Rate at which Sun converts mass to radiation

[BRN]

Hi folk, i need an help for this esercise.

1. Homework Statement
Determines the rest mass lost by the sun every second for emission of radiation. It assumes that the surface temperature of the sun is 5700 K and the diameter $D_s=1.4 × 10^9 m$.

The Attempt at a Solution

The solar nuclear reactions generates helium fron Hydrogen. Part of the mass involved in the reaction is lost by eletromagnetic radiation emission. Energy is emitted in the form of gamma fotons and fast neutrinos.
The amount of energy at all wavelengths that at each second strikes perpendicularly a m² of a surface exposed to solar radiation, takes the name of the solar constant and his value is $C_s=1.36*10^{3}[W/m^2]$

If I consider the sun as a sphere of radius $D_s/2$, its surface is:
$S=4\pi(\frac{D_s}{2})^2=6.1575*10^{18}[m^2]$

The total energy emitted fron the sun in all directions is:
$E=C_s*S=8.3742*10^21[W/s]=8.3742*10^21[J]$

Then, I determine the rest mass lost by Einstein reletion:
$E=mc^2 \Rightarrow m=\frac{E}{c^2}=9.3177*10^{4}[kg]$

but the solution is $4.10×10^9 [kg/s]$

Someone could help me? Thanks!

EDIT: The post title is wrong. How can I change it?

 

[Carlos PdL SdT]

I think you used the solar constant on Earth when you should have been using the one on the surface of the sun which is way higher. I would suggest you work with the 5700K and work out the Energy from there

 

[gneill]

EDIT: The post title is wrong. How can I change it?

What would you like it changed to?

 

[vela]

What would you like it changed to?

I changed it yesterday after I moved the thread; it used to be about the "cohesive energy of argon."

 

[Cutter Ketch]

They gave you the temperature expecting you to treat the sun as a blackbody. You have probably been taught a formula for the radiance of a blackbody.

Looking up the solar constant probably wasn't the expected approach, but you can get there that way. The solar constant is the radiance at 1 AU. If you know how the radiation falls off with distance (and I bet you do) then it is simple to scale that value back to the surface of the sun.

 

[BRN]

Sorry for the delay, but were difficult days...

Yes, I verified, the solar radiance constant is referred at 1AU radius, and using this data instead of the radius of the sun all adds up.
If, however, consider the sun like a blackbody, the irradiance is:
$R=\frac{c}{4}u(\epsilon ,T)$

with $u(\epsilon ,T)$ states density.

But the density of states as it is calculated?

If, I use the balckbody states density:
$u=\frac{1}{V}g_sg_{ph}(\epsilon)[n_E]_B\epsilon=\frac{1}{\pi^2\hbar^3c^3}\frac{\epsilon^3}{e^{\frac{\epsilon}{k_BT}}-1}$

with $g_{ph}(\epsilon)=\frac{V}{\pi^2\hbar^3c^3}\epsilon^2$

I get $R(\epsilon, T)=0$...

 

[mfb]

How do you get this?

There is an easier formula for the total electromagnetic emission. You don't have to integrate anything that way.

 

[BRN]

I simply reported blackbody equations that I have on my book.

My reference book is the "Manini - introduction to the physics of matter" and I admit that it is the worst book in which to investigate these things ...

What formula are you speaking?

 

[mfb]

The Stefan–Boltzmann law is sufficient here.

 

[BRN]

I don't have Stefan - Boltzmann law in my book...

At any way, if i use this law, I have:

$q=\sigma T^4=5.9855*10^7 [W/m^2]$

with Stefan - Boltzmann costant $\sigma=5.6703*10^{-8}[W/m^2K^4]$

I'm confused...

In the exercise test, there is the sun diameter too. What is this good for?

 

[gneill]

In the exercise test, there is the sun diameter too. What is this good for?

Looking at the units of a given constant can be insightful.

Note the m² in the denominator of the units for σ. The implication is that it involves a measure of area. So the constant relates the energy emitted in Watts from a surface of some area at a given temperature... Now, how might you determine the area of the surface in question?

 

[BRN]

Ok, the Stefan - Boltzmann law is referred at $1 m^2$. With the units I have no problems, but with the values ...

From Stefan - Boltzmann law I get the emittance for $m^2$ unity:
$q=\sigma T^4=5.9855*10^7 [W/m^2]$

The Sun surface is:
$S=4\pi(\frac{D_s}{2})^2=6.1575*10^{18}[m^2]$

then,

$E=q*S=3.6855*10^{26}[W]=3.6855*10^{26}[J/s]$

From Einstein equations I have:
$E=mc^2 \Rightarrow m=\frac{E}{c^2}=1.2293*10^{18}[kg/s]$

I miss a pass, maybe from q to E.

 

[gneill]

I miss a pass, maybe from q to E.

Try from E to m. You didn't square c.

 

[BRN]

OH DAMN! When I make these mistakes would kill me!

Now it's all ok!

Do you recommend a good book for studying the physics of matter? (not too difficult )

 

[gneill]

Do you recommend a good book for studying the physics of matter? (not too difficult )

I don't have any particular recommendations myself, but you can ask in the Science and Math Textbooks forum.