Coated (e. g. enamel, PTFE, ceramic)
I can't answer in general, but that one's easy. Sudden thermal shock causes strain in a material by unequal expansion, either in the same material by high thermal gradients, or in interfaces between materials with different coefficients of thermal expansion. The strain in this case (two different materials) can be very high. If the material in question is not elastic (e.g. enamel + ceramic; I would think PTFE is different, but I'm not sure), then the bonds between the coating and the metal would be severely strained and it would likely crack and chip.
I can tell you from personal experience that I have actually used this to my advantage:
In the spring, I produce a small quantity of maple syrup by boiling sap in an uncoated stainless steel pan. On rare occasions, accompanied by the release of many expletives, I have let the syrup boil down too far, at which point it burns and seems to coat the bottom of the pan with a thin but hard and very resilient layer of carbon black. The trick to removing this stuff is to get some kind of stress crack started, e.g. by scrubbing w/ steel wool or a copper pad, and then what I do is I put the pan on the stove for a while to let it heat up hot (but not red hot), and then bring it over to the sink and spray cold water on the inside pan bottom where the carbon black has stuck to. After a few times, the carbon black will start to flake off and then it becomes easier to remove by a combination of abrasion and thermal shock. (The two pans I've done this on have been fine; both are stainless steel with a thick (>8mm) bottom, and I've put them through at least 30 or 40 thermal cycles of this type.)
edit re: general topic:
Wikipedia says this:
The robustness of a material to thermal shock is characterized with the thermal shock parameter:
where
- k is thermal conductivity,
- σT is maximal tension the material can resist,
- α is the thermal expansion coefficient
- E is the Young's modulus, and
- ν is the Poisson ratio.
Higher thermal conductivity means it's more difficult to get a large thermal gradient across the material (less prone to shock); higher thermal expansion means more strain (more prone to shock), and higher Young's modulus means more stress for a given strain (more prone to shock).
So theoretically you could compare the different materials. (exercise for the reader ;) Most likely copper would be more resilient than the other metals, because of its higher thermal conductivity and higher ductility.
Thermal conductivity k: Copper = 401, Aluminum alloys = 120-180, stainless steel = 12-45 (units = W/m*K)
σT: no idea:
Coefficient of thermal expansion α: Copper = 17, Aluminum = 23, iron = 11.1, stainless steel = 17.3 (units = 10−6/°C)
Young's modulus E: Copper = 117, Aluminum = 69, iron/steel = around 200 (units = GPa)
Poisson's ratio ν: Copper/stainless steel/aluminum are all around 0.3-0.33, cast iron = 0.21-0.26
So stainless steel is worse than aluminum or copper (much lower thermal conductivity, higher Young's modulus).
Best Answer
I was looking at this same question awhile back when I got an IR thermometer, what I found was 2 things. The first thing is that there are too many variables to have a single answer that would work. The pan temperature you want at the start depends on a number of factors including:
So you would have to develop all these numbers for each pan on each burner, with different heat settings through experimentation. Other people's numbers would mean nothing unless you are using the same equipment.
Another reason is that the pan's starting temperature isn't actually that important. When you put food in a pan the temperature is going to drop, how quickly depends on the factors above, however the total amount of heat it will take to cook your food and the rate you want to apply heat remains about the same no matter which pan and material you use. It's a question of heat supply to the pan over time, all you really need to know is that your pan is hot enough to start. It's the temperature of the pan while the food is on it that is important. You can measure it quite a bit but at the end of the day all you conclude is you need to have your burner set to a certain level.