I know the inverse square law means that radiation intensity decreases in proportion to the radial distance from the source (I is proportional to 1/r^2).
Does this apply to the temperature increase of an object with increasing distance from the heat source? So, if an object is 10 times farther away from a heat source than an identical object, will it experience 1/100th the temperature increase? Would it be possible to calculate the temperature of a heat source from the inverse square law?
I'm confused because the Stefan-Boltzmann law says radiation emitted from a blackbody is proportional to the fourth power of temperature.
Best Answer
Yes, it applies, and it's not really related to the Stefan-Boltzmann law.
The energy radiated from a blackbody at temperature $T$ does indeed scale like $T^4$. Any object (blackbody or not) can absorb radiated energy, and that is the part which increases the temperature.
The inverse square law is a statement about the density of radiation (or intensity, in units of $W/m^2$) from a point source, not about either the source or receiving blackbody itself. If at a distance of one meter from a point source an object receives 1 $W/m^2$ of radiative energy, then at a distance of 2 meters the same object will receive 0.25 $W/m^2$ of energy. That's true for a monochromatic point source as well as for a blackbody, and comes exclusively from geometry.
I won't say that it maps directly on to temperature rise of a receiving object (just because that also depends on heat capacity and re-radiation and such) but a rigorous statement is that the rate of thermal energy absorption follows the inverse square law.
Regarding the question of determining the temperature of a blackbody heat source by the inverse square law, the answer is "not really". If you know the surface area of the heater, then measuring the radiated intensity at one distance is all you need. If you don't know the area, measuring the radiated intensity is only going to give you a number proportional to $AT^4$, where $A$ is the heater surface area. You can get more information about temperature by looking at the spectrum of the emitted radiation and using Wien's law: $\lambda_{max} = b/T$, where $b\approx$ 2.9 mm K and $\lambda_{max}$ is the most intense wavelength.