what are the best metals for infrared absorption for heating application?
89873
Best heat transfer should occur with a good thermal conductor like copper, to carry the absorbed heat away, coated with a thin heat-conducting layer of flat black, may be an oxide of copper.
1101939
A thermocuple might be the one you seek. \n\nThe seebeck effect on thermocouple uses this theory, as there are two different material soldered together to make a single V or U shaped bar, there is a point where those two metals met. When we heat that point, the heat will move the electron, and since one metal has different coefficient than the other, so the electron will move from one metal to the other, thus an electric current will be created.\n\nPeltier effect, uses the vice versa theory, thus if we put electrical current through that thermocouple, then the thermocouple will need to absorb heat from it surroundings, thus, it will lower the ambient temperature.\n\nI haven't got any clue on how effective does this "refrigerant" work, but I think It might, might, might, work..
431802
You are not going to be able to buy nitric acid above 70% pure.  Nitric acid pretty much dissolves anything it comes in contact with which causes for its low purity.  I would not suggest heating concentrated acid since the fumes it will produce are very dangerous.  The fumes at room temperature will even cause burns. Depending on what you are trying to do, the acid will need some oxygen to dissolve metals.  With stainless steel, a small scratch on the surface will be enough.
311737
infrared and ultraviolet!
809561
malleable and ductile lustrous conduct heat and electricity high melting points almost all are solids at room temp except mercury
46078
copper is one of many "transiton metals" which means it is one of the elements in the middle-lower part of the periodic table - groups 3 to 12, also called the d-block because they have electrons that occupy d orbitals.  other transition metals include zinc, magnesium, mercury, iron, gold, silver - altogether there are 40 of them.\n\ncopper is shiny, red-orangish in color, and is a very good conductor of both heat and electricity.
907332
Different types\nA variety of thermocouples are available, suitable for different measuring applications (industrial, scientific, food temperature, medical research, etc.).\n\nType K (Chromel (Ni-Cr alloy) / Alumel (Ni-Al alloy)) \nThe "general purpose" thermocouple. It is low cost and, owing to its popularity, it is available in a wide variety of probes. They are available in the −200 °C to +1200 °C range. The type K was specified at a time when metallurgy was nowhere near as advanced as today and consequently characteristics vary considerably between examples. There is another problem in that one of the consituent metals is magnetic (Nickel). The characteristic of the thermocouple undergoes a step change when a magnetic material reaches its Curie point. This occurs for this thermocouple at 354°C. Sensitivity is approximately 41 µV/°C. \n\nType E (Chromel / Constantan (Cu-Ni alloy)) \nType E has a high output (68 µV/°C) which makes it well suited to low temperature (cryogenic) use. Another property is that it is non-magnetic. \n\nType J (Iron / Constantan) \nLimited range (−40 to +750 °C) makes type J less popular than type K. The main application is with old equipment that cannot accept modern thermocouples. J types cannot be used above 760 °C as an abrupt magnetic transformation causes permanent decalibration. Type J's have a sensitivity of ~52 µV/°C \n\nType N (Nicrosil (Ni-Cr-Si alloy) / Nisil (Ni-Si alloy)) \nHigh stability and resistance to high temperature oxidation makes type N suitable for high temperature measurements without the cost of platinum (B, R, S) types. They can withstand temperatures above 1200 °C. Sensitivity is about 39 µV/°C at 900°C, slightly lower than a Type K. Designed to be an improved type K, it is becoming more popular. \n\nThermocouple types B, R, and S are all noble metal thermocouples and exhibit similar characteristics. They are the most stable of all thermocouples, but due to their low sensitivity (approximately 10 µV/°C) they are usually only used for high temperature measurement (>300 °C).\n\nType B (Platinum-Rhodium/Pt-Rh) \nSuited for high temperature measurements up to 1800 °C. Unusually type B thermocouples (due to the shape of their temperature-voltage curve) give the same output at 0 °C and 42 °C. This makes them useless below 50 °C. \n\nType R (Platinum /Platinum with 13% Rhodium) \nSuited for high temperature measurements up to 1600 °C. Low sensitivity (10 µV/°C) and high cost makes them unsuitable for general purpose use. \n\nType S (Platinum /Platinum with 10% Rhodium) \nSuited for high temperature measurements up to 1600 °C. Low sensitivity (10 µV/°C) and high cost makes them unsuitable for general purpose use. Due to its high stability type S is used as the standard of calibration for the melting point of gold (1064.43 °C). \n\nType T (Copper / Constantan) \nSuited for measurements in the −200 to 350 °C range. The positive conductor is made of copper, and the negative conductor is made of constantan. Often used as a differential measurement since only copper wire touches the probes. As both conductors are non-magnetic Type T thermocouples are a popular choice for applications such as Electrical Generators which contain strong magnetic fields. Type T thermocouples have a sensitivity of ~43 µV/°C \n\nThermocouples are usually selected to ensure that the measuring equipment does not limit the range of temperatures that can be measured. Note that thermocouples with low sensitivity (B, R, and S) have a correspondingly lower resolution.
62935
sure, what you want is an IR filter, which you can pretty easily get from any optics supplier.  here are a wide range of choices from melles-griot, but they're kinda expensive.  however they are quality professional-grade filters.\n\nhttp://shop.mellesgriot.com/products/optics/optics.asp?plga=051408&CatID=10222&mscssid=&Page=All\n\nhere are some cheaper versions with slightly less clean spectra, but probably still functional depending on your needs:\nhttp://shop.mellesgriot.com/products/optics/optics.asp?plga=085680&CatID=1023&mscssid=\n\nyou can probably get some for cheaper if you hunt around on the web.  just be careful you aren't getting the opposite of what you want - a stop-visible/pass-IR filter, which is used a lot in infrared photography.
821260
The transmittance of good quality glass 1cm thick ranges from 99.6% to 99.9%, so the absorption is less than .4% in the visible spectrum.  It increases in the ultraviolet.  But if you have glass covering a box, nearly all the light passes through and heats whatever is in the box.  The infrared energy radiated is mostly blocked by the glass.  That's what heats a greenhouse.
1108936
Palladium (Pd) should not show up on an IR trace of any sort.\n\nIR spectroscopy is looking at the interatomic bonds of mostly organic compounds.  Palladium metal would not have any bonds of the appropriate energy (quantum) levels to absorb in the IR region.\n\nIs there a specific Pd-containing compound you are looking at?  Perhaps I could help with some further details.
237020
Gold AU\nSilver AG\nLead PB\nCarbon C\nNickle NI\nIridium IR\nZR Zirconium\nCU copper\nLi Lithium
632924
good conductors are material with very tight atomic structure. Remember you are talking about electron flow here. they flow best and easies through material with very tight atomic make up like lead gold copper aluminum silver so forth. Most good conductors are non ferrous metals. This means If you heat them up to melting point they will not get red hot first, you will see very little color change in them from solid to liquid state.
1185872
I can think of two ways to measure the absorption and reflection coefficient of a material.  The first would only work with a polished surface.  First measure the intensity of a laser beam with a photodetector, then shoot the laser beam at the surface and aim the reflection at a photodetector.  Coefficient of reflection = transmitted intensity/initial intensity  (coefficient of absorption =1-reflection).  The other, harder way, which would work with any material, is to measure the heat gain in the target.\n\nFor this you're gonna want as small a piece of the target material you can get so that the laser still completely hits it.  Now attach a thermistor to the back of the target (again the smaller and thinner the target the better, ideal would be a foil just thick enough to be completely opaque).  Now shoot the laser at the target and measure the temperature as a function of time  It should be linear up until it starts radiating nearly as fast as it absorbs energy, so take points until it starts to level off and drop points after that.  Now find the specific heat of the thermistor-target apparatus by dipping it in water with a thermometer in it, using the initial and final temperature of the water, the volume of the water, and the initial temp of the apparatus, determine the heat it absorbed per degree it's temperature changed (this will actually give you specific heat*mass which is fine).  Now convert your temperature vs. time data to heat vs. time using the specific heat you determined (temp*(specific heat*mass)).  The slope of the line of best fit for this data will give you the rate it absorbed energy in watts.  Divide this by the lasers output in watts and you get the absorption coefficient.
140510
What will work for you depends a lot on what the material to be coated is, on the application, time spent at temperature, heat cycling, peak temperatures, etc.\n\nHere's a non-ablative coating for metals that can withstand sustained temperatures of 850ºC (1560ºF) and thermal shocks from 1030ºC (1885ºF) immediately to a cold water bath:\n\nhttp://www.indestructible.co.uk/cat2/PL134-R1.html
748284
copper, silver, aluminum, lead, magnesium, platinum and tungsten
768677
Gallium based semi-conductor right now is too expensive to deploy. \n\nIf you want completely out of the left field, I'd go for optical processing.
363070
Copper, Iron, Chromium, Aluminium