Astronomy Question: Est. Sun's Power & Temp

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The forum discussion addresses a two-part astronomy homework question regarding the Sun's energy generation rate and surface temperature. The user initially calculated the Sun's luminosity as 2.0 x 1014 watts using a ratio based on a 100-watt light bulb, which is significantly lower than the actual value. For the second part, the user derived the Sun's surface area and attempted to calculate its temperature using the Stefan-Boltzmann law, resulting in an incorrect temperature of 5.1 K instead of the expected 5000 K. The user ultimately resolved their confusion and requested to disregard their initial inquiry.

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[SOLVED] Astronomy Question

Homework Statement


This is a two part question. I believe the problem I'm running into is coming from the answer to the first question.
Q1: Hold your hand 7.5 cm from a 100 watt light bulb and you will feel the same heat sensation that you feel when your hand is pointed towards the Sun on a nice clear day. Assuming that your hand is a perfect total energy flux sensor and knowing that the sun is 150,000,000 km away, estimate the Sun's energy generation rate in watts.

Q2: Knowing that the Sun has an angular diameter of 0.5 degrees in the sky, calculate the surface temperature that is required for it to radiate the power that you determined in question #1. For this problem assume that the Sun is a perfect black body that it radiates power per unit area from all of it's surface at a rate of σT^4 (where σ is the Stephan Boltzmann constant.


Homework Equations





The Attempt at a Solution



For the first question, i figured the watts of the sun by setting up the ratio 100 watts/7.5 x 10^-5 km = x watts/1.5 x 10^8 km
solving for x = 2.0 x 10^14 (which is way less than the sun's actually luminosity, why I'm figuring this is incorrect)

So for the second question i used the angular diameter, and the distance from the Earth to the Sun, to find the diameter of the sun tan(0.5/2)2(1.496 x 10^11)=1.306 x 10^9 m
Then use this to find the surface area using 4πr^2, and the answer I got was 5.35 x 10^18.
From the problem the equation i derived was Power/area = σT^4, solving for T, T = (Power/area*σ)^(1/4).
((2.0 x 10^14)/(5.35 x 10^18)(5.67 x 10^-8))^(1/4)=5.1 K
Should be over 5000K.

Please help because I can't see my error in what I'm doing :(

 
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I figured this out. Please disregard.

Thank you :)
 

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