skip to main |
skip to sidebar
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?
I tried 8200K, 6800K, 6200K, 5800K and 4100K.
ReplyDeleteAt 8200K,wavelength is 353.4nm.
At 6800K, the wavelength is 426.1nm.
At 6200K, the wavelength is 467.4nm.
At 5800K, the wavelength is 499.6nm.
At 4100K, the wavelength is 706.8nm.
As the temperature decreases, wavelength increases. At higher temperatures, the peak has a wavelength which gives violet. At lower temperatures it is becoming red.
At 8200K, flux is 2.56*10^8 Watts per unit area. At 4100K, flux is 1.6*10^7 watts per unit area. When we double temperature, wavelength decreases to half, and flux increases by 16.
Division of 2.56*10^8 by 1.6*10^7 gives 16. Which means that energy per unit time per unit area is multiplied by 16.
At 6400, the peak wavelength is 452.8 nm. At 5800 K, the peak wavelength is 499.6 nm. At 4600 K, the peak wavelength is 629.9 nm. As temperature increases, the peak wavelength decreases. Also, as the temperature increases, the intensity of every wavelength emitted also increases. This means that as an object warms, the peak frequency shifts into the lower wavelength part of the spectrum, but that it also starts glowing more brightly overall. Since the energy emitted is proportional to the 4th power of temperature, doubling the temperature would result in 16 times the energy being emitted. When I highlighted the area of the curve, I divided the area of the curve of 6000 K by the area under the curve of 3000 K, and got 16.01.
ReplyDeleteI used the simulator to check the effect of doubling the temperature. For T=6000K, the area under the curve is 7.35E7 W/m^2 and for T= 12000K the area is 1.18E9 W/m^2. Taking the ratio of these two area, I got a ratio of 16. This is true for Stefan-Boltzmann law which states that the energy is directly proportional to the temperature raised to the 4th power. The energy will be 16x greater if you double the temperature
ReplyDeleteAs I move the slider, I notice that the higher the temperature, the shorter is the wavelength and the greater is the area under the curve (energy). And the lower the temperature, the higher is the wavelength and the lesser is the area under the curve.
The peak shifts to the right at a higher temperature (towards blue) and shifts to the left at a lower temperature (towards red) A hotter blackbody will appear to be bluer and a cooler blackbody will appear redder.
An increased in temperature will result in a decrease in the peak wavelength and more radiation closer to the blue spectrum will be emitted. On the other hand, a decreased in temperature will result in an increase in the peak wavelength and lesser radiation closer to the red spectrum will be emitted. This is actually similar to the statement of Wien’s law
ReplyDeleteA cool star will appear red and a hot star will appear blue to our naked eye.
Using the simulator, I noticed that if you double the temperature the energy will be 16x as great. For T= 20000K, the area under the curve is 9.07 x 10^9 W/m^2 and for T= 10000K , the area under the curve is 5.67 x 10^8 W/m^2. Taking the ratio of the areas, I got a factor of 16. This result satisfies the Stefan-Boltzmann law which states that the energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature (2 raised to 4 is 16)
Using the simulator, I observed that as the temperature increases, the peak wavelength decreases. A decreased in temperature results in an increase in peak wavelength. This is a direct result of Wien's Law which states that as the temperature of the blackbody increases, more radiation with wavelengths closer to the blue part of the spectrum will be emitted and a lower temperature will have wavelength closer to the red part of the spectrum. This is why the cooler a star will appear redder and hotter star will appear bluer.
ReplyDeleteUsing the idea about the Stefan-Boltzmann law, if you double the temperature the energy of that blackbody will be 16 times as great. The simulator also demonstrate Stefan-Boltzmann. At T= 25000K the area under the curve is 2.21E10 W/m^2 and T= 12500K , the area under the curve is 1.38E9 W/m^2. Comparing these two areas, I got a factor of 16.014. This means that a 25000K blackbody will have an energy 16 times as great than the blackbody with a temperature of 12500K.
The peak energy for the sun was at 499.6 nm which surprises me as I expected this value to fall within the yellow area of the color spectrum. However, looking at the distribution of the area under the curve there is more energy within the yellow range than the blue. Highest energy just above 2.5 x 10^13.
ReplyDeleteWhen the temperature of the star doubles that of the sun (11600 K) the peak shifts to the shorter wavelength’s past the blue end of the spectrum. The peak has shifted to 249.8 nm (half that of the sun’s peak). This fits with the expectation that a hotter star will be blue in color. Highest energy at 8.5 x 10^14
When I divide 8.5 x 10^14 by 2.5 x 10^13 I get an answer of 34. If my reasoning is correct that means that the star twice as hot will have 34 times the energy of the cooler star. This answer surprises me… However, in looking at the plot of the curve for both stars on the same graph it is very obvious that the hotter star has a significant greater area under the curve than the cooler star, which supports my answer.
The next change I made was to triple the sun’s temperature (16,500K). In this case the curve became much more sloped with the peak at 175.6 nm and peak energy at 5.0 x 10^15. The change in energy in this example is 200 times greater than that of the sun.
The peak wavelength varies at different temperatures. At a high temperatue the peak moves toward the blue part of the spectrum and a low temperature the peak moves toward the red spectrum.
ReplyDeletefrom these observations, I may say that blue stars are hotter than red stars (hot stars would be more blue and cold stars would be more red from the naked eye)
When I double the temperature, it gives me an energy which is 16 times as great. Using the simulator, the area under the curve which is the energy of the blackbody changes a factor of 16 as I double or reduce the temperature into half. The simulator demonstrates the Stephan-Boltzmann law which states that the energy of a blackbody is directly proportional to the temperature raised to the 4th power. When you double the temperature, the energy will be 16x as great, if you tripled the temperature, the energy will be 81x as great
As I moved the slider, the peak wavelength shifted toward the blue part of the spectrum when T was maximum and it moved toward the red part of the spectrum when T was minimum. I could say that the higher the temperature, the greater is the area under the curve, the greater the intensity and the shorter is the wavelength. And the lower temperature, the smaller is the area under the curve and greater is the wavelength.
ReplyDeleteThe simulator demonstrates the Stefan Boltzmann law. I used the following pairs of temperature to check the validity of the simulator with regards to the Stefan-Boltzmann law. For T= 3000K and 6000K, I noted the areas under the curve which are actually the energies of the blackbody for the two temperatures. Then I took the ratio of the areas and I got a ratio of 16. Then I compared the energies at T=3000K and T=9000K. The ratio of the areas was 81. Lastly, I compared the area at T=3000K and T=12000K. Then I got a ratio of 256. These were true for any blackbody. The energy of the blackbody is proportional to T^4
Temperature of 5000K has a peak wavelength of 579.6 nm and are undercurve of 3.54E7 W/m^2
ReplyDeleteTemperature of 10000K has a peak wavelength of 289.8 nm and are undercurve of 5.67E8 W/m^2
Temperature of 20000K has a peak wavelength of 144.9 nm and are undercurve of 9.07E9 W/m^2
Comparing the area of the curves of T= 10,000K and 5,000K. I got a ratio of 16 Then I compared the ratio of the area undr curves of T= 20,000 K and T= 10000 K. I got the same ratio of 16. If you double the temperature the energy will be 16 times more. This is a direct result of stefan-Boltzmann law. This law relates the proportionality of energy to the T^4 of the blackbody.
As I moved the slider, I noticed that the spectrum moves towards the blue part of the spectrum when I set the temperature from lower to higher temperature. The peak wavelength decreases and the area under the curve incrases. When I do the opposite, the peak wavelength moves to the right towards the red part of the spectrum, the area under curve decreases and the wavelength increases.
The sun produce a white light because its peak is at the center of the spectrum. Hotter star looks bluer and cooler star looks redder in our naked eye
Using the Blackbody Curves and Filters Explorer, we see that:
ReplyDeleteAt a wavelength of 400 nm the flux is 1.56 x 10^8 W/m^2
At a wavelength of 800 nm the flux is 9.74 x 10^6 W/m^2
If we combine S-B and Wein’s laws, we note that the longer the wavelength the lower the power emitted, or flux. In fact F = ( 2.89 × 10-3 K m /)^(1/4).
Doubling the temperature results in sixteen times the power emitted.
Defining the flux of a blackbody, F1 at temperature t and the flux of another blackbody, F2 at twice the temperature, 2t. Applying the Stefan-Boltzmann law, then:
F1 = t)^4 and F2 = t)^4, then F2/F1 =t)^4/t)^4 = 16t^4/t^4 = 16.
So a blackbody that is twice the temperature is emitting energy at a rate sixteen times that of the cooler blackbody. Using the Blackbody Curves and Filters Explorer we will look at an object with t = 3000 K and one that is 2t = 6000 K. The cooler object is then emitting F1 = 4.59 x 10^6 W/m^2 and the hotter is emitting F2 = 7.35 x10^7 W/m^2. Then:
F2/F1 = 7.35 x10^7 W/m^2/4.59 x 10^6 W/m2 = 16.01
The following are data I collected doing the simulations:
ReplyDeleteTemp. SED shape Wavelength Intensity
3000 wide no peak 965 nm 1.1 x 10**12
6000 less wide 483 nm 3 x 10**13
9000 less wide 321 nm 2.5 x 10**14
12000 less wide 241 nm 1 x 10**15
Conclusion: Increasing temperature results in a change in the shape of the SED, peak wavelength decreases and intensity increses.
Item 2 When the temperature is doubled the flux increases by a factor of 16.
F = b (T)**4
F = b (2T) **4
F = b 16 T**4
Item 3 At 9470 K the r-filter and b-filter peak at the same height.
Item 4 At 10,300 K and greater the peaks are U>B>V>R.
Item 5 The star B is brighter in the red-filter images. Star A is brighter in the blue-filter image.
I used two sets of temperatures: 3000K and 6000K, 4430K and 8860K
ReplyDeleteFor T= 3000K and 6000 I found the flux to be 4.59 x 10^6 W/m^2 and 7.35 x 10^7 W/m^2 respectively. Comparing these two areas under the curve (larger/smaller), I got a ratio of 16.013
For T=4430K and 8860K I found the flux to be 2.18x10^7 W/m^2 and 3.49x10^8 W/m^2. Comparing these two values, I got a ratio of 16.009.
This is a direct numerical proof of Stefan-Boltzmann law. The energy is directly proportional to the T^4. In case of T=3000K and T=6000K, taking the ratio of 6000K^4 to the 3000K^4. I got a ratio of 16….so doubling the temperature will give us an energy of 16x as great
As I moved the slider, the peak shifted to the right for a lower temperatures and it shifted to the left for a higher temperatures. The higher the temperature, the lower is the peak wavelength and the lower the temperature, the higher is the peak wavelength.
There is an inverse relationship between temperature and and energy.
ReplyDelete3000K = 956.9nm
6000K = 483.0nm
12000K = 241.5nm
As I double the temperature (K) the energy (frequency of wavelengths) is halved.
The temperature of our sun, 5800K emits energy at a wavelength of 499.6nm which falls in the blue spectrum.
According to this observation, energy sources that have shorter wavelengths (450-500) fall in the blue spectrum and are burning at hotter temperatures. Energy sources that have longer wavelengths (600-650) fall in the red spectrum and are burning at "cooler" temperatures.