Question

Calculate the ratio of the total intensity of radiation from the sun's surface, T=6000K as compared to Earth's surface T=288K

I know the formula is ¡=eøT^4
e=emissivity
ø= 5.67x10^-8W/m^2

Therefore, I got 7.34X10^23

Can anyone help me check my answer?

Answer 1

Since you're looking at the ratio, all of the constants cancel and you're left with
\[ \frac{T_{sun}^4}{T_{earth}^4} \]
Check your answer against that; it looks far too large.

Answer 2

hmmm?

Answer 3

so it's 6000^4÷288^4?

Answer 4

Yes. \[{\epsilon \sigma T_{sun}^4 \over \epsilon \sigma T_{earth}^4} = {T_{sun}^4 \over T_{earth}^4}\]

We can treat the sun and the earth as a blackbody, which means that \(\epsilon = 1\). Unless your problem states otherwise.

Answer 5

But even if the emissivity \( \epsilon \) is not 1, we can assume it is the same in both cases and therefore cancels.

Answer 6

Oh ok, thanks so much guys! Are you guys physic teachers?

Answer 7

Why can we assume they are both the same? Every object has a different emissivity. The sun is obviously a blackbody, but since water and land have different emissivity values and the fact that the atmosphere scatters radiation, the earth is not a blackbody. I'm curious as to your rational on this. I'm not trying to start a debate.

Answer 8

In general, no you're right. I'm just saying for the purposes of this problem where no other information is given, we should assume they are the same. You are probably also right that the sun is very close to perfect blackbody and for all purposes \( \epsilon = 1 \).

Answer 9

*for all practical purposes \( \epsilon_{sun} = 1 \)

Answer 10

Yeah, sorry guys for no further information...it was just a plain question...

Answer 11

Makes sense. I always complicate things and sometimes forget that not everyone is taking an upper level heat transfer class. :-)

No problem Unam. We were just discussing the finer points of radiation energy transfer.