Direct way of measuring in kelvin the radiation emitted by bodies

 

Author; Rogelio Perez C

Summary; Stefan-Boltzmann's law states that the total radiation emitted by a body, E, is proportional to the temperature T, raised to the fourth power of its temperature, (E=σ*A*T⁴), but in order to measure the incoming radiation to a body, and to know the amount of heat it emits, it was not possible to do it directly, so throughout the history of physics we have worked with a theoretical model known as the body black. This model consists of a theoretical object that absorbs all radiant light, for the study of heat emission, this is the same model on which the theory of climate change is based for climate projections, which according to the model will continue to increase, but as the story must be known not to repeat it, it turns out that more than 100 years ago this model produced an error in the temperature measurement results, Known in science as ultraviolet catastrophe, the error consisted of a failure of the classical theory of electromagnetism to explain the electromagnetic emission of a body in thermal equilibrium with the environment, but the error in the model was corrected by the German physicist Planck, Max Karl Ernst Ludwig, with the law that bears his name, but apparently this model has again erred in the thermal balance of the planets and their environment, by presenting different temperatures between those that are actually absorbed and those that emit the planets, to maintain the thermal balance, this work presents not a correction of the model, but a way to be able to measure light directly without the need for the theoretical model of the black body, and it presents the differences between the temperature that the planets absorb directly and the one that they should absorb according to the established model.

Introduction;

Stefan-Boltzmann's law which states that the total radiation emitted by a body, E, is proportional to the temperature, T, elevated to the fourth power of its temperature, (E=σ*A*T⁴) this work based on this law, Provide a new command to measure the thermal radiation of the bodies directly with this formula T=⁴√E/A σ, because for the study of radiation measurements,, could not be measured directly, that is why the concept of black body was applied that constitutes an idealized physical system for the study of the emission of electromagnetic radiation, Knowing also that in ancient times this model presented some serious problems, known as the catastrophe of the ultraviolet, which consists of a failure of the classical theory of electromagnetism, explaining the electromagnetic emission of a body in thermal equilibrium with the environment, the solution to this problem was raised by Max Planck in 1900, now known as Planck's law. That moment is considered as the principle of quantum mechanics. This law describes electromagnetic radiation emitted by a black body in thermal equilibrium at a defined temperature, with this work we see that there are also differences between the energy that absorbs a black body, and the energy it emits, so it is necessary to use the new direct method of measuring electromagnetic radiation.

Theory;

The Stefan-Boltzmann law states that a body emits thermal radiation with a total hemispherical emissive power (W/m2) proportional to the fourth power of its temperature.1.

The law is very accurate only for ideal black objects, the perfect radiators, called black bodies; it works as a good approximation for most gray bodies.

The law was deduced in 1879 by the Austrian physicist Jožef Stefan (1835-1893) on the basis of experimental measurements made by the Irish physicist John Tyndall. Stefan published this law in the article "Über die Beziehung zwischen der Wärmestrahlung und der Temperatur" (on the relationship between thermal radiation and temperature) in the Vienna Academy of Sciences Session Bulletin.

The law was derived in 1884 from theoretical considerations by Ludwig Boltzmann (1844-1906) using thermodynamics. Boltzmann considered a certain ideal thermal motor with light as a source of energy instead of gas.

A black body is a theoretical object that absorbs all the light and all the radiant energy that affects it, constituting an idealized physical system for the study of the emission of electromagnetic radiation.2

Classic and quantum black body models

The physical principles of classical mechanics and quantum mechanics lead to mutually exclusive predictions about black bodies or physical systems approaching them. Evidence that the classic model made predictions of emission at small wavelengths in open contradiction with what was observed led Planck to develop a heuristic model that was the germ of quantum mechanics, the contradiction between classic predictions and empirical results at low wavelengths is known as ultraviolet catastrophe.

Planck's law describes electromagnetic radiation emitted by a black body in thermal equilibrium at a temperature definida.3

The ultraviolet catastrophe is a failure of the classical theory of electromagnetism to explain the electromagnetic emission of a body in thermal equilibrium with the environment. According to predictions of classical electromagnetism, an ideal black body in thermal equilibrium was to emit energy in all frequency ranges; so the higher the frequency, the higher the energy. ​This was demonstrated by Rayleigh and Jeans, for whom the ultraviolet catastrophe is also known as the Rayleigh-Jeans catastrophe. According to the law they stated, the density of energy emitted for each frequency should be proportional to the square of the latter, which implies that emissions at high frequencies (in the ultraviolet) must carry enormous amounts of energy, so much so that when calculating the total amount of radiated energy (i.e. the sum of emissions in all frequency ranges), it is apparent that this is infinite, a fact that puts the energy conservation postulates at risk.4

The law of energy conservation states that the total amount of energy in any isolated physical system (without interaction with any other system) remains unchanged over time, although that energy can be transformed into another form of energy. In short, the law of energy conservation states that energy is not created or destroyed, only transformed, 5

 

How would we explain the temperature of the sun, according to the movement of its molecules?

The Sun is made up of gaseous hydrogen. When the atoms of this gas collide with each other due to their movements, they cause such high pressures and temperatures; they fuse to form another chemical element: Helium. This fusion reaction releases large amounts of energy in the form of light and more heat.

How do we know the speed of molecules in the sun?

First we have to find the speed of the atoms that make up the sun, at the temperature of 5778, this we do with the formula of average square velocity.

Vcm=√ (3 *R *T)/M=

Sun.

R= 8.31 J/mole .k= kg·m²/s².k

T= 5778 k

M = 1,98E+30 kg

Vcm= √ (3 *8, 31j/mole/k *5778k)/ 1,98E+30kg =

Vcm= √ 144045,54 kg·m²/s²/ 1,98E+30kg=

Vcm= √7,27503E-26m²/s²= 2,69723E-13 m/s

The mean square velocity (CMV) of a body with mass m=1.98E 30 kg and temperature T=5778 is equal to 2.69723E-13 m/s

Now knowing the average square velocity of the mass of the sun, the kinetic energy of the sun can be known.

Ec= ½ mv²                       

Ec or Ek. SUN:

M=1,98E+30kg/mole

Vcm= 2,69723E-13 m/s

Ec= ½ 1,98E+30 kg/ mole*(2,69723E-13 m/s)²

Ec or Ek= 72022.77 Joule = kg*m²/s²

The kinetic energy (E) of a body with mass m =1,98E+30 kilograms and velocity v = 2,69723E-13 m/s equals 72022,77Joule.

Now we will calculate the temperature of the molecular motion of the sun based on the following;

Of the two expressions for the pressure of a gas, one derivative of macroscopic experimental data, P· V = k ·T and other derived from Newton's laws, P = m*(v2)pr / 3V. If both describe the same reality, then it should happen that k·T = m*(v2)pr / 3. It follows that the temperature, T = 2/(3K)*m*(v2)pr /2, i.e. the temperature of a gas is proportional to the average kinetic energy of its molecules.

T = 2/ (3k) · m · (v2) pr /2,

T = 2/(3R 8,31 J/mole .k) · m ·(Vcm m/s)² /2.

T=2/(3*8.31 J/mole .k). 98E+30 kg *2,70E-13 m/s /2     

T=2/ (24.93 J/mole .k)* 98E+30 kg *7,28E-26 m²/s²/2

T=5778 K

The sun is the main source of energy that reaches the planet.

The Stefan-Boltzmann law says that the total radiant heat energy emitted from a surface is proportional to the fourth power of its absolute temperature. The law was formulated by Josef Stefan in 1879 and then derived by Ludwig Boltzmann. The formula E = σT4 is given, where E is the radiant heat emitted from one unit area per unit time, T is the absolute temperature and σ =5.670367 × 10−8 W · m2⋅K4 is the Stefan-Boltzmann constant.

What is the total radiation value emitted by the sun

Formula; E = σ * A * T⁴

E = sun-radiated heat (Luminosity).

Σ= Stefan-Boltzmann constant.

A =Area of a body (Sun)

T⁴=Temperature raised to 4 power.

Result;

σ =    5,67E-08   W/(M²K4)

A=     6,06679E+18     m²

T⁴=   1,11458E+15     Kelvin4

E =    3,834E+26         Wm²

 

An area 6.066C10 18m² with a temperature of 5778°k at the fourth power, multiplied by the Stefan-Boltzmann constant, has an energy of 3.834E 26wm²

 

Mathematical formula for knowing the kelvin degrees of energy in watts

To find out how many kelvin degrees are equivalent to all these energy values in watts, which are related to the temperature of the Sun and the planet, I use the following mathematical formula from Stefan-Boltzmann's law;

E= A* σ * T⁴

A* σ * T⁴=E

σ * T⁴=E /A

T⁴= E /A* σ

T=⁴√E/A σ

Where;

E = heat radiated by the one body (Luminosity).

σ = Stefan-Boltzmann constant.

A =Area of a body

T⁴=Temperature raised to 4 power.

Conversion of watts to kelvin:

Kelvins for the total value of watts emitted by the Sun

σ =    5,67E-08   W/(M²K⁴)

A=     6,06679E+18     m²

E = 3,834E+26 W=Joule /s

Results.

T=⁴√E/A σ

Te=⁴√3,834E+26 W= / 6,06679E+18 m²*5,67E-08 W/(M²K⁴)

Te=⁴√3,834E+26 W / 3,43987E+11 W/K⁴)

Te=⁴√1,11458E+15k⁴ = 5778 kelvin.

An area 6.06679E 18 m² with energy of 3.834E 26 Wm² has a temperature equal to 5578 k

How much energy of what it sends from the sun is absorbed by the planets?

The amount of energy from the sun that the planets receive is given by the solar constant for the planet and the area of the solar disk that receives it, however, not all radiation intercepted by the planets is absorbed; a fraction of the incident energy is reflected back into space, known as the albedo.

 

It must be recognized that the rotation of the planets must play an important role in the temperature of the planets, because it is not the same as the planet Venus takes almost 1 year to absorb light throughout the planet, as Jupiter, every 9 hours it bathes in the energy of the sun throughout its sphere.

To know how much energy each planet absorbs, as they are supposed to be in thermal equilibrium, the energy they emit with which they absorb is then used the following formula:

The temperature calculated in this way is known as the effective temperature of the planets, TE, which only depends on the distance of the earth to the sun (i.e. the solar constant) and the albedo. This can be interpreted as if the sun-heated planets acquire an effective temperature, which for the earth is 255°K (-18°C).

1-Effective temperature for the planeMercury;

Ks = solar constant (9040 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.058

Te=Effective temperature kelvin

 

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√9040 W/M²(1-0.058) /4*5,67E-08 W/(M²K4)

Te=⁴√9040 W/M²(0.7) / 2,268E-07 W/(M²K4)

Te=⁴√ 37547089947=440.1 kelvin.

 

2-Effective temperature for the planet Venus;

Ks = solar constant (2610 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.71

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√2610 W/M² (1-0.71) /4*5,67E-08 W/(M²K4)

Te=⁴√2610 W/M² (1-0.71) / 2,268E-07 W/(M²K4)

Te=⁴√ 2,268E-07=240.3 kelvin.

 

3-Effective temperature for the planet Earth;

Ks = solar constant (1364 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.3

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√1364 W/M²(1-0.3) /4*5,67E-08 W/(M²K4)

Te=⁴√1364 W/M²(0.7) / 2,268E-07 W/(M²K4)

Te=⁴√4197530864=254,54 kelvin.

 

4-Effective temperature for the planet Mars;

Ks = solar constant (590 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.17

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√590 W/M²(1-0.58) /4*5,67E-08 W/(M²K4)

Te=⁴√590 W/M²(0.83) / 2,268E-07 W/(M²K4)

Te=⁴√ 2159171076=215.1 kelvin.

 

5-Effective temperature for the planet Jupiter;

Ks = solar constant (50 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.52

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√50 W/M²(1-0.52) /4*5,67E-08 W/(M²K4)

Te=⁴√50 W/M²(0.48) / 2,268E-07 W/(M²K4)

Te=⁴√ 105820106=101.4 kelvin.

6-Effective temperature for the planet Saturn;

Ks = solar constant (15 w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.47

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√15 W/M²(1-0.47) /4*5,67E-08 W/(M²K4)

Te=⁴√15 W/M²(0.53) / 2,268E-07 W/(M²K4)

Te=⁴√ 35052910,05 =76.95 kelvin.

7-Effective temperature for the planet Uranus;

Ks = solar constant (3.7w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.51

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√3.7 W/M²(1-0.51) /4*5,67E-08 W/(M²K4)

Te=⁴√3.7 W/M²(0.49) / 2,268E-07 W/(M²K4)

Te=⁴√ 7993827,16=53.17 kelvin.

8-Effective temperature for the planet Neptune;

Ks = solar constant (1.5w)

σ = Stefan-Boltzmann constant. 5.67E-08 W/(M²K4)

a = Albedo 0.411

Te=Effective temperature kelvin

Mathematical formula for effective temperature;

Te=⁴√ Ks(1- α) / 4 σ  

Te=⁴√1.5 W/M²(1-0.41) /4*5,67E-08 W/(M²K4)

Te=⁴√1.5 W/M²(0.59) / 2,268E-07 W/(M²K4)

Te=⁴√ 3902116,402 = 44.45 kelvin.

Te=4√ So(1- α)/4 σ
Planet	K (W/m²) So	albedo	(1-a)	4 σ= Ksb*4	So(1- α)/4 σ	Te  Sun K
Mercury	9040	0,058	0,942	2,3E-07	3,75E+10	440,19
Venus	2610	0,71	0,29	2,3E-07	3,34E+09	240,35
Earth	1360	0,33	0,7	2,3E-07	4,2E+09	254,54
Mars	590	0,17	0,83	2,3E-07	2,16E+09	215,56
Jupiter	50	0,52	0,48	2,3E-07	1,06E+08	101,42
Saturn	15	0,47	0,53	2,3E-07	35052910	76,95
Uranus	3,7	0,51	0,49	2,3E-07	7993827	53,17
Neptune	1,5	0,41	0,59	2,3E-07	3902116	44,45

 

To know how much temperature of the sun each planet absorbed, we must know how much energy the planet received, and subtract the albedo.

The energy of the Sun received by each Planet is;

Er=R²*ksolar
Planet	Radio	Radio	R²	π pi	K Solar	Er W/m²
Mercury	2,44E+06	2,44E+06	5,95E+12	3,1416	9040	1,6904E+17
Venus	6,05E+06	6,05E+06	3,66E+13	3,1416	2610	3,003E+17
Earth	6,37E+06	6,37E+06	4,06E+13	3,1416	1360	1,7342E+17
Mars	3,39E+06	3,39E+06	1,15E+13	3,1416	590	2,1245E+16
Júpiter	6,99E+07	6,99E+07	4,89E+15	3,1416	50	7,68E+17
Saturn	5,82E+07	5,82E+07	3,39E+15	3,1416	15	1,59796E+17
Uranus	2,54E+07	2,54E+07	6,43E+14	3,1416	3,7	7,47687E+15
Neptune	2,46E+07	2,46E+07	6,06243E+14	3,1416	1,5	2,85686E+15

The energy received from the sun by each planet, minus the albedo is;

 

To know the value of the temperature that comes from the sun, and is absorbed by the planets is done directly as follows;

Conversion of watts to kelvin with the following mathematical formula:

To find out how many kelvin degrees are equivalent to all these energy values in watts, which are related to the temperature of the Sun and the planet, I use the following mathematical formula from Stefan-Boltzmann's law;

E= A* σ * T⁴

A* σ * T⁴=E

σ * T⁴=E /A

T⁴= E /A* σ

T=⁴√E/A σ

Where;

E = heat radiated by the one body (Luminosity).

σ = Stefan-Boltzmann constant.

A =Area of a body

T⁴=Temperature raised to 4 power.

Conversion of watts to kelvin with the following mathematical formula:

T=⁴√E/A σ

Where;

σ = Stefan-Boltzmann constant = 5.67E-08 W/(M²K

A= area = 1.27E 14 m²

E = energy to convert 1,21687E+17Wm²

Result;

T=⁴√E/A σ

Te=⁴√1,21687E+17Wm² = / 1,27E+14 m² *5,67E-08 W/(M²K⁴)

Te=⁴√1,21687E+17Wm² / 7,23E+06 W/K⁴)

Te=⁴√ 1,68E+10 k⁴ = 360.23°k (87.08°C).

An area 1.27E 14 m² with energy of 1,21687E+17Wm² has a temperature equal to 360.23°k (87.08°C).

Te=4√ p/A*σ

Planet	Ea= Et(1-a) watt	A, área M²	σ =Ksb W (m2 K4)	A*σ	p/A*σ	Te=⁴√p/A*σ kelvin
Mercury	1,59E+17	1,87E+13	6E-08	1,06E+06	1,50E+11	622,53
Venus	8,71E+16	1,15E+14	6E-08	6,52E+06	1,33E+10	339,91
Earth	1,21E+17	1,28E+14	6E-08	7,23E+06	1,68E+10	359,97
Mars	1,76E+16	3,60E+13	6E-08	2,04E+06	8,64E+09	304,85
Jupiter	3,69E+17	1,54E+16	6E-08	8,71E+08	4,23E+08	143,44
Saturn	8,47E+16	1,07E+16	6E-08	6,04E+08	1,40E+08	108,82
Uranus	3,66E+15	2,02E+15	6E-08	1,15E+08	3,20E+07	75,20
Neptune	1,69E+15	1,90E+15	6E-08	1,08E+08	1,56E+07	62,86

Now we will know the difference in the temperatures absorbed by the planets, between the indirect method for knowing the kelvin value of the absorbed radiation, and the direct method;

 

Conclusion:

The electromagnetic energy that comes from the sun to the bodies no longer needs a theoretical model like that of the black body, in order to measure the heat that this energy that falls on the surface of the bodies, in other words now all bodies will now behave like a perfect "radiation absorber" because this work has presented a new way to achieve these values directly, in addition, this new way of measuring electromagnetic energy is of great importance not only for climate science, but also for other physics sciences such as quantum which is the product of the way electromagnetic radiation is measured, and also for some measurements in astronomy.

The way in which the relationship between the absorption and emission of electromagnetic energy known as the black body is measured, has to be rechecked, because the data measured directly of the temperature that the planets should absorb, present a difference of 29.29% in all the effective temperature that has been measured for the planets.

Bibliography

1-https://es.wikipedia.org/wiki/Ley_de_Stefan-Boltzmann

2-https://es.wikipedia.org/wiki/Cuerpo_negro

3-https://es.wikipedia.org/wiki/Ley_de_Planck

4-https://es.wikipedia.org/wiki/Cat%C3%A1strofe_ultravioleta

5-Física Volumen 1. Escrito por Víctor Campos Olguín., p. 159, en Google Libros