How accurate is this infrared thermometer?

January 4, 2007 10:03 AM

I got a small infrared thermometer as a gift, and I'm trying to understand how well it takes temperatures of different materials.

The specs says it has a preset emissivity of 0.95. There's no other adjustments or controls on the unit, other than the trigger.

Is it correct to assume that if I point it at materials with that same emissivity, like concrete (from this table), it'll be pretty accurate? So what happens when I try measuring temperatures of materials with emissivity that's vastly different than 0.95? Like gold or silver (both 0.05)? Will the temperature I measure be far off the mark? Can I calculate a corrected value based on the emissivity of the target material?

posted by jaimev to technology (3 comments total)

No, it won't be accurate for such a material. These things work by measuring the surface flux at a given wavelength from the object you're pointing it at, which for an object with emissivity 1 is given by the black-body spectrum. If the object is emitting less infrared than it "should", because of a lower emissivity, the thermometer is going to think that the material is colder than it actually is.

You can, in principle, extrapolate what the real temperature is if you know the real emissivity of the object, the emissivity that the thermometer expects, and the wavelength of the radiation that the thermometer is looking at. I did a little bit of work with this kind of pyrometry in a lab class a few years ago; my recollection is that this reverse-engineering isn't simple, but it's not that complex either. Let me see if I can dig it up.
posted by Johnny Assay at 11:01 AM on January 4, 2007

Oh, on more thing: emissivity is frequently wavelength-dependent. So if you're reverse-engineering things, you need to know the emissivity at the thermometer's wavelength.
posted by Johnny Assay at 11:04 AM on January 4, 2007

Here (PDF) is the appropriate page from my old lab report. For your purposes, a₀ is the emissivity of the target material divided by 0.95 (since your thermometer is calibrated to that emissivity, not an emissivity of 1 like mine was.) You'll have to look up what the appropriate wavelength is for your thermometer; but as long as it's below about 10 microns and you're not measuring stuff that's much above room temperature, the approximation I made to go from equation (19) to the simpler equation (20) should still be valid.

Oh, and my apologies if this has all gone entirely over your head. I get technical and wonkish when someone asks a question that I've had the opportunity to think about & solve in the past.
posted by Johnny Assay at 11:35 AM on January 4, 2007