The Planet Mean Surface Temperature Equation: Tmean=[Φ(1-a)S(β*N*cp)¹∕ ⁴ /4σ]¹∕ ⁴

The Environmental temperature - page in progress

In progress...

We know what temperature is - it is what we measure with thermometers.

Before the age of thermometers' measurements there were only the cold and the hot. Cold and hot it were what our senses told us.

The Celsious scale of temperature - the sentigrade scale.

The Zero is for the melting ice and the Hundred is for the boiling water.

And that was all. When mercury in the thermometer's tube was expanded to show 20 it was 20 oC. When mercury was expanded more it may show 23 oC, then 25 oC and then 30 oC.

But it is not the case with environment.

When it is 20 oC means there is a constant energy supply to support the environment's infrared emisssion.

σ = 5,67*10⁻⁸ W/m²K⁴, the Stefan-Boltzmann constant

Let's transform in Kelvin:

20 + 273 = 293 K

J20 = J293K = σ*T⁴ = 5,67*10⁻⁸ W/m²K⁴ 293⁴ = 5,67*10⁻⁸ (W/m²K⁴) 7.370.050.801 K⁴ = 5,67*73,7 W/m² = 417,88 W/m²

25 + 273 = 298 Κ

J25 = J298K = σ*T⁴ = 5,67*10⁻⁸ W/m²K⁴ 298⁴ = 5,67*10⁻⁸ (W/m²K⁴) 7.886.150.416 K⁴ = 5,67*78,86 W/m² = 447,14 W/m²

30 +273 = 303 Κ

J30 = J303K = σ*T⁴ = 5,67*10⁻⁸ W/m²K⁴ 303⁴ = 5,67*10⁻⁸ (W/m²K⁴) 8.428.892.481 K⁴ = 5,67*84,29 W/m² = 477,92 W/m²

35 +273 = 308 Κ

J35 = J308K = σ*T⁴ = 5,67*10⁻⁸ W/m²K⁴ 308⁴ = 5,67*10⁻⁸ (W/m²K⁴) 8.999.178.496 K⁴ = 5,67*89,99 W/m² = 510,25 W/m²

 

J35 - J30 = 510,25 - 477,92 = 32,33 W/m²

J30 - J25 = 477,92 - 447,14 = 30,78 W/m²

J25 - J20 = 447,14 - 417,88 = 29,26 W/m²