Herb Callen problems 3.6-1, 3.6-2, and 3.6-3

(Revised 5 Aug 2021.  See my previous post for notes and comments on Callen's Section 3.6.)

3.6-1   The universe is considered by cosmologists to be an expanding electromagnetic cavity containing radiation that now is at a temperature of 2.7 K. What will be the temperature of the radiation when the volume of the universe is twice its current value?  Assume the expansion to be isentropic (this being a nonobvious prediction of cosmological model calculations).

So the entropy will be the same when the volume V₁ becomes V₂  and when  T₁ becomes T₂ :

S(U, V)   =  constant =  (4/3) (bV₂)^1/4 (U₂ )^3/4   =   (4/3) (bV₁ )^1/4 (U₁ )^3/4 , 

where  U₂ = bV₂ (T₂ ) ⁴   and  U₁ = b V₁ (T₁ ) ⁴ .  

Canceling the 4/3 and b factors on each side of the above entropy equation and putting in temperature T in place of the energy U gives (exponent on V's  becomes 1/4 + 3/4 = 1, and exponent of T's becomes 4 x 3/4 = 3)

V₂ (T₂ )³  =   V₁ (T₁ )³  

(T₂ /T₁ ) ³  =  V₁/V₂  =

(T₂ /T₁ ) ³  =  1/2    

T₂  =  (1/2)^1/3 T₁    =     0.79 (2.7 K)  =   2.1 K .

Question for further research: in how many years will this doubling in size occur? The accepted answer to this question is not the same now as it was in 1985 when Callen’s book was published, due to the discovery in 1998 of the acceleration of the expansion (dark energy). 

 

3.6-2 Assuming the electromagnetic radiation filling the universe to be in equilibrium at  T = 2.7 K,  what is the pressure associated with this radiation?  Express the answer in both pascals and atmospheres.

From Stefan-Boltz law, we have U/V = bT⁴ , and the pressure equation for isotropic radiation is P_R = U/3V.  So P_R = (b/3)T⁴

= (7.56x10^-16 Joule · meter^-3 · Kelvin^-4))(2.7 K)⁴/3 

= 134 x 10^-16  Newton/meter²

P_R =  1.34 x 10^-14 Pa

P_R =  1.32 x 10^-19 atm.

 

3.6-3  The density of matter (primarily hydrogen atoms) in intergalactic space is such that its contribution to the pressure is of the order of 10^-23 Pa. 

(a) What is the approximate density of matter (in atoms/meter³) in intergalactic space?

(b) What is the ratio of the kinetic energy of matter to the energy of radiation in intergalactic space? (Recall Problems 3.6-1 and 3.6-2.)  

(c)  What is the ratio of the total matter energy (i.e., the sum of the kinetic energy and the relativistic energy mc²) to the energy of radiation in intergalactic space?

 

(a)  We want to calculate N/V from P_M, and, assuming (maybe incorrectly) the hydrogen gas in intergalactic space is in equilibrium with the CMB radiation, we use the ideal gas law, P_MV = NkT, where T = 2.7 K.  Rearranging gives

 N/V  =  P_M/kT  =  10^-23 Pa/[(1.38 x 10^-23 J/K) · 2.7K]

=  (1/3.7) atoms/meter³ 

=  0.27 atoms/meter³ 

^{The answer is actually about one per cubic centimeter, or one million per cubic meter, alas. That means the temp of the very diffuse hydrogen gas is about 4 million degrees Kelvin.  Here's the first paragraph from a good discussion of the subject:}

There are some missing details that are often skipped over in popular science articles/documentaries, which can make this confusing. First: the temperature of space, 2.7 K or -270 C, refers to the remaining radiation from the Big Bang, not to the temperature of any matter. If you were in space and somehow prevented any matter from touching you, you would still absorb microwave radiation equivalent to being surrounded by matter with a temperature of 2.7 K (since all matter with temperatures above 0K emits radiation). Of course, with all the stars and such, you also receive a lot of other radiation, so you wouldn't necessarily cool down to 2.7 K. That temperature refers specifically, and only, to the background microwave radiation left over from the early history of the universe.

^--------

   (b)  For our intergalactic atomic hydrogen gas, the only energy is kinetic energy (well, ignoring the spin flipping the electron can undergo relative to the spin of the proton, resulting in the famous 21 cm spectral line).  (Also ignoring, until part (c), the relativistic mass-energy.)  So we have (KE)_H = (3/2) NkT = (3/2)P_MV for the kinetic energy of matter, which we want to divide by the U_rad =  bVT⁴  = 3P_RV of radiation:

 

(KE)_H / U_rad  =  (3/2) P_MV / 3P_RV

=  P_M / 2P_R

=  10^-23 Pa / 2(1.34 x 10^-14 Pa)

=   3.7 x 10^-10 

 

(c) The relativistic energy is U_rel = Mc², where M is the mass of all the hydrogen atoms in intergalactic space. We could do this calculation by estimating the size of the universe which we’d multiply by the density of hydrogen atoms. But volume cancels out in a ratio of energy density calculations, and energy density is the determining factor for the total energy of radiation, of hydrogen atoms’ KE, and of relativistic energy. And we've calculated number density of H atoms, so we should use it. We divide by total volume, V, of the universe to get density:

U_rel/V = (Mc² /V) =  m(N/V)c²

= (1.67 x 10^-27 kg for one H atom) (0.27 H atoms per meter³)(3 x 10⁸ m/s)²

=  9 x 10¹⁶ · 0.45 x 10^-27 joule/meter³

=  4 x 10^-11 joule/meter³

Now I need to divide this by U_rad/V for radiation, and add the result to the answer to (b). I didn’t do the U_rad/V calculation separately for radiation energy density in part (b)—but we know U/3V = P for the radiation pressure, so we take 3P as our radiation energy density

U_rad/V = 3P = 3 x 1.34 x 10^-14 Pa = 4 x 10^-14 joule/meter³

so

U_rel/U_rad  =  1000.

The KE/U_rad  ratio of 3.7 x 10^-10 is negligible in comparison with the factor-of-1000 ratio of relativistic-to-radiation energy. No point in adding them! But we’d expect the relativistic energy to be much greater than the total KE of atoms in a gas at temperature 2.7 K.

Questions for further research:

At what temperature does the KE of the H atoms equal their mass-energy?

At what temp does the radiation energy equal the H atoms mass-energy?

Does the mass-energy of all the loose H atoms in the universe have any significance beyond the fact that it has to be included when we account for all the energy released (created) in the Big Bang?

The Stefan-Boltz law as a thermo equation of state

In this discussion, the Stefan-Boltzmann law will be written in terms of the energy contained in a "trapped" electromagnetic field, rather than being written in terms of the power of an "escaping" electromagnetic field (as I discussed for sunlight in a recent post).  An energy-versus-temperature equation is one example of a thermodynamic equation of state. The well-known energy-versus-temp relation for a monatomic ideal gas, for instance, is U=3NkT/2. I'll look at many more examples of equations of state, and at how they are related to the "fundamental relations" of thermodynamics in a future post.

Our main man Herb Callen has a good discussion (Section 3-6 in his 2nd edition text) on the thermodynamics of radiation in an empty box: 

If the walls of any 'empty' vessel are maintained at a temperature T, it is found that the vessel is, in fact, a repository of electromagnetic energy. The quantum theorist might consider the vessel as containing photons, the engineer might view the vessel as a resonant cavity supporting electromagnetic modes, whereas the classical thermodynamicist might eschew any such mechanistic models.

The empty vessel could be almost anything. A metal breadbox, a wooden breadbox, a closed cardboard box, a spherical Christmas tree ornament, or Santa Claus ornament, as long as it's hollow. Yes, it's generally going to have air in it, which is a point that doesn't get discussed much. That's not to say it will be inflated with air like a tennis ball, football or basketball.  But it could be, and there's an electromagnetic field in there with that compressed air. We can just let these "vessels" be at room temperature. Otherwise we have the problem of keeping the walls uniformly at the same high or low temperature. 

Even if a box has walls that are at different temperatures, there will be some sort of radiation spectrum in the box.  Let's say one wall is heated so it's emitting most of the radiation in the cavity, presumably with a different emission spectrum near its surface than the spectra at the surfaces of the other walls.  But is this even possible? Can the spectrum of the radiation in the box be different on different sides of the box?

Well, what we would have is the usual linear superposition (addition) of the amplitudes of the all the waves/photons in the box.,* resulting in constructive and destructive interference, which results in an equilibrium energy-versus-frequency spectrum.  If we have a rectangular or square box, we can ask what the role of the radiation is in changing the temperature distributions in the five walls of our box that are not directly heated. This changing of the temps is a temporary, or transient, behavior. An equilibrium temperature distribution will come into being after the transient behavior is over, due to heat conduction in the walls and radiation from the heated wall of the box to the other walls of the box.

But the  radiation in the box won't have a black-body spectrum unless the walls of the empty vessel are maintained, as Callen says, at the same temperature, which gives equality of absorption and emission rates for every part of the walls. At least this is the desired black-body cavity experimental set-up. The other thing we need for the usual, real, set-up is a little peek-a-boo hole in the box to observe the spectrum of the radiation. 

But we aren't going to try to observe it here. We just want to ask, for a given temperature how much electromagnetic energy is in the vessel? The answer is given by the Stefan-Boltzmann law in yet another form, different from the three ways I wrote it in my February post. Instead of flux (power per unit area) or intensity, the total energy of the radiation is given as proportional to Kelvin-temperature-to-the-fourth-power. Callen writes this as

U = bVT⁴ 

where V is the volume of the empty vessel and b = 4σ/c = 7.56x10^-16 J/m³K⁴, where σ is the Stefan-Boltzmann constant and c is the speed of light. (See my 12 November 2020 post, where the integral of the Planck spectral energy density function at temperature T is shown symbolically and the factor of  c/4 is used.  Here, we're getting rid of that factor. If you divide both sides of the above equation by V, the result is energy density.) Sooooooo, the above equation really comes from integrating the Planck spectral energy density formula over all frequencies while holding temperature constant.  

What, then, is the difference between the views of the quantum theorist, the engineer, and the thermodynamicist in Callen's description above? The photons of the quantum theorist are after all normal modes of the electromagnetic field (see page 97 of these notes on the EM field), and a normal mode is a standing wave that has one of the resonant frequencies of the box. The best brief reference on this subject is Rodney Loudon's two-page discussion, "Introduction: The Photon," at the beginning of his book The Quantum Theory of Light, 3rd ed., and the five references therein.

The quantum theorist is not so much interested anymore in cavity radiation by itself as he or she is in cavity quantum electrodynamics, when there are a controlled number of atoms in a very small cavity (and no air, unless the "atoms" in the cavity are a few oxygen and nitrogen molecules) . An electrical engineer who wants to use the "vessel" as a resonant chamber for electromagnetic radiation is just like an acoustical engineer who wants to use an acoustic chamber to enhance certain frequencies and suppress others. (Here we can think about the usual "box of gas" as having acoustical modes of vibration.) Microwave ovens possess electromagnetic resonant cavities to enhance the microwave cooking frequencies, while musical instruments, for example, enhance the desired harmonics of sound waves.

But I'm getting off the subject here. The subject, which I haven't fully explained yet, is the equations of state of blackbody radiation, and how they can be used to find the entropy of blackbody radiation in terms of U and V. This is a "fundamental relation" in thermodynamics, and would be something the thermodynamicist is interested in, particularly the fact that N--whether he or she thinks of it as the number of photons in the box or number of normal modes--isn't an independent variable along with U and V. (See the quote from Callen below, and note he uses an exclamation mark.)

Besides the above equation of state for the total energy of trapped blackbody radiation in terms of temperature, there's a well-known relation between radiation pressure P, total energy U, and volume V,

P = U/3V

which you can copy and paste into an internet search box to find out more about. "It will be noted," says Callen, "that these empirical equations of state are functions of U and V, but not of N. This observation calls our attention to the fact that in the 'empty' cavity there exist no conserved particles to be counted by a parameter N. The electromagnetic radiation within the cavity is governed by a fundamental equation of the form S = S(U, V) in which there are only two rather than three independent extensive parameters!"

The relevant equation for finding the entropic fundamental relation is 

ST  =  U  +  PV ,

or 

S  =  U / T  +  PV / T.

From the two equations of state, we have 

1/T   = b^1/4 (V/U)^1/4

and 

P/T   = (1/3) b^1/4 (V/U)^1/4

so that the entropy fundamental equation is 

S(U, V)   = (4/3) (bV)^1/4 U^3/4

One of the things I'm interested in is how this expression for the entropy of black-body radiation in terms of U and V compares with the logarithmic expression found by Planck, but I'll come back to that later. First, I'll solve the three problems in this section of Callen's book (Section 3.6), then I'll post some detailed typed handouts from a Statistical Physics class I took in 1995 at Texas State University that used Callen's book as the text (thank you,  Mark Loewe!).

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*Professor Hazeltine said in a junior-level electrodynamics class I took at the UofTexas at Austin 30 years ago that "photons don't interact." I raised my hand and said, "What about interference and diffraction?" Hazeltine explained that "interacting" has to do with things that exert forces on each other, and photons don't exert forces on each other. I asked, "Could we still say they interact linearly then?" I don't recall his response, but he stuck with the "photons don't interact" paradigm. It's not a good paradigm, because photons, being bosons, are gregarious and will indeed interact in a certain way (a sort of attraction) that fermions don't (they sort of repulse each other). We don't call mutual boson interactions and mutual fermion interactions by the name "forces," but maybe we should, or maybe we should modify our idea of forces.

Trinity bomb test described as ammo dump explosion

 (From p.1 of the El Paso Herald-Post, Monday afternoon, July 16, 1945)

Army Ammunition Explosion Rocks Southwest Area

Great Blast Near San Marcial Lights El Paso

            An ammunition magazine containing a considerable amount of high explosives and pyrotechnics exploded at 5:30 a.m. in the New Mexico desert near San Marcial on a remote section of the Alamogordo Air Base reservation.

            No one was hurt.

            The blast was seen and felt throughout an area extending from El Paso to Silver City to Gallup, Socorro and Albuquerque.

            Many persons saw a flash light up the sky, like daylight, and felt earth tremors.  They thought an earthquake had struck.

            William O. Eareckson, commanding officer of the Alamogordo Air Base, released the following statement:

            “Several inquiries have been received concerning a heavy explosion which occurred on the Alamogordo Air Base reservation this morning.

            “A remotely located ammunition magazine containing a considerable amount of high explosives and pyrotechnics exploded.

            “There was no loss of life or injury to anyone, and the property damage outside the explosives magazine itself was negligible.

            “Weather conditions affecting the content of gas shells exploded by the blast may make it desirable for the Army to evacuate temporarily a few civilians from their homes.”

            El Paso men going to work at 5:30 a.m. said the flash illuminated Mt. Franklin.  E.R. Carpenter, Louie Ratliff and Jack Coulehan, riding down Alabama avenue, said the whole sky was ablaze with light.  Mr. Carpenter, mechanical superintendent of the Newspaper Printing Corporation, said many persons called the newspaper’s composing room to report seeing the flash and hearing the explosion.  Callers asked if a meteor had fallen.

Big Light in Sky

            L.R. Lessell, Gila forest headquarters supervisor, said rangers reported the shock was felt throughout the Mogollon mountains.

            Rangers at Chloride reported the blast lighted the sky brightly in the area of San Marcial.  The blast was followed by a terrific explosion, like a detonation, rangers said.

            Forest rangers, believing that an earthquake had struck, checked with  Alfred E. Moore at the Smithsonian Observatory on Burro mountain.  The observatory is near Tyrone and 800 feet high.

            Mr. Moore confirmed the flash, saying it was distinctly visible at the observatory, but the shock was unlike earthquake vibrations.  The observatory head has experienced earthquakes in South America and Mexico.

Silver City Shaken

Silver City residents reported three distinct blasts were felt there.  The shock cracked plate glass windows in downtown buildings.

            “The blasts sounded like heavy claps of thunder,” Don Lusk of Silver City, said.  “Houses shook.  People were roused out of sleep from the noise and tremors.” 

            Mrs. H.E. Wieselman of 901 North Ochoa street saw the explosion as she crossed the Arizona-New Mexico state line.  She was enroute to El Paso from California.

            “We had just left Safford, and it was still dark,” Mrs. Wieselman said.  “Suddenly the tops of high mountains by which we were passing were lighted up by a reddish, orange light.

            “The surrounding countryside was illuminated like daylight for about three seconds. 

            “Then it was dark again.

            “The experience scared me.  It was just like the sun had come up and gone down again.”

Front Seat at Sky Show

            Ed Lane, Santa Fe railroad engineer, was at Belen, N.M., when the blast occurred.

            He said he had a front seat to the greatest fireworks show he had ever seen.  The blast was in the direction of San Marcial and seemed to be only a few miles from Belen.
            “I was coming to El Paso,” Mr. Lane said.  “My engine was standing still.  All at once it seemed as if the sun had suddenly appeared in the sky out of darkness.  There was a tremendous white flash.  This was followed by a great red glare and high in the sky were three tremendous smoke rings.  The highest was many hundreds of feet high.  They swirled and twisted as if being agitated by a great force.  The glare lasted about three minutes and then everything was dark again with dawn breaking in the east.”

I copied this article sometime in the early 1990s by hand (pencil and paper) from the original July 16, 1945 El Paso Herald-Post in the bound newspaper archives of the Barker Center for Texas History at the University of Texas at Austin.  In May1994, I used it as an appendix to a paper I wrote ("Zero Patience for Zero Hour") for a History of the Atomic Bomb class taught by Bruce Hunt. Here are some other front-page Herald-Post headlines from July 16, 1945:

450 Super-Forts Heap 2500 Tons of Fire on Japan

Woman Happy as She Gets Back Lost $1480

Ice Cream Firm Owner Freezes to Death in Plant

Capitalizing on Prejudice, Nationalist Party Drives for Control of Government

B-29 Crew Throws Man to Safety from 9000 Feet

Temperature, thermal conductivity and heat capacity

In regard to my previous post, I'll remind you of this: an object's temperature doesn't by itself determine how hot or cold it feels when you touch it. For a given temperature, when the object (or surface) is in contact with your skin, how fast how well it conducts heat is the primary property that determines how hot or cold it feels. This property is known as thermal conductivity. 

One of the reasons I wanted to write the previous post on the temperature of an asphalt road in the summertime sun is to compare how hot the asphalt feels with how hot an iron manhole cover feels. The question is, which surface will burn you faster?

Another factor that comes into play in this regard is heat capacity, also called specific heat capacity, or just specific heat, which is usually described in terms of how much the temperature of a given mass of a certain substance increases when a given amount of heat is supplied to it.  Turn that heat transfer process around and the question becomes how much heat (internal energy) is available to be transferred to a cooler object or surface (your skin, for instance) from the heated substance.

Now we have two questions concerning asphalt and iron: which will burn you faster, and which will give you a more serious burn due to the amount of heat it can transfer. These are related, but exactly how, I don't know.  There is (or was) a spirited debate about this relationship at the physics "stack exchange" website.

We're not considering the emission of energy by radiation now.  We're considering the transfer of energy by conduction, because we have two surfaces at different temperatures in contact with each other: the asphalt (or iron) and your skin. The role of the radiation is to bring the asphalt and the manhole cover to their Stefan-Boltzmann-law temperatures, as discussed in my previous post.

Are the pavement and the manhole cover even going to be at the same temperature in the direct sunlight?  If we treat them both as black-body surfaces we are assuming they reach the same temp once equilibrium between absorption and emission of radiation has been established in each material.

Even when we don't treat them as black-bodies, if they are both equally good absorbers-emitters of radiation, we expect them to reach the same temperature.  That's the subject of this post:  studying the Stefan-Boltz law as it applies to non-black-bodies.

I mentioned in my previous post that the two problems I solved from the Milonni/Eberly textbook Lasers are elementary problems (2nd semester calculus-based freshman physics level), and in fact I found good explanations of how to use the Stefan-Boltzmann law for non-black-bodies in two freshman-level textbooks: University Physics by Sears, Zemansky and Young (SZY), and Principles of Physics by Frank Blatt. 

The Stefan-Boltzmann law as Milonni/Eberly write it, 

 I_total = σT⁴,

is applicable only to idealized surfaces or objects that are assumed to be emitting and absorbing a Planck black-body spectrum. But the fact that real objects aren't perfect emitters and absorbers can be taken into account if their emissivity, ε, is included in S-B law, which SZY's text does, using R instead of I_total for the radiative flux:

R = εσT⁴ 

Prof. Blatt takes this a step further and explicitly shows how the surface area of an object is used to calculate the amount of energy per unit time it’s emitting:

P = AεσT⁴ 

where A is the surface area of the object and P  is the power (joules per second or watts in our MKS case). This is a helpful equation when it comes to finding the radiant energy emitted by an object (the sun for instance) as opposed to just a surface, where only the power per unit area (flux) can be calculated (the asphalt surface or manhole surface, for instance). 

Emissivity is defined as the "ratio of the energy radiated from a material's surface to that radiated from a perfect emitter, known as a blackbody, at the same temperature and wavelength and under the same viewing conditions. It is a dimensionless number between 0 (for a perfect reflector) and 1 (for a perfect emitter)". 

Based on "too-much-information" I found at various websites, I'll estimate the emissivities of asphalt and of oxidized cast iron to both have a value of about 0.9, so they will both be heated by the sunlight to about the same temperature. Given the 756 watts/meter² of solar radiation flux I used in my previous post, how much does a change from ε = 1 (black-body) to ε = 0.9 affect the temperature as calculated by the above equation?

Well, yeah, right, by a factor of the fourth root of 0.9, which is .974, which we might as well call 1, which means we might as well say the temp of both the asphalt and the manhole cover will be 153° F (67° C), as I previously calculated.

Now we need to know the thermal conductivities of asphalt and cast iron.  I'm going to use 1.3 watts per meter per kelvin (1.3 W/m/K) for asphalt and 56 W/m/K for cast iron. Based on these numbers, the manhole cover is capable of burning your skin 40 times faster than asphalt.

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I crossed out the words "how fast" in the first paragraph above, but it's actually a good descriptor because time is implicitly present in the units for thermal conductivity: a watt is a joule per second. So how quickly (or slowly) something that's hot can burn you is indeed taken into account by its thermal conductivity.

If you reach into an oven where a loaf of bread has just finished baking, and accidentally let the metal oven rack touch your skin briefly, you'll get burned at least a little. You can touch the bread, however, without getting an immediate burn, and the air in the oven, which is at the same temp as the bread surface and the metal rack, would take even longer to burn you.  In our asphalt and iron manhole cover exercise, the manhole cover is like the oven rack--yeah, both are metal--and the asphalt is a little like the bread.  For much more about the thermodynamics of baking bread, see The Fresh Loaf  for a discussion on relevant heat transfer mechanisms. Radiation is one of them.

Now what about the heat capacity of asphalt versus the heat capacity of cast iron?  Heat capacity, as its name implies, is about the ability of a substance to store energy.  This is the currently missing middle component in the energy conservation process we're considering here:  Absorption of energy, storage of energy, and emission of energy. 

First we have solar radiation as the medium for the heat absorbed by the asphalt and the iron manhole cover. Second--since we assume no radiation is being reflected or transmitted--we have energy being stored in the material as kinetic energy of the atoms, molecules, and free electrons (if there are any) that make up the material. And third we have emission of heat from the surface of the material.  This was being done solely by radiation emission (so far we're ignoring the role of the heated air above the surfaces) in an equilibrium situation with absorption and storage of energy--until you came along and put one hand on the manhole cover and the other on the pavement to find out which feels hotter.  Or maybe it was me who did that. 

No matter who did it, that little one-second personal experiment would show the manhole cover feels hotter. You or I might not even be able to keep a hand on the manhole cover for the length of time it takes to say "Mississippi-one".

Why, then, when we use a thermometer to measure the temps of both the asphalt and the manhole cover do we not predict the manhole cover will be at a higher temp? It's a matter of perception, which is something discussed in a typical Coen-Brothers' way in A Serious Man.  Our un-serious man, Prof. Gopnick, remarks after smoking pot and drinking iced tea with his next-door neighbor Mrs. Samsky, that maybe everything is just a matter of perception as the junior rabbi, Rabbi Scott, had suggested to him.  But Rabbi Scott had said "perspective" not perception. 

Our perception of temperature depends on the rate at which energy is conducted to or away from our skin. Thermometers register this same effect by absorbing or losing energy (from a hotter object or to a cooler object, respectively) either rapidly or slowly while coming into thermal-conductance equilibrium with the object or surface they are in contact with. How quickly or slowly this equilibrium happens depends on how well the object or surface conducts heat, and also on how well the thermometer itself conducts heat.  I recall one of my favorite experiments in first-semester physics was called "The Time Constant of a Thermometer."

Now back to heat capacity. The heat capacity of asphalt is about twice that of cast iron:  900 J/kg/K for asphalt versus 460 J/kg/K cast iron. I merely leave this to the reader (and writer) as something to think about, that asphalt stores energy better than iron but conducts it about 40 times more slowly. One thing to think about in this regard is a block of iron in contact with a block of asphalt, where each is at a different temperature to begin with. Another thing is this: what is the heat capacity of a black-body? Einstein in 1907 used Planck's 1900 relation of discrete energy levels for continuously distributed frequencies to model the vibrations of solids, which are all about the heat capacity of solids, but heat capacity seems to be ignored when it comes to discussions of black-body radiation.

The "Pavements" website of the U.S. Federal Highway Administration is where I got the asphalt heat capacity number. The full name of this website is 

Pavement Thermal Performance and Contribution to Urban and Global Climate

The info there is relevant to most of this blog post of mine, and also mentions "albedo" in contrast to emissivity, although the relation between the two isn't simple. Well-written, check it out. I'm outta here.