Calculating Star Surface Emission at Different Temperatures | Physics Question

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The discussion revolves around calculating the power emitted by a star's surface as its temperature decreases from 1000 Kelvin to 900 Kelvin, using the Stefan-Boltzmann Law. The original poster is confused about applying the law and seeks guidance on how to proceed with the calculations. Forum members emphasize the importance of demonstrating effort and understanding rather than simply asking for answers. The conversation also notes that multiple postings by the same user have been merged for clarity. Overall, the focus remains on understanding the application of the Stefan-Boltzmann Law in this context.
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If a star has a surface temperature of 1000 degrees Kelvin and emits 500,000 watts of each square-meter of surface area, how much power does each square-meter of its surface area emit after it has cooled to 900 degrees Kelvin?
 
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Welcome to the Forums,

Have you any thoughts on the question? Which law do you think we should use?
 
If a star has a surface temperature of 1000 degrees Kelvin and emits 500,000 watts of each square-meter of surface area, how much power does each square-meter of its surface area emit after it has cooled to 900 degrees Kelvin?
 
Stephen-Boltzmann's Law i guess, but i couldn't figure out how? please help if u have any idea
 
Well you could start by writing down the Stephen-Boltzmann law and identifying what all the symbols mean...
 
Double posting isn't going to get your thread answered any quicker, I've already answered your other thread.
 
Thanks, am a kind of confused.
 
it may be easy for u to answer, so please help me?
 
Yarka said:
it may be easy for u to answer, so please help me?
I am trying to help you. The guidelines here at PF prevent us from providing solutions to problems, they also require that you show your efforts and that we guide you through the problem rather than just handing you the solution on a plate.
 
  • #10
ok. this is what i have= v/t=pi*P*D^4/128*L*Visc.

so 500,000=3.14*100*1000/128*L
Am i right?
 
  • #11
Two different threads have been merged because the OP made multiple postings.

Zz.
 

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