The class material briefly covers the Stefan - Boltzmann law. In this context we were discussing the amount of energy emitted by the surface of a hot (blackbody) object. It simply states that the energy an object radiates (per unit area) each second follows this function:
We get more precise now. Let's make sure we understand that unit area is only a small section of the object such as a square meter .... abbreviated m2 (or cm2 or ft2). This, technically, is known as energy flux and is expressed more precisely as:
E = σT4
where σ (sigma) is known as the Stefan-Boltzmann constant which depends on the units chosen.
Usually the units for length is measured in meters, temperature is measured in Kelvin and energy flux is given in Watts so the value of σ = 5.67×10−8 W m−2 K−4 If you are unfamiliar with scientific notation, this value is .0000000567 Note: Energy flux is really a unit of power (not energy) since we are talking about how much energy comes off (each unit area) of the object per second.
Question A - Suppose a blackbody has a temperature of 300 K. What is the energy flux?
Answer: E = σT4 = (5.67 x 10-8) (3004) = 459 Watts
Luminosity means the total energy radiated by a body per second. Luminosity is L= Energy Flux x Surface Area of a body. Luminosity is usually expressed in Watts.
Question B - Imagine a blackbody in the shape of a cube and suspended in space. This cube measures 1 meter on each side. Calculate the luminosity of this cube if it radiates at 6000K.
Answer: The cube has a surface area of 6 m2 (six sides). L = σT4 x area = (5.67 x 10-8) (60004) (6) = 441,000,000 Watts = 4.41 x 108 Watts
Stars behave much like blackbodies so these same laws work quite well.
Question C - Calculate the luminosity of our sun. Hint: You can use 5778 K as the temperature. You will need to determine the surface area of our sun. The formula for surface area of a sphere is 4πr2
... where r is the radius. You will need to look up the radius of our sun in meters.
Answer: The radius of the sun is 696,000,000 meters or 6.96 x 108 m L = σT4 x area = (5.67 x 10-8) (57784) (4)(π)(6.96x108)2 = 3.85 x 1026 Watts
Question D - Suppose a star is 100 time larger than our sun but only 3000K. Compare the luminosity of this star to our sun.
Answer: You can calculate an answer and do a division but let me show you an easier way. Just make a ratio and a lot of the items cancel out.
L (star) / L (sun) = σT(star)4 (area star)/ σT(sun)4 (area sun) = σT(star)4 (4 π r2 star)/ σT(sun)4 (4 π r2 sun) = (T star/T sun)4 (r star/r sun)2 = (3000/5778)4 (100)2 = 727 times the output of the sun