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Module 2, Assignment 3
Describe your best simulations and your interpretations on the Blog. Also demonstrate the Stefan-Boltzmann Law by answering the following: when you double the temperature, by how much does its total energy increase?
Spectrum analysis
ReplyDeleteAt a temperature of 3000K the power was 4.59 X106 W/m2
At a temperature of 6000K the power was 7.35 X107 W/m2
As predicted by Stefan-Boltzman Law the power increases by 16 times (24) when you double the temperature
Spectrum peak at different temperatures
Temperature peak
3000K infrared
4440K red
5800K(Sun) yellow
6000K blue
8000K UV
This table shows that cool stars are predominately red and hot stars are blue.
Fun Simulator – this would take me hours to graph manually.
ReplyDeleteTemperature of 800K gives a peak of greatest intensity of light at 362.2 nm, in the ultraviolet range. Temperature of 7000K gives a peak of greatest intensity of light at 414 nm, in violet visible light.
Temperature of 6530 gives a peak of greatest intensity of light at 443.8 nm, in blue visible light. Temperature of 5800 (that of the Sun) gives a peak of greatest intensity of light at 499.6. nm, in the blue-green area of visible light.
Temperature of 5000 gives a peak of greatest intensity of light at 579.6 nm, in yellow visible light.
Temperature of 4000 gives a peak of greatest intensity of light at 724.4 nm, in infrared light.
So what I have discovered is that as the temperature of the star decreases, the wavelength increases. Hot stars have lower wavelengths (blues) and cooler stars have higher wavelengths (reds) as the most intense frequency of light .
To describe The Stefan-Boltzmann law I used 4000K and 8000K for temperature and used the simulator to calculate area under the curve to find corresponding values of 1.45 x 107 and 2.32 x 108. For an increase of a factor of 16 – or 24 since you double the temperature which is raised to the 4th power in the equation.
I've posted a "carrot" under assignment 2 this week! We are delighted to see all of you posting your answers. If you have problems, you can always email Chris or me...
ReplyDeletekaty
It was fun experimenting on the blackbody simulator. The lowest temperature I set was 3000K , the peak of its wavelength is at 965.9 nm and the area under curve is 4.59 x 10 6 W/m2. The highest temperature the simulator could get is 25000, the peak of wavelength is at 115.9 nm and the area under curve is 2.21 x 10 10 W/m2. I also tried the sun’s surface temperature at 5800K, the peak of wavelength is at 499.6 nm and the area under curve is at 6.42 x 10 7 W/m2. I continue to adjust the simulator from lowest to the highest temperature and I discovered that as the temperature increase the peak of the wavelength in nm decreases while the area under curve increases. It also explains the Stefan-Boltzmann Law that if you double the temperature of an object it will emit 16 times more radiation ( for example 3000K the power has 4.59 x 10 6 W/m2 =4590000 x 16 = 73440000 W/m2; 6000K the power has 7.35 x10 7 = 73500000 W/m2).
ReplyDeleteIt also proves the Wiens Law that cool stars have the peak of long redder wavelengths and as the temperature increases the peak of its continuous spectrum shifts to shorter bluer wavelengths. Hot stars appear predominantly blue while cool stars seem red.
I notice that as the temperature of the star increases, the wavelength of peak emission shortens. This satisfies Wien's Law (The temperature of the blackbody increases, more radiation with wavelengths closer to the blue part of the spectrum will be emitted and I when I tried to decrease the temperature of the slider, the wavelength lengthens and more radiation with wavelength closer to red.
ReplyDeleteThis means that hotter stars will appear bluer and cooler stars will appear redder when we look at them with our eyes.
At T = 6000K, the wavelength is 483nm and the area under the curve is 7.35E7 W/m^2 and at T = 3000K, the wavelength is 965.9 nm and the area under the curve is 4.59E-6W/m^2. Comparing the two temperatures and the area under the curve, at T=6000K the energy is almost16x as great. This satisfies Stefan-Boltzmann law which states that if you double the temperature the energy of that blackbody will be 16 times as great.
Hot stars appear blue because the peak wavelength is shorter, making the flux peak in the bluer region of the electromagnetic spectrum. Decreasing the temp not only lengthens the peak wavelength, but the overall curve indicates less flux. So the blackbody with a longer peak wavelength is in a redder region of the spectrum and will appear red.
ReplyDeleteDoubling the temperature increases the total energy by 16 since L is proportional to T^4.
As mentioned above, with higher T, there is a shorter wavelength and higher flux. There's less area under the curve. I think that's intensity. The sun has a peak wavelength of 497.9, which is in the blue-green section of the visible spectrum. The sun's blackbody curve includes most of the electromagnetic spectrum, which is pretty neat.
ReplyDeleteHotter stars have more intensity at the wavelength corresponding to blue. Cooler starts have a higher intensity at wavelenghts corresponding to red.
When you double the temperature the energy (the area under the curve?) increases by almost 16 times, which demonstrates the Stefan-Boltzman Law.
I would like to read more about this topic. I'm not sure I completely understand. Can anyone suggest some online resources to help me out? A Google search just contributed to my confusion!
The hotter a blackbody the shorter its wavelength of peak emission and the colder the blackbody the longer its wavelength. The wavelength of peak emission is simply the wavelength at which a blackbody emits most of its radiation.
ReplyDeleteHotter stars emit most of their radiation at shorter wavelengths that is why they appear bluer .
Cooler stars emit most of their radiation at longer wavelengths and they appear redder
The Wien Law gives the wavelength of the peak of the radiation distribution, while the Stefan-Boltzmann Law gives the total energy being emitted at all wavelengths by the blackbody (which is the area under the Planck Law curve in the simulator).
The Wien Law explains the shift of the peak to shorter wavelengths as the temperature increases while the Stefan-Boltzmann Law explains the growth in the height of the curve as the temperature increases.
As I move the slider I notice that the change in height is very abrupt, since it varies as the fourth power of the temperature
I studied 3 temperatures ( 3000K, 6000K and 12000K) and I took the ratio of the areas under the curve. For T = 3000 K and 6000K, the ratio is equal to 16 and for T = 6000K and 12000K, the ratio is equal to 16.1.
These results satisfy the Stefan-Boltzmann law
From John Webster
ReplyDeleteThe simulator did a good job of showing the relationship between Temperature and wavelength and the resulting color.
Using Wien’s Law of T= 2.897 x107 K Å/ λ, you can calculate the temperature for red light at 6500 Å at 4457° K. For blue light at 4750 Å, the temperature is 6099 ° K. showing that the hotter stars have a bluish color and the cooler stars are red.
Using the simulator, I got the Sun’s temperature of 5800°K , the wavelength is 499.6 nm and the area under the curve of 6.42 x 107 W/m2. To demonstrate the Stefan-Boltzmann Law, I looked at the temperatures pairs of 4000 and 8000°K, and 5000 and 10000°K. In both cases when you double the temperature, the wavelength is cut in half, and the area under the curve is 16 times greater.
-The sun’s temperature spans the entire visible spectrum—makes sense—and also produces UV and IR radiation.
ReplyDelete-Hotter stars appear blue because at that temperature they are emitting more of that wavelength of light, while cooler stars are emitting more red light (lower energy).
-Stefan-Boltzmann law: I started the temperature at 6000 K and noted the peak wavelength of light—483 nm. Then I doubled the temperature. At 12000 K the peak wavelength is 241.5 nm. Since this wavelength is half of the original, the energy is double the original.
-Since black holes emit X-rays, does this mean they are super hot?
-B = R at about 9640 K. It looks like the lowest temperature that U>B>V>R is about 10100 K.
Busy week, but finally getting to my homework: Since the maximum wavelength is inversely related to the temperature in Kelvins according to Wein’s Law, our sun with a temperature of 5800 K has a peak wavelength of 499.6 nm, which is in the bluish are of the spectrum. Doubling the temperature to 11600 K results in a peak wavelength of 249.8 nm, which is half of our sun’s peak wavelength and well into the ultraviolet region. It is clear that stars with higher temperatures emit more light in the blue-violet regions of the spectrum, while cooler stars emit more light (and a broader curve) toward the red end of the spectrum (to peak in the red region, a star has a temperature of about 4400 K).
ReplyDeleteWhen we double the temperature of the star, the flux changes by 2^4 according to the Stefan-Boltzmann Law, for an increase of 16 (at 5800 K, F = 6.42 x 10^7 W/m^2, at 11600 K, F = 1.03 x 10^9 W/m^2).
The simulator demonstrates Planck law. It gives a distribution of the peaks at a certain wavelength, the peak shifts and the area under the curve that grows rapidly with increasing and decreasing temperature.
ReplyDeleteAll heated objects emit a characteristic spectrum of electromagnetic radiation, and this spectrum is concentrated in higher wavelengths for cooler bodies and shorter wavelength for hotter objects. The cooler a star is, the more radiation it will emit toward the redder part of the spectrum and the hotter the star is, the more radiation it will emit toward the bluer part of the spectrum. This means that hotter stars will appear bluer and cooler stars will appear redder when we look at them with our eyes.
The full spectrum is at the center of the peak of the emitted radiation of the sun (T=5800K). That is why we are seeing a continuous spectrum of the sun (combination of all colors or white light)
I examined three temperatures:5800K, 11600K and 23200K. I noticed that if you double the temperature of the blackbody, its energy (area under the curve) increases 16 times. This is a direct result of Stefan-Boltzmann Law
The “blackbody simulator” is a neat tool to illustrate and really get a feel visually for how the wavelength of the light changes when the temperature changes.
ReplyDeleteIt was apparent that the hotter the star, the lower the wavelength (on the ultra-violet side of the spectrum). The cooler the star, the longer the wavelength (on the infrared side of the spectrum). Inputting the Sun's temperature (5800 K), I got a wavelength of 499.6 nm and the simulator computer the area under the curve as 6.42 x 107 W/m2.
In summary, hotter stars are on the violet side of the spectrum and therefore appear blue and cooler stars are on the red side of the spectrum and therefore appear red.
When I use the Stefan-Boltzmann law and double the temperature from an initial value (T), I find that the the total energy (which is proportional to T4) increases by (2T)4 = 16T4 or 16 times the initial value.
The energy of the blackbody (area under the curve)increases and the peak of the curve moves to shorter wavelengths with increasing temperature and the energy decreases and the curve moves to larger wavelength with decreasing temperature.
ReplyDeleteThis explain why stars have different colors:
Cool stars emit most of their energy in the red and infrared region of the electromagnetic spectrum and they appear red.
Hot stars emit mostly at blue and ultra-violet wavelengths, making them appear blue.
I noticed that the area of the radiation curve increases 16 times when I double the temperature from 4000K to 8000K and from 8000K to 16000K. These results satisfy Stefan-Boltzman Law
Josie, you are correct.. It isn't so much that the black hole is hot (we don't see any energy/radiation from it), but as gas swirls into the black hole it is heated to extraordinarily high temperatures, and this is what we detect.
ReplyDeleteTest of Wein's law: The peak does indeed move to different wavelengths at different temperatures. At a temperatue of 12,900 K the peak is at approximately 224.6 nm. At 3930 K the peak is located at 737.3 nm and when the Sun's surface temperature of 5,810 K is inputted the peak is at 498.8 nm. All of these answers are within a couple of nanometers of what Wein's law predicts, so I'd say that's a reasonably good approximation.
ReplyDeleteSince the wavelength decreases (becomes more blueish) as the temperature increases it makes sense to me that hot stars would be more blue and cold stars would be more red.
The Stephan-Boltzmann law also appears to be obeyed. Changing the temperature has a dramatic influence on the total power. Interestingly as stars get hotter and hotter their maximum gets narrower and narrower. Though as I say that I don't want to imply that they are emitting less over any one particular wavelength, the peak is narrowing but the total flux is always increasing. When I put more than one curve on the graph the curve at a lower temperature is always completely covered by the curve at the higher temperature.
Using two temperatures where one was double the temperature of another did produce a difference in maximum peak power of about 16x.