Two ideal, black-body radiators have temperatures of 275 K and 1100 K. The rate of heat transfer from the latter radiator is how many times greater than the rate of the former radiator?
This is College Physics Answers with Shaun Dychko. The rate of heat transferred due to radiation is Stefan-Boltzmann's constant times emissivity times surface area times temperature in absolute terms to the power of four. So, we're going to divide these two rates of heat transfer due to radiation. And, their formulas are going to be the same for each of them except in the first case, we'll have temperature one. In the second case, we'll have temperature two. And so, all these other factors cancel. And so, the ratio of the rate of heat transfer is the ratio of the temperatures to the power of four. So, we have 1100 Kelvin divided by 275 Kelvin to the power of four is 256. So, that's how many times greater the rate of the former is to the latter. That's how much greater this one is than this one. This being the latter, this being the former.