How do you determine the temperature of radiation?
80719
You can however determine the temperature of the body that emits the radiation. The higher the energy of the photon ,gamma rays being the highest, the higher the temperature of the body. This is how we determine the temperature of stars.
278956
This has been asked and answered many times before:\n\nhttp://answers.yahoo.com/search/search_result;_ylt=Amylv82zdJRUVkmYew.javjpy6IX?p=temperature+of+the+sun
307190
temperature itself is a property of matter. so, in space, where there is a near perfect vacuum, there is no temperature.\n\ntemperature of stars / dark matter in space can be measured by studying the radiation that is emitted / absorbed by it.
535160
what do you want to calculate? gravitation? emission? something else?
734519
(1) Take a container and measure the length of the container. Let us say the length of the container is L. \n\n(2) Take a beam of light (preferably a monochromatic source like laser beam) and shine it through the empty container. \n\n(3) Measure the intensity of the light at the end where the beam has gone through the empty container. Let this be Io. \n\n(4) Now fill the container with the liquid whose molar extinction coefficient is unknown. Now measure the light intensity as the beam travels the length L of the container. The measured intensity is I.\n\n(5) Beer-Lamberts law is I=Io exp(-a L) where all the quantities have been experimentally determined except a which is the extinction coefficient. \nRearrange the above formula and we get, a=(-1/L)*log(I/Io) where log is to the base e.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
171995
3K background radiation measurement instruction:\n\n\n(1) take a full spectrum snapshot of the sky sphere.\n\n(2) substract all known active radiation sources (stars)\n\n\n(3) take the resulting frequency spectrum I(n) and\n\n(3.1) assume that you know allmost each existing active radiation source\n\n(3.2) assume that the universe doesn't radiate outwards and nothing radiates into it\n\n(3.3) then you can speak of a radiation field equilibrium and are allowed to apply 'blackbody radiation' theory\n\nhttp://en.wikipedia.org/wiki/Black_body\n\n\n(4) apply 'blackbody radiation' theory to the frequency spectrum: I(n) of step (3)\n\n(4.1) fit Planck's function of blackbody radiation: I(T,n) by variation of the remaining variable T\n\n\n(5) if you get a proper fit, call the temperature T, the temperature of the universe.\n\n\nuntil now, all such measurements led to T = 3 Kelvin.\nBut this doesn't mean that empty space would have a temperature itself.
322590
You do not possibly have enough information to answer this question properly, you need to know, where it is at, how far from the sun, like the others said, is there a planet or any obstacle between it and the sun, what is it's current temp, so on and so on....here is something about tempature in space though\n\nEmpty space itself cannot have a temperature, unless you mean some abstruse question about quantum vacuums.\n\nHowever, if you put a physical object into space, it will reach a temperature that depends on how efficiently it absorbs and emits radiation and on what heating sources are nearby.  For example, an object that both absorbs and emits perfectly, put at the Earth's distance from the Sun, will reach a temperature of about 280 K or 7 C. If it is shielded from the Sun but exposed to interplanetary and interstellar radiation, it reaches about 5 K.  If it were far from all stars and galaxies, it would come into librium with the microwave background at about 2.7 K. Spacecraft (and spacewalking astronauts) often run a bit hotter than 280 K because they generate internal energy.
487729
http://www.wisdomportal.com/Temperature.html\n\nThat website could help you out....
526508
It depends on what part of the sun.\n\nThe core where fusion occurs is about 15 million Kelvin while the darkest sun spots are only several thousand Kelvin.\n\nThe photosphere (layer in which the optical depth is 1 which makes it the part of the sun we see when we look at it) is 5800 K with the parts of the sun below it being hotter and the Chromosphere above cooling at first and then reaching about 7000 K before the Corona which reaches temperatures usually about 1 million Kelvin but going up to 3 million Kelvin.\n\nSunpots are a couple of thousand Kelvin cooler than the surrounding photosphere.\n\nMeasuring the temperature of the Sun is usually done by looking at the spectrum of the sun and seeing where the peak is.  Since the sun is approximately a black body by looking at where the peak is you can determine how hot it is.\n\nThe temperatures underneath the Photosphere have to be calculated from what we know about the sun since we can't directly measure them.
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
278728
If it is for a science project, web sites may or nmay not have the answers....that is all part of a project, to figure it out for yourself, develop your own theories, hypothesis, etc.\n\nI will say it all depends on the temperature....beyond that you're on your own.
163823
Q=mc(Tf- Ti)\nWhere Q is the heat added\nm is the mass\nc is the specific heat capacity\nTf is the final temperature\nTi is the initial temperature
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
424227
take 32 degrees off the F, then divide by 1.8.  So for example lets say you have a F temp of 50.  Take 32 off that, which gives you 18.  Divide by 1.8, which gives you 10.  So 50 F is the same as 10 C.
893197
Try a histogram and use a range of temperatures.
100101
Use a calculator, add all three together (be sure to include the negatives) and then multiply by the number of temperatures.