The energy received from the Sun on Earth in m² should not be divided into 4, because the result will not be the temperature of an area m².

 

Author Rogelio Perez C

Summary;

When studying the climate of the planet, the energy of the sun that enters the earth by area, is divided into 4 parts because the earth is a sphere, which should not be so, since the earth in its sphere, receive the same amount of energy due to the rapid rotation that the earth has. Knowing that all bodies emit electromagnetic radiation due to their temperature, But Stefan-Boltzmann's law explains that a hemispheric body (area in m²) emits energy at 4 powers of its temperature, then when is divide the energy emitted by a hemispheric body (area m²) into 4 parts, what are doing is the opposite of Stefan-Boltzmann's law, therefore the result is not the energy of a hemispheric body (m²), nor will it be the temperature of a hemispheric body (area m²).

In case this division is due to the energy received from the sun, is the energy emitted by its sphere, at the distance from the earth, which is given by the solar constant, it must be said that the solar constant has the same energy value, when it is emitted by the area of the circle of the sun or by the area of its sphere. Then the energy and temperature per 1m² should be the same in all areas of its body.

Introduction;

Incandescence is an emission of light by heat. A body exposed to sufficient heat emits electromagnetic radiation in the visible spectrum from a certain temperature.

The quality of the light emitted directly depends on the temperature at which the body is located, a slightly warm body (around 1600°C) emits red-orange light, while a very hot body (around 5000°C), It emits very white light and can even reach bluish white at extreme temperatures (8000-9000°C).

To measure the incandescent state of an object or describe the light it emits, Stefan-Boltzmann's law is used;

Establishes that a body emits thermal radiation with a total hemispherical emissive power (W / m2) proportional to the fourth power of its temperature;

It's mathematical expression;

Formula; E = σ * A * T⁴

E = sun-radiated heat (Luminosity).

σ = Stefan-Boltzmann constant.

A =Area of a body

T⁴=Temperature raised to 4 power.

So we have that all bodies emit electromagnetic radiation because of their temperature, but when is apply the Stefan-Boltzmann law, to a hemispheric body it times at the 4 power of its temperature, this is because the Stefan-Boltzmann law, is a measure of the body in area m².So when we divide the energy of an area into 4, what we're doing is the inverse of the Stefan-Boltzmann law, causing an error that will lead to lower energy of the temperature of the body under study. Therefore, when is divide the energy of this body in m² into 4, the result will not be, nor the energy of a body in m², because this is divided into 4, and the temperature will not be that of a body measured in m², because it is the temperature of only a part of its energy.

This work will present the decrease in the temperature of a body such as the sun, from 5778 ° k to 4085 ° K, When the energy of the area of the circle of this is distributed in the total area of its sphere, it will also be done by dividing the temperature absorbed by the earth, going from 360 ° k to 255°k , when the energy that enters its area is divided, by the whole area of its sphere, although what is really done is to take the solar constant subtract the albedo, and take out the fourth power of its temperature, mathematically so;

Finally, the value of the solar constant will be known when it is emitted by the area of the sun's sphere, and when it is emitted by the area of its circle.

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

Incandescence is an emission of light by heat. A body exposed to sufficient heat emits electromagnetic radiation in the visible spectrum from a certain temperature.

The quality of the light emitted directly depends on the temperature at which the body is located, a slightly warm body (around 1600°C) emits red-orange light, while a very hot body (around 5000°C), It emits very white light and can even reach bluish white at extreme temperatures (8000-9000°C).

Hence the expression red hot or red alive to describe the state of a metal in the forge, Hence also the brightness of the lava of the volcanoes (the molten rock is so hot that it emits light).

To measure the glow state of an object or describe the light it emits, a theoretical model called the black body is used.6

Developing

 

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

What is the total radiation value emitted by the area 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=     1,5167E+18m²,  m²

T⁴=   1,11458E+15     Kelvin4

E =    9,58501E+25 wm²,     Wm²

 

An area 1,5167E+18m², with a temperature of 5778°k at the fourth power, multiplied by the Stefan-Boltzmann constant, has an energy of 9,58501E+25 wm²,  

 

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

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 6.06679E18m² with energy of 3.834E26Wm² has a temperature equal to 5578 k

 

Conversion of watts to kelvin for area:

Kelvins for the total value of watts emitted by the Sun

σ =    5,67E-08   W/(M²K⁴)

A=     1,5167E+18 m²

E = 9,58501E+25 W

Results.

T=⁴√E/A σ

Te=⁴√9,58501E+25 W= / 1,5167E+18 m²*5,67E-08 W/(M²K⁴)

Te=⁴√9,58501E+25 W / 8,60E+10 W/K⁴)

Te=⁴√1,11458E+15 k⁴ = 5778 kelvin.

 

An area 1,5167E+18 m² with energy of 9,58501E+25 Wm² has a temperature equal to 5578 k

 

Then we have that the energy emitted by the sun in its area is equal to, 9,585011E 25 Wm², if we divided this energy among the whole sphere of the sun, 6,066ritE 18 m², we would get the effective temperature of the sun, it would be;

 

Conversion of watts to kelvin:

Kelvins for the total value of watts emitted by the Sun

T=⁴√E/A σ

σ =    5,67E-08   W/(M²K⁴)

A=     6,06679E+18 m²

E = 9,58501E+25 W

Results.

T=⁴√E/A σ

Te=⁴√9,58501E+25 W= / 6,06679E+18 m²*5,67E-08 W/(M²K⁴)

Te=⁴√9,58501E+25 W / 3,44E+11 W/K⁴)

Te=⁴√2,79E+14  k⁴ =  4.085,66 kelvin.

An area 6,06679E+18 m² with energy of 9,58501E+25 Wm² has a temperature equal to 4.085,66 kelvin.

 

Now the same operation for the Land;

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).

Conversion of watt into kelvin for every sphere of the Earth:

T=⁴√E/A σ

Where;

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

A= area = 5,10E+14 m²

E = energy to convert 1,21687E+17Wm²

 

Result;

T=⁴√E/A σ

Te=⁴√1,21687E+17Wm² = / 5,10E+14 m² *5,67E-08 W/(M²K⁴)

Te=⁴√1,21687E+17Wm² / 28905994,02 W/K⁴)

Te=⁴√ 4209733669 k⁴ = 254.72°k (-18°C).

An area / 5,10E+14 m² with energy of 1,21687E+17Wm² has a temperature equal to 254.72°k (-18°C).

What is the value of the solar constant for the area of a sphere 4πr²?

The radiant energy that reaches us from the sun will decrease inversely to the area formed by a sphere of radius of 1 U.A., that is:

An astronomical unit =1.496E 11M

Astronomical unit in M²=

m²=1.496E 11M x 1.496E 11M= 2,238022E 22m²

Astronomical unit for the radius of a sphere;

=2,23802E+22m² x  4π 3,14 =2,23802E+22m² x π 12.56=

=2,81095E+23.m²

 

Total energy from the sun by the area of a sphere 4πr²;

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

Then;

The value of the solar constant= to the energy emitted by the sun in the area 4πr² of a sphere= 3.834E 26w/m², between the astronomical unit in the area of a sphere 4πr²=2, 81095E+23m²

3.834E 26w/m² /2, 81095E+23 = 1363.953709w

The value of the solar constant is equal to 1364wm²

 

What is the solar constant value for the area of a circular disc πr²?

An astronomical unit =1.496E 11M

Astronomical unit in M²=

m²=1.496E 11M x 1.496E 11M= 2,238022E 22m²

Astronomical unit for the radius of a circle πr²;

=2,238022E 22m² x π 3.14 = 7,027E 22m²

Energy emitted by the sun for the area of a circle πr².

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=     1,5167E+18m²,  m²

T⁴=   1,11458E+15     Kelvin4

E =    9,58501E+25 wm²,

An area 1,5167E+18m², with a temperature of 5778°k at the fourth power, multiplied by the Stefan-Boltzmann constant, has an energy of 9,58501E+25 wm²,  

Then;

The value of the solar constant= to the energy emitted by the sun in the area 4πr² of a sphere= 9,58501E+25 wm², between the astronomical unit in the area of a sphere 4πr²= 7,027E 22m²

9,58501E+25 wm²/ 7,027E 22m² = 1363.953709w

Value of the solar constant for the circle of the sun is equal to 1364wm²

Conclusion;

1- When the temperature of the area of the sun circle, was divided over the whole sphere of the sun, then the temperature of the sun, decreased by 29% (going from 5778°k to 4085°k), which shows that the emitted energy of 1m² of a body cannot be divided into 4m², because you would no longer get the temperature that this body emits because of its temperature.

2- The solar constant has the same energy value of 1364wm², when measured by taking the area of the sun's sphere or the area of its circle.

3- The mathematical formula from which the specific temperature of the earth originates, 255°K -18°C, can be read as the temperature emitted by a non-hemispheric body (not area in m²), because its energy is one quarter that of a hemispheric body measured in m², and the temperature will also be that of a body with one quarter of the energy of a m².

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

6- Dionysius Lardner (1833). Treatise on Heat. Longman, Rees, Orme, Brown, Green & Longman. "The state in which a heated body, naturally incapable of emitting light, becomes luminous, is called a state of incandescence."

John E. Bowman (1856). An Introduction to Practical Chemistry, Including Analysis (Second American edition ed.). Philadelphia: Blanchard and Lea.

William Elgin Wickenden (1910). Illumination and Photometry. McGraw-Hill