Emissivity and IR Thermometers

  • Context: Undergrad 
  • Thread starter Thread starter ap12
  • Start date Start date
  • Tags Tags
    Emissivity Ir
Click For Summary
SUMMARY

The discussion centers on the concept of emissivity in relation to infrared (IR) thermometers and blackbody radiation. A blackbody with an emissivity of 0.97 emits approximately 97% of the radiation of a perfect blackbody. When set to 50°C (approximately 323K), the effective temperature reading from an IR thermometer, without any adjustments, would be around 47.6°C due to the emissivity factor. The conversation highlights the importance of understanding that radiation scales with the fourth power of temperature and that the wavelength distribution is also affected.

PREREQUISITES
  • Understanding of blackbody radiation principles
  • Knowledge of emissivity and its impact on thermal readings
  • Familiarity with temperature scales, particularly Kelvin
  • Basic concepts of infrared thermometry
NEXT STEPS
  • Research the principles of blackbody radiation and emissivity
  • Study the effects of temperature on radiation using the Stefan-Boltzmann Law
  • Learn about the calibration and adjustment of IR thermometers
  • Explore the relationship between wavelength distribution and temperature in thermal radiation
USEFUL FOR

Researchers, engineers, and technicians involved in thermal imaging, infrared thermometry, and those studying the principles of heat transfer and emissivity.

ap12
Messages
1
Reaction score
0
Hi

I am trying to get my head around Emissivity and was wondering if anyone could help.

There is a "blackbody" with emissivity of 0.97.
We want to use it to see whether some IR thermometers are giving suitable readings. Unfortunately the emissivity of the thermometer is unknown and cannot be altered so we will only be able to estimate their accuracy.

So if the "blackbody" is set to a temperature say 50°C, will it emit 97% of this temperature? So the IR thermometer should give a reading around 50*0.97 = 48.5° if it does not include a correction?

I would be grateful of any help. I have done some reading on the subject but I am getting more confused as most websites say about adjusting the IR thermometer but this is not possible in my situation.

Many thanks
 
Science news on Phys.org
First: The absolute temperature scale is Kelvin, anything proportional to temperature has to be expressed in Kelvin. "97% of the Celsius-value" is meaningless, 97% of -100°C would be -97°C and therefore hotter?

A blackbody with an emissivity of 0.97 will emit 0.97 of the radiation of a perfect blackbody. 50°C correspond to ~323K. As radiation scales with the temperature to the 4th power, 3% less radiation means ~3/4% less temperature, which would correspond to ~323*0.9925=320.6K or ~47.6°C. By coincidence, this is not so far away from your value, but this has two errors which partially cancel here. Do not use my values, they are just estimates.

Keep in mind that radiation does not simply scales with temperature, the wavelength distribution gets shifted, too. A perfect blackbody with a real temperature of 47.6°C will emit the same power, but slightly shifted towards longer wavelengths.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 15 ·
Replies
15
Views
10K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
8K