The temperature of the Earth is not equal to the Temperature it absorbs from the Sun.
 

Author; Rogelio Pérez C

Abstract

When we take the energy emitted or absorbed by an area of a sphere, and distribute it throughout the sphere, the result will be a temperature 29% lower than the temperature of the entire sphere. For example; The earth has an input energy of 1.73E 17w, if we subtract from this value the albedo of the earth which is 0.3, the net input energy of the earth will be 1.21E 17wm², which is equivalent to 360 kelvin, but when this energy is divided into the entire sphere the temperature then is 255k, which is 29.29% lower temperature, than the one received in the area, now if we do the same with the Sun, taking the energy emitted by an area of the sun that is 9.58501E 25 wm² , and we divided it into the entire sphere of the sun its temperature would be 4,085.66 kelvin and not 5778 kelvin, which would be 29.29% lower.

Introduction;

The sun can be taken as a perfect example, to prove that the energy absorbed by an area of a sphere, and distributed throughout the area of the sphere, decreases the temperature by 29%, When we measure the sun as a sphere 4Πr²; we get that its area is 6.06679E 18m², its energy is; 3,834E 26wm², and its temperature is; 5778 kelvin, Now if we measure the energy emitted by the area (Πr²) of the sun of 1.5167E 18m², we see that it emits energy equal to; 9.58501E 25 wm², and its temperature is; 5778 kelvin

But if we wanted to know the temperature of the sun, as the of earth is known; we would get the following results; net radiation emitted (emitted radiation is equal to that absorbed) by the area (Πr²) of the Sun, 9.58501E 25 wm², and divided it into the area (4Πr²) of the sun's sphere 6.06679E 18 m², we would obtain that the temperature of the surface of the Sun, would be, 4,085.66 kelvin a difference of 1,692.34, kelvin less temperature, or 29.29%.

But if we would like to know the temperature of the sun, obtaining it as it is obtained from earth; we would obtain the following results; the net radiation emitted by the area (Πr²) of the Sun is; 9.585011E 25 wm², and we divided it into the area (4Πr²) of the sphere of the sun 6.066Chas 18 m², we would obtain that the temperature of the surface of the Sun, would be, 4,085.66 kelvin a difference of 1,692.34 kelvin  of lower temperature, of the 5778 kelvin that it actually has, (-29.29%).

Now on Earth

Energy sent by the sun for the area (Πr²) of the earth of 1.27451E 14m², is equal to 1.73E 17w, to this we subtract the albedo which is 0.3 (1-0.3)= 0.7= 1.73E 17w*0.7= 1.21E 17wm².

Who we use the formula of finding the radiation in kelvin, we have that the energy of the sun absorbed by the area of the earth is 1.21E 17wm². And we divided it by the area that received it 1.27451E 14m², we would get an effective earth temperature of 360.23°k (87.08°C).

Now when we divide the energy absorbed by the earth from 1.21E 17wm², and distribute it over the entire area (4Πr²) of the earth sphere which is equivalent to 5.10E 14 m², we obtain an effective temperature of 255k or (-18°C). That's 105.5 kelvin temperature less than the 360 kelvin, which represents the absorbed energy, or 29.29%

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

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 kelvin

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).

Conclusion:

Taking the temperature absorbed by the area of an sphere, and dividing it into the area of the whole sphere, results in a temperature value 29.29% lower than that which was absorbed, when this same procedure is done with the sun, we get the same results, it is therefore concluded that the temperature absorbed by the earth is 29.29% much higher than the value of 255 kelvin, which shows the data.

Knowing that the temperature of the earth is 288 kelvin, we can affirm, that the temperature of the earth is not equal to the energy it absorbs from the sun.

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