Calculate the Average Surface Temperature of Earth

[lasisdabomb]

Homework Statement

The Earth receives on average about 390 W m−2 of radiant thermal energy from the Sun, averaged over the whole of the Earth. It radiates an equal amount back into space, maintaining a thermal equilibrium that keeps the average temperature on Earth the same. Assuming the Earth is a perfect emitter of radiant energy (e = 1), estimate the average surface temperature of the Earth in °C

Homework Equations

P = eσAT^4
P ∝ T^4
λmaxT = 2.898 × 10−3 m K
3. Attempt
I've tried all these formulas, but I'm not getting anywhere. Am I supposed to know the surface area of the Earth because I feel like it's impossible without it

 

[Qwertywerty]

Assume the area of the Earth to be $A$, a variable. What, then, is the total power incident on the Earth's surface?

 

[lasisdabomb]

Assume the area of the Earth to be $A$, a variable. What, then, is the total power incident on the Earth's surface?

Is the 390 m^-2 relevant for the equation. Should that be stated as the P value or should it just be 390W

 

[Qwertywerty]

Do you know what intensity is? It's formula? It's unit?

 

[lasisdabomb]

Do you know what intensity is? It's formula? It's unit?

I've realized how to do it. The Power is average for each m^2 of earth. That means the surface area should be 1m^2 and the Power should be 390.
By subbing everything in, you get an average surface temperature of 288 Kelvin or 15°C

 

[Qwertywerty]

That means the surface area should be 1m^2 and the Power should be 390.

More appropriate wording would be - an average of 390W of power is incident per sq.m on the surface of the earth. This is what you mean, right?

And congratulations, on having solved the problem

 

[haruspex]

More appropriate wording would be - an average of 390W of power is incident per sq.m on the surface of the earth. This is what you mean, right?

And congratulations, on having solved the problem

Yes, that's the right answer with the given information, but 390 W/m² is too high. Should be more like 340.
Taking the 30% albedo into account as well would bring the temperature down to 255K, which is the standard result.

 

[Qwertywerty]

Yes, that's the right answer with the given information, but 390 W/m² is too high. Should be more like 340.
Taking the 30% albedo into account as well would bring the temperature down to 255K, which is the standard result.

Thanks for the info!