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Emissivity & internal blackbody

  1. Feb 23, 2015 #1
    Please confirm my understanding and inquiry..

    in material without transmission... 100% = Emissivity + Reflectivity...
    when using thermal camera (I own one).. we have to adjust the emissivity

    In two objects that is both 30 Celsius.. one a near black object with emissivity nearly 1 (black body) and another a reflective aluminum. We know the aluminum reflects more hence the thermal camera would see less heat if emissivity not adjusted in the camera.

    my inquiry. even if the reflective aluminum has less emissivity.. it is both 30 Celsius as a near black object beside it. Does it mean the molecules vibration inside both are hiding same black body spectrum? with the reflective object not emitting it outside? I mean.. I know the atoms vibrate at IR range and it is producing the frequency producing the heat.. so one atom vibrates at certain frequency producing certain photons.. and different atoms vibrating at different frequency producing certain photons? Or do the atoms vibrate and redistribute the frequencies via normal modes? If so how come reflective aluminum has less emissivity.. meaning it radiates less frequencies? Are the frequencies just hidden inside the atoms due to the fact that reflective object reflects the IR from the surrounding. I mention hidden because the reflective object and black object are both 30 Celsius even if one has less emissivity (in other words, what goes on internally in the molecules of objects.. one with less emissivity versus one with more.. yet both having same temperature.. why are they still same temperature if the IR radiation is responsible for the temperation.. unless the conduction, convection can still produce the black body spectrum that hidden inside both materials explaining why they are same temperature?).. are my questions clear.. i'll rephrase if no one still understands what i'm asking or saying).. thank you..
     
  2. jcsd
  3. Feb 24, 2015 #2
    You may be discussing a grey body, which has an emissivity <1, for all wavelengths, or a real body where emissivity depends upon the wavelength. A black body has emissivity = 1 for all wavelengths.

    If both of your bodies, ie the black body and the grey body, are at a temperature of 30 Celsius, the grey body will output less radiation than the black body.
     
  4. Feb 24, 2015 #3
    But what is going on inside the molecules of both the black body and grey body.. would they have similar internal black body making them produce same temperature. If not. What can create similar temperature yet the frequency spectrums inside the black and grey body are dissimilar?
     
  5. Feb 24, 2015 #4
  6. Feb 24, 2015 #5
    It is mentioned in the above that "By definition, a black body has scattering matrix [PLAIN]http://sciencewise.info/media/libs/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/120/0053.png[PLAIN]http://sciencewise.info/media/libs/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/120/003D.png[PLAIN]http://sciencewise.info/media/libs/MathJax/fonts/HTML-CSS/TeX/png/Main/Regular/120/0030.png [Broken] , because all incident radiation is absorbed. If the absorption is not strong enough, some radiation will be transmitted or reflected and http://sciencewise.info/media/libs/MathJax/fonts/HTML-CSS/TeX/png/Math/Italic/120/0053.png will differ from zero. Such a ``grey body'' can still be in thermal equilibrium, but the statistics of the photons which its emits will differ from the negative-binomial distribution"...

    But why is both grey body and black body still have both 30 degree Celsius when the grey body ims missing some radiation? Maybe other part of the radiation compensate with stronger strength or something? Also please confirm if the vibrating molecules distribute the frequency in the normal modes or whether individual part of the molecules vibrate at different frequencies to produce the spectrum.
     
    Last edited by a moderator: May 7, 2017
  7. Feb 24, 2015 #6
    I have to bug out for a quite a while, so someone should help you out more, we hope.
     
  8. Feb 24, 2015 #7

    Bystander

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    Rate of radiation does NOT define temperature. Kinetic energy of molecules if we're dealing with matter, or the spectrum of radiation in a cavity if we're not, defines temperature.
    No.
     
  9. Feb 24, 2015 #8

    DrDu

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    Take a piece of glas. It is certainly not a blackbody. Nevertheless it can be in thermal equilibrium with a blackbody of the same temperature as it both absorbs less of the energy emitted by the blackbody and emits itself less radiation.
     
  10. Feb 24, 2015 #9
    But in a thermal imager.. it detects IR frequencies that corresponds to certain temperature. Also note that as object gets hotter, it begins to emit visible spectrum like red.. so the kinetic energy is connected to the frequencies.. maybe faster molecular vibration can produce higher frequencies that shift the black body spectrum to smaller wavelength?
     
  11. Feb 24, 2015 #10

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    The peak of the Planck radiation intensity for a black body occurs at a specific wave length for a specific temperature. For gray bodies, the radiation intensity as a function of wave length does not match Planck's radiation law, and IR thermometers, and other radiant temperature methods approximate temperature by looking at total radiation in some IR bandwidth, and applying a "fudge factor" in the form of an emissivity for the source for which temperature measurement is desired, or by fitting intensities of several wavelengths of radiation to Planck's law.

    In the sense of the Planck radiation law,
    yes.
     
  12. Feb 24, 2015 #11
    A mercury thermometer can't look into some IR bandwidth.. so since there is less radiation in gray body (bec of missing wavelength) than black body, then a mercury thermometer would have less reading for gray body? Or maybe there is more energy in other bandwidth of the gray body to compensate (producing same mercury thermometer reading)?
     
  13. Feb 24, 2015 #12

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    Conventional thermometers operate by applying the so-called "Zeroth Law of Thermodynamics." If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system C if the two are brought into contact; system B can be called a thermometer. It is in actual contact with the other two systems, rather than loosely coupled to them through radiation exchange. A mercury thermometer will read the same temperature for both bodies --- if they are both at the same temperature, by being immersed in a thermostat, or being in contact with one another.
     
  14. Feb 24, 2015 #13

    DrDu

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    Temperature is only defined for the whole spectral range. A thermometer in equilibrium with the full thermal radiation of a grey body will show its temperature. Equilibrium means here that it receives radiation from all sides, too. However, if you filter out a certain frequency range, it won't have a definite temperature any more.
     
  15. Feb 24, 2015 #14
    does the object temperature reach the thermometer in equilibrium by conduction, convection or ir radiation?
     
  16. Feb 24, 2015 #15

    DrDu

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    That s up to your experimental setting. If you bring the thermometer inside an evacuated cavity of given temperature, then only radiation will be relevant.
     
  17. Feb 24, 2015 #16
    How come I couldn't find any Gray Body Radiation Spectrum graph at goggle or anywhere? I'd like to compare it to a Black Body Radiation Spectrum to see what frequencies are missing in the Gray Body let's say for an emissivity of 0.5 and a temperature of 85 Celsius. Thanks.
     
  18. Feb 24, 2015 #17

    DrDu

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    There are tons of spectral emissivity curves outside in the net. The corresponding greybody spectrum can then be obtained by multiplying the spectrum of a blackbody of the same temperature with these spectral emissivity values for each frequency.
     
  19. Feb 24, 2015 #18
    Thanks for the words to search, so its "spectral emissivity" rather than "gray body spectrum". Well in reflective aluminum, it is logical that certain frequencies are reflected bec the surface are reflective, but how come the molecules spectral em cant produce those same frequencies from the molecular vibration and convertion to em waves itself?
     
  20. Feb 25, 2015 #19
    I'm looking for what material features suppress and make it hard to emit all the em waves produced by the molecular vibrations. When you read explanation of black body and even grey body, you would hear how black body has all radiation absorbed and emitted. But when you have certain fuses made of certain material in the circuit breaker and it gets hot, you won't get the right temperature if you don't adjust the emissivity setting in the thermal scanner. But the fuse gets hot by conduction or direct initiation of heat from the electrical current. Note the heat did'nt get to the fuse from the environment by ir radiation! So what's the connection with the explanation of black body absorbing all IR and radiating it. In the fuse, it heats up by resistive or conductive. So when the molecules vibrating fast and em waves are produced by normal modes distributed in equilibrium theorem following planck curve, how come certain frequency got suppressed and can't radiate and this coincides with the frequency that is reflected off from the surface. And also in a shiny metal, what particular frequency are reflected and why those specific frequency? Anyone?
     
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