# Emissivity & Thermal Transmission

1. Jul 20, 2011

### riezer

Stainless steel has emissivity of 0.12. It means it only absorbs 12% of radiation and reflect the rest of the 88% as well as tranmitting the 12%. Now if you heat the stainless steel to say 300 Celsius. It won't transmit the 300 Celsius but only a little of it. This is why thermal imagers can see it colder than it actually is (and you have to adjust the emissivity setting). But if you paint the stainless steel black, the thermal imager would see its true temperature. Now question, how come mere painting the surface black can make it transmit 96% of the heat radiation? We know that painting it black can no longer make it reflective. But heating it is an internal process within the steel.. so how come without black pain, it can only transmit a fraction of the heat. With it painted black, it can transmit majority of the heat? What has the surface reflectivity got to do with the internal events of the sample?

2. Jul 20, 2011

### Naty1

Thermal heat transfer depends on conduction, convection and radiation.

Painting a suface any color significantly changes the first and the last. Heating is not JUST an internal process it is also primarily a surface process...if the impinging radiation wavelengths are absorbed by the exposed surface electrons, they become excited (energized) and that serves to excite adjacent internal electron's.

NASA uses real thin, flexible, reflective coatings on both the inside and outside of spacesuits.

I think you'll find the the lattice structure of materials also affects heat characteristics.. because such structure affects the degrees of freedom of bound electrons...

3. Jul 20, 2011

### riezer

You mentioned about "impinging radiation wavelengths". You didn't answer exactly my question. My question is. For a piece of metal like stainless steel coin with internal heat after it is baked in an oven. How come if you pant one half of it black. The black portion can be seen hotter in a thermal imager. Remember we are not talking about impinging radiation to to the steel coin but internal heat being transmitted outside. How come mere painting the surface black can make it transmit more.

Or let me rephase my question further. Our intuition tells us shiny surface can reflect more... say reflecting 80% of the impinging radiation. But when it is internal heat. It can only transmit it 20%. Nothing is being reflected so how come the surface can still maintain the 20% transmittance from the inside internal heat??

4. Jul 20, 2011

### Drakkith

Staff Emeritus
The black paint is in direct contact with the steel, which allows it to absorb heat by conduction. This allows it to absorb heat from the steel and emit it much quicker than steel itself emits it. The difference between the black paint and the steel surface for radiative transfer IS emissivity. Less emitted heat would be seen as a lower temperature possibly. (I am unsure how much has to do with intensity and how much has to do with the actual wavelengths emitted)

5. Jul 20, 2011

### riezer

Thanks, it solves the mystery if you are right. Anyway. Thermal imagers can detect in the far infrared 900–14,000 nanometers range. Now what exactly does this range represent? is 900 nanometer mean the heat is less intense than 14,000 nanometers? Or do they stand for something else like visible frequency colors where it's not related to heat?

6. Jul 20, 2011

### Drakkith

Staff Emeritus
It refers to the wavelengths of light that it can see. An object an any temperature above 0k emits EM radiation. Until an object is several hundred degrees minimum this is mostly in the infrared and below range.

7. Jul 20, 2011

### riezer

I know 9-14 um is the wavelength of light it can see. But does 9um compared to say 11 um is the intensity of heat in the object or other parameters?

8. Jul 20, 2011

### Drakkith

Staff Emeritus
It is the wavelength of the EM radiation. Infrared radiation is the part of the EM spectrum from about 1 micrometer to 1 mm. Similar to how Blue light is about 450 nm and Red light is about 700 nm. The intensity would be how many photons of each wavelength are being produced.

9. Jul 20, 2011

### klimatos

I think that you will find that it is the paint that is emitting the EMR and not the steel. It is my understanding that both electromagnetic radiation and electromagnetic absorption are essentially surface phenomena.

The steel may "transmit" enthalpy to the paint by conduction, but the release of photons into the atmosphere is accomplished only by the surface layer of paint molecules.

10. Jul 20, 2011

### riezer

The reason it's confusing is because thermal imagers depect the temperature by the colors. We know visible light has wavelength of 390-750 nm and the intensity is in each wavelength. So in the far infrared wavelength of 9-14 um. The temperature is the intensity of each wavelength. So the red versus blue color in the thermal imager image is not related to the wavelength but to the intensities of all wavelength from 9-14 um.

Ok. What would happen if you have 3 thermal imagers set to only 9-10um in the first imager. 11-12um in the second unit, and 13-14um in the third imager. What would each imager see say of your hand? I think it can record the same temperature of your hand but what other difference between their images since the wavelength sensitity is not the same?

11. Jul 20, 2011

### Drakkith

Staff Emeritus

Not only do you have higher frequency light being emitted at higher temperature, you also have more photons of those higher frequencies being emitted. So figuring the temprature based on thermal radiation is a combination of the range of emitted wavelengths and also the intensity of each one.

I'm not sure the imagers could determine your temperature with such a small wavelength range, but I really don't know.

12. Jul 21, 2011

### riezer

From the above site it says "At any given temperature, there is a frequency fmax at which the power emitted is a maximum."

This means that at 10 Celsius, there is more of 9um frequency out of the 9-14 um range while at 200 Celsius, there is more 14um frequency out of the 9-14 um, correct?

13. Jul 21, 2011

### Drakkith

Staff Emeritus
Pretty much. You have your wavelength and frequency confused though. If 10 Celsius peaks in the 14 um range, 200 Celsius would peak in a shorter wavelength range, such as 9 um. Shorter wavelength = higher frequency = more energy.

14. Jul 27, 2011

### riezer

In blackbody radiation. 900 Celsius can produce orange color. 1000 Celsius can produce yellow color and 1200 Celsius white color. In the case of 900 Celsius causes yellow color, it corresponds to visible light peak wavelength of 590nm (yellow). What is the formula that connects the peak wavelength and temperature? I'd like to understand how 0-250 celsius corresponds to 7.5-13um in the thermal imager.. whether a formula can show say 90 Celsius corresponding to peak 10um wavelength (in the blackbody curve).

15. Jul 28, 2011

### Drakkith

Staff Emeritus