• Support PF! Buy your school textbooks, materials and every day products Here!

Solar Energy and Green House Effect Question

  • Thread starter TFM
  • Start date
  • #1
TFM
1,026
0

Homework Statement



Negelcting internal heat sources, the surface Temperature of the Earth is calculated to be 255 K. Its actual mean surface Temperature is 288 K. The additional heating results from the greenhouse effect. Regarding this as an internalk heat source, calculate that heat source as a fraction of the energy absorbed from the Sun.

Homework Equations





The Attempt at a Solution



I am not sure what I am supposed to do exactly, do I just work out the differnect in Temperature and put it over the final temperature (ie [tex]\frac{33}{288}[/tex]) or do I need to work out the energy in Joules using equations?

TFM
 

Answers and Replies

  • #2
madmike159
Gold Member
369
0
That sounds right to me. The energy in joules would be very difficult to calculate and would be very large because the mass of the surface of the earth would be huge.
 
  • #3
454
0
I think you are supposed to work out how much extra the earth radiates because it's at 288 K and not at 255K, and divide this by the energy that the earth gets from the sun.

You can do everything in units of W/m^2.
The heat radiation from the earth for each square meter is Ct^4. (C an unknown constant)
The intensity of sunlight is 1366 W/m^2 but you can't use that because the sun isn't always directly overhead and it isn't always daytime and there are clouds. You can use an unknown constant for the amount of sunlight received per m^2. (lets call this E)

you know that C(255^4) = E
 
  • #4
TFM
1,026
0
Would this formula be of any use:

Amount of Radiation absorbed:

[tex](1-A)*L_{sun}\frac{\pi r^2}{4\pi D^2}[/tex]

Where A is the Albedo, r the radius of the Planet, D the distance from the planet to the Sun, [tex]L_{Sun}[/tex] is the Sun's Luminoscity.

?

TFM
 
  • #5
TFM
1,026
0
Is the Equation in the post above required for this question?

TFM
 
  • #6
454
0
Is the Equation in the post above required for this question?

TFM
I don't think so. You can use the incoming energy of the sun as an unknown constant. there are some simplifications made here, because the albedo does depend on having an atmosphere, as does convection, evaporation etc. Then there's the day-night and seasonal cycles. There is years of work here.

But I think the only thing you need here is the formula for black body radiation.
 
  • #7
TFM
1,026
0
Does this look right then?

[tex] I = \sigma T^4 [/tex]

where sigma is the Stefan Boltzmann Constant, 5.67*10^-8

[tex] I_{from sun} = 5.67*10^{-8) * 255^4 [/tex] gives 240

[tex] I_{from greenhouse} = 5.67*10^{-8) * 288^4 [/tex] gives 390

Would you then:

390-240 = 150

and thus make thge answer: 150/240?

TFM
 
  • #8
454
0
I think that would be it
 

Related Threads for: Solar Energy and Green House Effect Question

  • Last Post
Replies
4
Views
913
  • Last Post
Replies
7
Views
994
Replies
7
Views
704
  • Last Post
Replies
3
Views
581
  • Last Post
Replies
6
Views
436
  • Last Post
Replies
1
Views
336
  • Last Post
Replies
8
Views
6K
  • Last Post
Replies
1
Views
3K
Top