Would placing a polarized sunglass lens in front of an IR thermometer double its max temperature measurement? I have been doing some backyard metal casting, mostly aluminum.  My IR thermometer will only read up to 1000F. Aluminum has a melting point of around 1250F and should be poured around 1350F.  
I've found that since I made some changes to my forge I'm over heating my melt by quite a bit.  Could I take a good polarized sunglass lens, put it over the IR sensor on my contactless thermometer and effectively read temperatures up to 2000F?
It seems to make sense half the photons hitting the sensor but I'm not sure if IR light is polarized like visible light or if I'd be wasting my time trying this?
 A: No, it won't work. The temperature sensor probably uses wavelengths in the far infrared, near 10 μm. The plastic polarizer sheet is probably black there. You only measure the temperature of the polarizer.
I just tried this with my infrared camera. When I hold a polarizer in front of the lens I cannot see the candles that I lit. The polarizer is opaque at these wavelengths. This is an image where you can see the heat imprint of my fingers on the polarizer sheet, but not the heat of the candles behind it. (Only contours from the visible-light camera that the FLIR software mixes in.)

A: Almost certainly this will not work. Most modern pyrometers, which most digital infrared  thermometers are a kind of, are multi-color pyrometers, which means that they measure the ratios of powers received from the probed emitter at two or more wavelengths. The temperature is then inferred from the Planck radiation law from the ratios alone, under the assumption that the emissivity of the surface in question is the same at the wavelengths used in the measurement. Most modern pyrometers do not use the Stefan-Boltzmann law to infer temperature, because the emissivity of surfaces is highly variable and  dependent on unknown and uncontrollable factors, such as roughness. The idea of using ratios as measurands cancels the emissivity out of the measurement. It also cancels the overall source intensity out of the measurement.
Therefore, the putting of an attenuator in the path between the probed object and the pyrometer, which is what you are essentially proposing here, will make no difference to the ratios which are the basic measurands. It will only lower the signal to noise ratio, so the same temperature will be measured with the attenuator in place, but with larger uncertainty owing to the poorer signal to noise ratio.
A: You're right, IR light is polarized the same way as visible light. So in principle your method should work. Just a few points to consider: 
Conventional polarizers may be less effective for IR, so you would probably need to calibrate you're intensity ratio.
In case, the is already a polarizer built in, you wouldn't gain anything. Although I would not see an immediate reason, why this would be the case.
A: No, blackbody intensity does not scale linearly with temperature, but as the 4th power.
Assuming the sensor is using something near $1 \mu \text{m}$ as the sense wavelength, we can run it through a Plank's Law calculator and find the spectral radiance.  $1000F$ is $811K$.
$R(811K @1\mu \text{m}) => 2.3508 \text{W/(m}^2\text{-sr-}\mu\text{m)}$
Assume your polarizer blocks half the light, so this would allow double the radiance to be checked.   Run the calculation in the other direction
$4.7016 \text{W/(m}^2\text{-sr-}\mu\text{m)} => R(844K@1\mu \text{m})$
So the receiver would see double the radiation at only $1060F$.  The exact temperature would depend on the specific wavelength sensed, but that fourth power of temperature makes it difficult to do this trick.
