What would a normal thermometer read at the Sun's photosphere? I was reading about the Earth's Thermosphere, and I quote this:

"The highly diluted gas in this layer can reach 2,500 °C (4,530 °F) during the day. Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would read significantly below 0 °C (32 °F)."

So, while we know that the Sun's photosphere is about 6000 °K, it has very low density and should somehow behave similarly to the Earth's thermosphere.
So what would a thermometer read at the surface of the Sun?
 A: The specific heat capacity of mercury is 140 J/kg/K. Let's suppose you have about 1cc of mercury in your thermometer; let's assume this isn't going to boil and your thermometer isn't going to melt and let's suppose the mercury is in the form of a cube which is 1cm on a side. The density of the mercury is about 7.6 g/cc, so you have 7.6$\times10^{-3}$ kg of mercury.
Now put it in the radiation field at the Sun's photosphere. The flux is $\sigma T^{4}$, where $T=5770$K and $\sigma = 5.7\times10^{-8}$ in SI units. So the flux is $6.3\times10^{7}$ W/m$^2$. The cube has a face area of 1 cm$^2$. Let's assume that all the radiation incident upon it is absorbed (you can divide my timescale by whatever you think the reflectivity is if you like). i.e. The cube absorbs about 6300 J every second, sufficient to raise its temperature by $6300/(7.6\times 10^{-3} \times 140)$, which basically means it would heat up to the photospheric temperature in a second.
You can mess about with the assumptions, put some reflectivity in if you like, allow the mercury to radiate away some energy, but the basic point is that the Earth's thermosphere is nothing like the Sun's photosphere and if you make a thermometer that would survive, it would probably very quickly come into equilibrium with the Sun's radiation field.
