..
Q1: ‘What will be the resultant equilibrium temperature of the whole system (air and water)?’
A1: Indeterminate, because Kip has not specified the mass of air contained in the box. If he had specified it, we could have determined the equilibrium temperature (call it T) in the following way, starting with the following equation:
Heat lost by Air (HA) = Heat gained by Water (HW)
HA = Mass Air (MA) x Specific Heat of Air (SHA) x Temp. Change of Air
Let us call the initial temperature of the Air (i.e. 30⁰C, but let’s stick with the general symbols for now) TA. Then the temperature change of the Air is (TA – T) and the formula for HA becomes:
HA = MAxSHAx(TA – T)
Let us now use the corresponding letter-symbols for the formula for HW but with the W-suffix in place of the A-suffix we used for HA. In this case, the Water has an initial temperature TW (specified as 20⁰C) and its temperature change is (T – TW). Hence,
HW = MWxSHWx(T – TW)
But our first equation above told us that HA = HW.
Therefore MAxSHAx(TA – T) = MWxSHWx(T – TW).
Solving this equation for T gives:
T = (MAxSHAxTA + MWxSHWxTW)/(MAxSHA + MWxSHW).
In order to find the specific value of T from this formula now, we just need to plug the appropriate values for all of the terms into the formula and perform the calculation. However, we cannot complete the calculation in this instance because the value for MA (the mass of air in the box) is indeterminate and unknowable because Kip has not specified it. Therefore, the value of T is indeterminate, as I said at the start, and Kip’s question has no unique answer; it could be anything.
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You should be able to determine the mass of both the air and water given the parameters of the problem Kip presented.
To which specific parameters are you referring? Please point them out, because I can’t see them.
The mass of the water is easy to determine from its given volume (i.e. 1 cu.m) because water is an incompressible fluid (ideally) that maintains the same density regardless of its pressure.
But air is a compressible fluid whose density – and hence, its mass for a given volume (again, 1 cu.m) – will vary according to its pressure. And Kip has not specified its density or its pressure either.
So, how should we determine the mass of air from the given parameters of the problem? Please demonstrate.
Assume the standard atmosphere at 1 atm.
That is an assumption chosen arbitrarily by you, bdgwx, not a given parameter of the problem presented by Kip. He never mentioned the standard atmosphere or anything else pertaining to the air-pressure inside the box either. So, why are you telling me to assume it?
He is quite accommodating and threw you an irrelevant bone. Please see his preceding post.
The spirit of the thought experiment isn’t pedantry. It’s about the divergence of the trivial average temperature and the final equilibrium temperature given the large spread in heat capacity of the bodies like would be the case of the ocean and the atmosphere. The standard atmosphere is a natural choice, but you can use any reasonable assumption you want. The point is to provide a numerical answer.
‘The spirit of the thought experiment isn’t pedantry.’
I’m glad to hear it, although I can’t think why you felt a need to tell me that – unless, perhaps, you thought my original comment was excessively detailed or that I wrote it just to show off my mathematical skill with physics problems. Let me assure you that neither was the case. My comment actually contained the minimum possible detail that I could give it to make the points that I wanted to make. And I pitched it at the most elementary level of mathematical physics that I could too, so as to make it intelligible to the widest possible range of possible readers, many of whom might not possess more advanced skills in mathematics or physics like some others here do.
‘It’s about the divergence of the trivial average temperature and the final equilibrium temperature given the large spread in heat capacity of the bodies like would be the case of the ocean and the atmosphere.’
Yes, I already understood the allusion, thank you.
‘The standard atmosphere is a natural choice, but you can use any reasonable assumption you want.’
Yes, I know. But if I had assumed the standard atmosphere as you suggest (or any other specific value of pressure, for that matter) I would have had to digress from the points that actually I wanted to make and did make in fact, although you appear not to have noticed them. (Err, you have read my comment, haven’t you?)
‘The point is to provide a numerical answer.’
Is it? And what would be the point of my doing that? What would it achieve?
Q2: ‘And why will it NOT be the mathematical average (the mean)calculated as 20+30/2 = 25 °C?’
A2: If we could pack enough air into the box to make MAxSHA = MWxSHW, then we may see from the formula above that, in fact, the final temperature T WOULD be the simple arithmetic average of TW and TA, i.e. (20 + 30)/2. In that instance the formula for T becomes:
T = (MAxSHAxTA + MAxSHAxTW)/(MAxSHA + MAxSHA)
= MAxSHAx(TA + TW)/MAxSHA(1 + 1)
= (TA + TW)/2
Plugging in the given values for TA and TW gives us
T = (30 + 20)/2 = 25°C.
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Bonus Q: ‘If we raised the initial air temperature by 2°C, to 32°C, how much would it change the final answer of the system temperature at equilibrium?’
Bonus A: Again, the answer is indeterminate, but the formula for T derived above shows that it would depend on the value of MA – the mass of air in the box. The greater the mass of air, the greater would be the effect of raising the initial air temperature to 32°C.
However, if we put enough air in the box to make MAxSHA = MWxSHW as before, so that we can make use of the ‘simple average’ formula T = (TA + TW)/2, then we may see that raising the initial air temperature from 30°C to 32°C would result in a rise of the final temperature by 1°C to 26°C.
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Tricky, tricky numbers? Not really in this case. We just need to follow the standard mathematical rules conscientiously.
John Powers ==> As Wm Briggs says (over and over and over) “models only tells us what we tell them to tell us”.
So, yes, if we change the whole scenario, change the problem, so as to get the answer desired, then we would get the answer desired!
That’s CliSci!
It was interesting to see how many of the readers thought this was a problem that needed solving and that in solving it, there was something to be learned. The problem is one suitable for a high school physics class or basic freshman engineering 101. Perhaps there is something to be learned if you are completely unfamiliar with the concepts of mass, heat capacity and energy balances, but otherwise it is irrelevant to the question raised by Kip; that question being (I think) is it possible to learn the true average temperature of something by simply averaging temperature taken at different points in a physical system. It’s true that you can’t determine the average temperature of two things by a simple average if they are of differing mass and specific heat. However, it is possible if you are talking about a continuous mass of reasonably constant composition. That portion of the atmosphere at an elevation of two meters above the surface is in fact quite constant in composition, Determining the average temperature is no different than getting the average temperature of a swimming pool, You do not need to know the mass of water in the swimming pool, nor do you need to know any of its other physical properties. All you need to do is go around and get as many temperature points as you like and then calculate the arithmetic average. Your number will have an error due to measurement error and sampling error, but those are practical matters which can be addressed by practical means. There is, however, no theoretical problem with this approach in spite of what has been presented here and elsewhere.
Now try it with two swimming pools 50 miles apart. Will the combined average temp of the two pools tell you anything about each? Will the combined average temp tell you anything mid-way between the two swimming pools?
Obviously, if you have two objects of different size, then you must take that into account. The atmosphere on the other hand has a fixed size.
The atmosphere has continuously varying microclimates. It isn’t a matter of fixed size but a matter of varying factors like pressure, temperature, and humidity. There is no difference between two swimming pools 50 miles apart and two atmospheric parcels 50 miles apart. They *will* be different and an average of the two won’t tell you much about either one let alone someplace midway between them.
Except for the average temperature, you are correct.
The average temperature at a specific location. And even that is questionable because of uncertainty.
Consider this thought experiment. If you could adiabatically transfer samples of the atmosphere from far flung places into a single container and mix thoroughly, the mixture would have a single average temperature. Measuring and averaging the remote temperatures would produce the same answer.
Don’t think so. If the humidity of two parcels are different it will affect the average. If the atmospheric pressure is different then that will also affect what happens since the masses won’t be the same.
Adiabatic only means no change no transfer of heat or mass. It doesn’t mean the two parcels will be the same.
And you forgot to specify the volumes of each and the final volume.
You are determined to believe what you want to believe instead of what is already known to science. In my professional life, I used software like this to design, analyze, and optimize chemical processes. I no longer have access to it and no longer have a need for it, but if I did, I could show you in considerable detail that mixing air samples would result in a mixture of completely predictable temperature. This is proven science that is used to design and operate industrial processes. We know it works. It’s not guesswork. Aspen HYSYS | Process Simulation Software | AspenTech
I checked out the midstream component and it looks MUCH better than what we had 20+ years ago. Upstream engineers like me often got caught up in field level gas and condensate conditioning plant processes. We had software, but this seems much improved.
Malarky. This is just the argumentative fallacy of Argument by Dismissal.
You may be able to predict mixing temps IF you have complete knowledge of the samples. You didn’t give complete knowledge of the samples and now you are trying to avoid responsibility for that.
Even though I don’t have the high powered Hysys program at my disposal, I was able to download an Excel AddIn for psychrometric calculations. I did a calculation for mixing air at 50 deg F with air at 70 deg F where they both had the same absolute humidity (moisture content in lb of water per lb of dry air). The calculated enthalpy at 60 deg F (the presumed mix temperature) is equal to the average enthalpy of 50 deg F air and 70 deg F. This demonstrates what I was saying. However, you could have air of such differing humidity that when you mixed them, condensation occurred. This would result in a different temperature due to the latent heat of water. I suppose some would argue that this shows you can’t average temperatures, but this of course is not what was being argued. We are merely calculating the mathematical average of a bunch of temperatures which is by definition, the average temperature. It is the same as the brightness temperature that the satellites measure which follows the same general trend as terrestrial temperature measurements as well as radiosonde measurements. I leave it with you.
As a check I did another calculation where the air samples being mixed were of different humidity, but not sufficient to cause condensation. The presence of more moisture in one of the air samples does not appreciably change the adiabatic mix temperature.
Are you trying to say that adiabatic mixing happens in the atmosphere?
radiation acts at a distance. Temperature does not. Averaging atmosphere temperature at different locations is not the same thing as mixing gases in a box.
Of course, it was Kip who brought up mixing things in a box, but it is a way of looking at the problem. May or may not help understanding it. Radiation heat transfer is a function of temperature to the fourth power, so apparently temperature does act at a distance.
“ May or may not help understanding it. Radiation heat transfer is a function of temperature to the fourth power, so apparently temperature does act at a distance.”
Temperature doesn’t act at a distance. RADIATION does.
That is a reasonable way of explaining it. In the atmospheric science field they often refer to these samples as parcels. Parcel theory has a lot of practical uses including the analysis of final states after transfers or movements of parcels from one location to another. Averaging parcel properties occurs (sometimes prolifically) in nearly every atmospheric science text I have on my shelf.
The answer is in the region of 20.003 degrees.
It is not 25 degrees, because the heat capacity of the water is
(998kg/1.225kg) x (4.186J/K/kg/1.005J/K/kg) = 3400 times higher than the water.
Then 10 degrees/3400 = 0.003 degrees.
This is added to the start temperature of the water, i.e 20 + 0.003 = 20.003 degrees.
So this requires the best thermometers to measure this very small change.
However I don’t get what is the problem. The average temperature is the average temperature, not an average heat content.
At time zero (before the temperatures become equilibrated), the average temperature is indeed 25 degrees.
kzb ==> The general global warming hypothesis is that increasing GHGs (mostly CO2) are causing the Earth climate system to retain extra energy (as heat) — see Global Energy Budget.
The Average Temperature — as used in CliSci — is not a scientifically meaningful metric (see Parts 1 and 2 and 3). It is also not a measure of increasing energy (heat) in the climate system.
Since the hypothesis calls for increasing heat energy, it cannot be supported by temperature measurement, only measurements of energy (heat).
That is true, but the usual temperature averages we see are an average of air temperatures. Not an average of heat contents. Also, it is definitely air temperature and not water temperature or ground temperature.
kzb ==> Moat of the Global Average Surface Temperature metric are land/sea averages, often averaging sea skin surface temperature with land air 2-meter temperatures.
“NASA Goddard’s Global Surface Temperature Analysis (GISTEMP) combines land surface air temperatures primarily from the GHCN-M version 4 with SSTs of the ERSSTv5 analysis into a comprehensive global surface temperature data set ” https://climatedataguide.ucar.edu/climate-data/global-surface-temperature-data-gistemp-nasa-goddard-institute-space-studies-giss
I think Roger Pielke Snr used to make a similar point – a bit short of saying that temperature was meaningless, but he used to say that the only really informative parameter was change in heat, ie energy.
This was an entertaining and informative series of posts and gave rise to a lively and entertaining discussion. Thanks.
michel ==> You’re welcome — glad you enjoyed it. Hope your experience wasn’t marred by the quibblers….
Would you make the same case in reverse? If say, there had been no rise in average temperatures over the last 50 years, would you argue that it’s possible that internal energy might still have increased in line with rising CO2 levels?
It certainly shouldn’t be ruled out, but requires more information than just temperatures.
Answer: 25C, +/- 5C
Basic thermodynamics, actually. 1 m^3 of air has a mass of about 1.2 kg. 1 m^3 of water, about 1000 kg. Air has a heat capacity of 700 J/kg per degree C or K. Water, 4200 J/kg per C/K. The latent energy of the 20 deg water is 20 * 4200 * 1000 = 84,000,000 j. The air at 30 deg is 30 * 700 * 1.2 kg = 25,200 J more or less. The combined is additive, or 84,025,200 J. The trick is to divide that by the specific heats of both substances and their masses:
84025200 / (1000 kg * 4200 + 1.2 kg * 700) =
84025200 / 4200840 = 20.04 deg.
And that’s for a 10 degree difference, starting.
Adding 2 more degrees pushes is up another 0.01 degree C.
In equilibrium
Thermodynamics is fun!
GoatGuy
GoatGuy ==> Close enough for our purposes here….others have found slightly different final answers. Thank you.
You are DEFINITELY welcome. I wish I could edit the comment, I made an error. The actual ‘right calcs’ are 20.0020 for the 30 C air case, and 20.0024 for the 32 C case. I really should use spreadsheets first, instead of checking after posting! Also, as a general comment, I really appreciate your postings and running dialog.
I think I made a mistake in taking the specific heat capacities from other posters ! You’ve correctly used the specific heat of air at constant volume, rather than at constant pressure, since the experiment is in a closed box.
The specific heat for air at constant volume is close to 717 J/K/kg at 20 degrees C.
So if I make this correction I now obtain:
(998kg/1.225kg) x (4186/717) = 4756 (the ratio of water/air heat capacities)
Then, 10 degrees/ 4756 = 0.0021 degrees (C or K)
Then 20 + 0.0021 = 20.00210 degrees C
For an initial 12 degrees difference, that answer would be 20.00252 degrees.
Yeah. I was debating which one to use as well. I went with the isobaric option because I thought that better represented the spirit of the thought experiment. I also made a lot of simplifying assumptions. Anyway, I do agree that if the box is mandatory then the isochoric option is better. I’m not sure it matters much though since the spirit of the thought experiment was to demonstrate that the equilibrium temperature will not be everywhere close to the trivial average of the water and air.
GoatGuy ==> In a totally irrelevant aside, my wife was, for many years, known as “The Goat Lady” — she kept dairy goats to provide goat milk our babies (one after another) and then for neighbors whose babies had trouble with cow milk.
Glad you’re having fun and hopefully broadening you knowledge base — I know I get a dual benefit from all this: 1) The knowledge gained in the original research on a topic and 2) The additional viewpoints from readers, many of whom are top professionals in a wildly varying range of topics.
Mr. Hansen,
I really enjoyed this series by someone who correctly understands this subject.
I gave up on arguing this point twenty years ago, since only “oldsters” (I’m 73) seem to have been properly educated on this topic.
My high-school physics and chemistry teacher placed great emphasis on the difference between intensive and extensive properties of matter. Even to the point that she required us to differentiate between 10˚C (a temperature of 10 degrees Celsius) and 10C˚ (a temperature difference of 10 Celsius degrees) since a temperature difference is not itself a temperature. Fifteen years later, in engineering school, science professors (and texts) seemed to ignore the matter entirely.
As she would say, one may average all the telephone numbers on any page of the phone book, but the answer is not a phone number, and tells us nothing about phone numbers.