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XYLENE POWER LTD.

HEAT

By Charles Rhodes, P. Eng., Ph.D.

INTRODUCTION:
This web page reviews the concepts of heat, temperature and thermal radiation. Temperature is expressed in terms of radiation density and radiation spectral distribution.
 

HEAT:
A high school definition of heat is "energy related to molecular motion", but that definition is not helpful in addressing thermal radiation issues. A more general definition of heat is "thermal excitation energy". A change in temperature, together with mass and heat capacity indicate a change in contained thermal excitation energy, also known as a change in enthalpy.

The heat capacities of materials vary and are somewhat temperature and pressure dependent. Typical heat capacities and atomic weights for solid elemental metals are set out in the following table.

ELEMENT   HEAT CAPACITY (J / gm-C)   ATOMIC WEIGHT    ROW PRODUCT
Lithium3.5596.9524.735
Beryllium1.8259.01216.446
Sodium1.21422.99027.910
Magnesium1.04724.30525.447
Aluminum0.92126.98124.849
Potassium0.75339.09829.441
Calcium0.62840.07825.169
Nickel0.50258.69329.463
Iron0.46155.84525.744
Zinc0.37665.3824.582
Tin0.217118.71025.76
Lead0.126207.226.107

The heat added to an object is given by:
(Heat Added) = (Heat Capacity) X (Mass) X (Rise in Temperature)

The row products in the above table indicate that the product:
(Heat Capacity in J / gm-C) X (Atomic Weight in gm / mole) X (1 mole / 6.023 X 10^23 atoms)
is nearly constant.

Hence the amount of heat added to an object is approximately the product of:
(Number of Atoms present) X (rise in temperature).

Thus a rise in temperature causes a similar rise in thermal excitation energy in all atoms.

In general temperature is an indication of nearly equally distributed thermal excitation energy per atom.
 

THERMAL ENERGY:
1. Thermal energy consists of lattice thermal energy and photon thermal energy.

2. Generally the lattice thermal energy is large compared to the photon thermal energy.

3. The lattice thermal energy consists of molecular kinetic energy and changes in lattice potential energy that are closely coupled.

4. The photon thermal energy consists of photons and excited electrons that are closely coupled.

5. There is weak coupling between the two closely coupled energy classes. Lattice vibrations move the electron spheromaks into a slightly different external magnetic field environments which couples energy onto the electrons. Alternatively absorbed photon energy may excite electrons which couples energy back to the lattice. The atomic nuclei may also directly absorb a small fraction of externally supplied photons.
 

TEMPERATURE:
In reality in a solid there are two temperatures. There is the lattice temperature which indicates average kinetic energy of molecules and there is the photon temperature which indicates average electromagnetic energy of photons. Photons and molecules exchange energy primarily via electrons. At thermal equilibrium the lattice temperature equals the photon temperature. However, under nonequilibrium conditions these two temperatures may be different.

Lattice temperature is indicated by thermal expansion (eg a mercury thermometer). In this respect a convenient reference temperature is the freezing point of water (273.15 K, 0 C, 32 F). However, thermal expansion does not convey any concept of radiative heat transfer.

A small portion of the thermal excitation energy exists as photons. The photons exchange energy with electrons. A measurement of the thermal radiation spectrum indicates the photon temperature. The main issue in photon temperature measurement is design of the instrumentation so that it accurately measures the thermal emission radiation spectrum, which is typically in the far infrared.
 

THERMAL RADIATION:
Consider the following:
1. A vacuum isolated object is at the same temperature as its environment if there is radiative energy balance. That is if the change in the object's total energy Et with respect to time t is zero, or:
dEt / dt = 0

2. An object is at zero absolute temperature (0 degrees K) = zero photon temperature if it emits no radiation.

3. Hence temperature is an indication of an object's total energy with respect to the object's total energy when it emits no radiation.

4. Heat spontaneously flows from objects at high temperatures to objects at lower temperatures.

5. Heat can flow across a vacuum barrier confirming the existence of radiative heat transfer.

6. The rate of heat transfer across a vacuum barrier is less for reflective object and container surfaces than for black object and container surfaces.

7. Increasing temperature is indicated by increasing radiation emission.

8. Hence temperature can be expressed in terms of radiation emission.

9. The temperature of an object is independent of the number of atoms or molecules in the object.

10. For an object composed of only one atomic or molecular type, temperature is an indication of average excitation energy per atom or per molecule.

11. The photon temperature of an object can be precisely obtained by analysis of the object's emitted thermal radiation spectrum.

12. Thermal radiation primarily occurs when excited electron spheromaks emit radiation photons.

13. Thermal radiation also occurs when excited nuclei emit radiation photons.

14. Consider an evacuated container. When the evacuated space is at thermal equilibrium with the container walls the vacuum is filled with photons emitted from the container's inside walls. If a uniform spherical object is inserted into the container it reflects, absorbs and emits photons with spherical symmetry.

15. The thermal radiation has a temperature dependent energy density and pressure. Planck calculated the photon energy density and frequency distribution as a function of absolute temperature.

16. Electron and proton spheromak energies follow:
E = h F
where:
h = 6.62607004 X 10^-34 m^2 kg / s
is known as the Planck Constant.

17. Due to the properties of the spheromaks from which the radiation is emitted the energy Ep of each radiated photon is related to the photon frequency Fp via the expression:
Ep = h Fp

18. Bose-Einstein statistics give the relative number of photons at each energy level Ep as:
{1 / [Exp(Ep / K T) - 1]) = {1 / [Exp(h F / K T) - 1])
whete:
K = Boltzmann Constant
= 1.38064852 X 10^-23 m^2 kg s^-2 K^-1
and
T = absolute temperature

19. Thus although there is a high density of states at high photon energies (high photon frequencies} the probability of high energy (high photon frequency) states being occupied is very low, which limits the maximum radiation energy density at any particular temperature.

20. We need to portray the relastionship between temperature, radiation frequency and radiation energy density.

21. An advantage of expressing temperature in terms of photon radiation is that with suitable instrumentation radiation emitted by distant objects can be precisely measured.
 

PLANCK LAW:
Planck's law states that:
I(Fp, T) = {2 h Fp^3 / C^2} / {[Exp(h Fp / K T)] - 1}
where:
I(Fp,T) is the power (the energy per unit time) of frequency Fp radiation emitted per unit area of emitting surface in the normal direction per unit solid angle per unit frequency by a black body at temperature T, also known as spectral radiance;
h is the Planck constant = 6.62607004 × 10-34 m2 kg / s;
C is the speed of light in a vacuum = 299 792 458 m / s;
K is the Boltzmann constant = 1.38064852 × 10-23 m2 kg s-2 K-1;
Fp is the frequency of the electromagnetic radiation in Hz;
T is the absolute temperature of the body in degrees K.
Note that the Planck Law is only true for uniform temperature bodies and hence requires modification for planets and stars surrounded by atmospheres that are at a lower temperature than the planet or star surface. The Planck law is only valid for the temperature at the point of photon emission. If received photons originate from points at different temperatures analysis of the data becomes more complicated.

Photons emitted into space from Earth's atmosphere preferentially originate from locations in the atmosphere where the temperature is just below 273.15 K, the freezing point of water. The issue of the change in Earth's planetary Bond albedo (solar reflectivity) as this temperature rises has enormous practical significance.
 

DERIVATION OF PLANCK LAW:
The essence of the Planck Law is that at any particular temperature at thermal equilibrium there is a corresponding photon energy density and a corresponding photon spectral distribution. This web page is primarily concerned with finding that energy density and spectral distribution.

The presentation herein relies on the reader having some familiarity with Bose-Einstein statistics, the concept of statistical occupancy of different energy states, boundary conditions applicable to electromagnetic radiation in a conducting cavity, orthogonal Fourier functions, solid angle geometry, electron spheromaks, the Planck constant and vector notation.

Consider a cavity inside a solid. ASssume that the cavity has electrically and thermally conducting walls. The cavity is filled with electromagnetic radiation in thermal equilibrium at temperature T.

The total energy in the cavity with respect to the ground state can be found by summing over the contained energies of all allowed single photon states. This can easily be done because the mathematical formulation allows integration from photon energy Ep = 0 to Ep = infinity. To calculate the energy in the cavity we need to evaluate how many photon states there are at each photon energy. The probability of a photon state being occupied is given by Bose-Einstein statistics.

The photon energy density U(T) in a cavity of volume V is given by:
U(T) V = Integral over all possible energies of:
[Photon Energy Ep]
X [Number of photon states in the cavity between energy Ep and energy (Ep + dEp)]
X [Probability of photon state occupancy at Energy Ep]
= Integral from E = 0 to E = infinity of:
{[Ep] [g(Ep) dEp] / [Exp(E / K T) - 1]}

In this expression:
Ep = photon energy;
g(Ep) dEp = number of photon states with energies between Ep and (Ep + dEp);
{1 / [Exp(Ep / K T) - 1]} = Statistical mechanics Bose-Einstein probability of a photon energy state being occupied at temperature T.
 

DENSITY OF PHOTON STATES:
From special relativity:
Ep^2 = |Pp|^2 C^2 + Mo^2 C^4

For a photon Mo = 0

Hence a photon with energy Ep has momentum Pp given by:
|Pp| = Ep / C

However, Pp has many different possible propagation angles.

Assume that photon momentum states are uniformly distributed through positive momentum space and that the separation between adjacent momentum states is constant. This assumtion is equivalent to assuming that each momentum state contains an orthogonal Fourier component of the total photon momentum.
Px = momentum state spacing along the x axis;
Py = momentum state spacing along the y axis;
Pz = momentum state spacing along the Z axis.

Then a momentum state is defined by:
Pp(Nx. Ny, Nz) = Nx Px + Ny Py + Nz Pz
where:
Nx, Ny, Nz are positive integers;
and
|Px| = |Py| = |Pz| = |Po|

Ep(Nx, Ny, Nz) = |Pp(Nx, Ny, Nz)| C
= [Nx^2 Px^2 + Ny^2 Py^2 + Nz^2 Pz^2]^0.5 C
= [Nx^2 + Ny^2 + Nz^2]^0.5 |Po| C
where:
|Po| = Eo / C

Part of the adjacent momentum state separation is due to photon energy. The remaining momentum state separation is due to photon propagation angle.

Then the momentum state vectors Pp(Nx, Ny, Nz) take discrete equally spaced values. Then:
|Pp| = [(Nx Px)^2 + (Ny Py)^2 + (Nz Pz)^2]^0.5
= [Nx^2 + Ny^2 + Nz^2]^0.5 |Po|
= N |Po|
where:
N = [Nx^2 + Ny^2 + Nz^2]^0.5

The density of momentum states in momentum space is:
1 state / |Po|^3

Recall that:
Ep = |Pp| C
or
|Pp| = Ep / C

Define:
Eo = |Po| C

Assume that due to photon polarization there are 2 photon states per momentum state. Hence for every momentum vector Pp with integer components larger than or equal to zero, there are two photon states. This means that the number of photon states in a certain region of momentum space is twice the number of momentum states in that region.

The discrete states in a particular energy range in momentum space lie between |Pp| and (|Pp| + d|Pp|).

Because Nx, Ny, Nz are positive, this momentum space shell spans an octant (1 / 8) of a sphere. The number of photon states g(E) dE, in an energy range dE, is thus given by:
g(Ep) dEp = 2 (1 / 8) (4 Pi |Pp|^2 d|Pp|)(1 / |Po|^3)
= Pi (Ep / C)^2 d(Ep / C)[1 / (Eo / C)^3]
= Pi Ep^2 dEp / Eo^3

This is the density of photon states between energy Ep and (Ep + dEp)
 

CAVITY ENERGY:
The photon energy in the cavity with respect to the ground state is given by:
U(T) V = Integral from Ep = 0 to Ep = infinity of:
Ep [Pi Ep^2 dEp / Eo^3] { 1 / [Exp(Ep / KT) - 1]}
= Integral from Ep = 0 to Ep = infinity of:
[Pi Ep^3 dEp / Eo^3] { 1 / [Exp(Ep / KT) - 1]}

Define:
X = Ep / KT
= (h Fp) / (K T)
and
dX = dEp / KT

Then:
U(T) V = Integral from Ep = 0 to Ep = infinity of:
[Pi Ep^3 dEp / Eo^3] { 1 / [Exp(Ep / KT) - 1]}
 
= Integral from X = 0 to X = infinity of:
[K T]^4 [Pi X^3 dX / Eo^3] { 1 / [Exp(X) - 1]}

This expression has units of energy. That energy is distributed over a cavity volume V.
 

CAVITY VOLUME:
Eo = h Fo = h C / Lamdao
or
Lamdao = h C / Eo

The cavity wall boundary conditions on the thermal equilibrium standing wave photon require that cavity length L satisfy:
L = Lamdao / 2 = h C / 2 Eo

The cavity volume V is:
V = L^3
= [h C / 2 Eo]^3
 

CAVITY ENERGY DENSITY:
Hence the average energy density in the cavity is:
U(T) = [U(T) V] / V
= Integral from X = 0 to X = infinity of:
[K T]^4 [Pi X^3 dX / Eo^3] {1 / [Exp(X) - 1]} / [h C / 2 Eo]^3
 
= {8 [KT]^4 Pi / (h C)^3}Integral from X = 0 to X = infinity of:
{X^3 dX / [Exp(X) - 1]}
 
= {8 [K T]^4 Pi / (h C)^3} {Pi^4 / 15}
 
= 8 [K T]^4 [Pi^5 / 15] / (h C)^3

This equation gives the cavity energy density as a function of temperature.

Recall that:
U(T) V = Integral from E = 0 to E = infinity of:
[Pi E^3 dE / Eo^3] { 1 / [Exp(E / KT) - 1]}
and
V = [h C / 2 Eo]^3 Hence:
U(T) = [U(T) V] / V
= Integral from E = 0 to E = infinity of:
[Pi Ep^3 dEp / Eo^3] { 1 / [Exp(Ep / KT) - 1]} / [h C / 2 Eo]^3
 
= Integral from Ep = 0 to Ep = infinity of:
8 [Pi Ep^3 dEp] { 1 / [Exp(Ep / KT) - 1]} / [h C]^3

Substitue:
Ep = h Fp
and
dEp = h dFp
to get:

U(T) = = Integral from Fp = 0 to Fp = infinity of:
8 [Pi (h Fp)^3 h dFp] { 1 / [Exp(h Fp / KT) - 1]} / [h C]^3

Hence:
dU = 8 [Pi Fp^3 h dFp / C^3] { 1 / [Exp(h Fp / KT) - 1]}

This equation gives the cavity energy density as a function of frequency and temperature.
 

CAVITY ENERGY EMISSION:
Consider energy emitted via a small hole in the cavity. The spectrum of this emitted radiation is known as "Black Body Radiation" because the spectrum is independent of the material surface reflectivity. If all the exiting energy propagated forward the radiation intensity in the aperture would be:
(radiation density in cavity) C.

However, in reality the radiation propagates omni-directionally so that the forward component of the radiation is:
I(Fp, T) dFp = (radiation density in cavity) C / 4 Pi sterradians
= dU(Fp, T) C (1 / 4 Pi)
= 8 [Pi Fp^3 h dFp / C^3] { 1 / [Exp(h Fp / KT) - 1]} C (1 / 4 Pi)
= {2 h Fp^3 dFp / C^2} / {[Exp(h Fp / K T)] - 1}

This equation is known as the Planck Law. It has units of watts / m^2-sterradian

The Planck Law is:
I(Fp, T) = {2 h Fp^3 / C^2} / {[Exp(h Fp / K T)] - 1}

By definition the Stefan Boltzmann constant is:
Cb = [2 Pi^5 / 15] [K^4 / C^2 h^3]
= 5.670400 X 10^-8 J / s-m^2-K^4

Hence U(T) can be written more compactly using the Stefan–Boltzmann constant Cb, giving:
U(T) = 8 [K T]^4 [Pi^5 / 15] / (h C)^3
= [4 Cb / C] T^4

The constant:
[4 Cb / C]
is sometimes called the radiation constant. This constant allows easy calculation of contained radiation energy at a particular temperature.
 

NUMERICAL EXAMPLE:
Find the cavity radiation energy density at 300 degrees K:
[U|T = 300 deg K] = [4 Cb / C] T^4
= [4 X 5.670400 X 10^-8 J / s-m^2-K^4 / (3 X 10^8 m / s)] X [300 K]^4
= [4 X 5.670400 X 10^-8 J / m^3] X [3]^3
= 6.124032 X 10^-6 J / m^3

This radiation energy density does not seem large but the associated radiant power flux can be substantial because this radiation energy potentially propagates away at the speed of light.
 

WIEN'S DISPLACEMENT LAW:
Wien's displacement law shows how the spectrum of black-body radiation at any temperature is related to the spectrum at any other temperature. If we know the shape of the spectrum at one temperature, we can calculate the shape at any other temperature. Spectral intensity can be expressed as a function of wavelength or of frequency.

A consequence of Wien's displacement law is that the wavelength Lamdam at which the intensity per unit wavelength of the radiation produced by a black body is at a maximum is a function only of the temperature:
Lamdam = {b / T}
where the constant b, known as Wien's displacement constant, is:
b = 2.8977729(17) × 10-3 K m.

Planck's law was also stated above as a function of frequency. The frequency Fm of maximum intensity is given by:
Fm = T X 58.8 GHz / K
 

STEFAN BOLTZMANN LAW:
The Stefan–Boltzmann law states that the power emitted per unit area of the surface of a black body is directly proportional to the fourth power of the body's absolute temperature:
J = Cb T^4,
where:
J is the total power radiated per unit area,
T is the absolute temperature
and
Cb = 5.670400 X 10-8 W m-2 K-4
is the Stefan–Boltzmann constant
.
 

STEFAN BOLTZMANN LAW DERIVATION:
Recall that Planck's Law is:
I(Fp, T) = {2 h Fp^3 / C^2} / {[Exp(h Fp / K T)] - 1}

The Stefan Boltzmann law follows from integrating I(Fp ,T) over frequency and solid angle:
J = (Integral from Fp = 0 to Fp = infinity) dF (Integral dOmega cos(theta) I(Fp, T))
where:
(Integral dOmega cos(theta)
= (Integral from Phi = 0 to Phi = 2 Pi) d(Phi)) (Integral from Theta = 0 to Theta = Pi / 2) d(Theta) sin(Theta) cos(Theta)
= Pi

The cos(Theta) factor appears since we are considering the radiation in the direction normal to the surface.
The solid angle integral extends over the full Phi = 0 to Phi = 2 Pi in azimuth and over the polar angle Theta = 0 to Theta = Pi / 2.

I(Fp,T) is independent of angles theta and Phi and passes through the solid angle integral. Inserting the formula for I(Fp ,T)} gives:
J = {[2 Pi (K T)^4] / C^2 h^3]} {Integral from X = 0 to X = infinity of {X^3 dX / [(Exp X) - 1]}}
where:
X = h Fp / K T
is unitless.

The integral over X has the value Pi^4 / 15}, which gives:
J = Cb T^4
where:
Cb = [2 (Pi^5 / 15)(K^4)] / (C^2 h^3)
= 5.670400 X 10^-8 W / m^2-K^4

 

This web page last updated May 13, 2017.

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