torsdag 17 november 2011

Radiative Cooling

How quickly will a hot (black) body cool off if all exterior heat sources are shut off and the
heat energy stored in the body radiates from the body? How quickly does the Earth surface cool off during night?

There are different answers:

1. Mathematical Physics of Blackbody Radiation gives the following cooling curve for the temperature T(t) as function of time t starting at t=0:
  • T(t) = T(0)/(1 + C t)^0.5
as a consequence of the fact that the stored heat energy E scales like T^2 while the heat loss scales like T^4 by Stefan-Boltzmann's Law which also defines the constant C.

2. The standard answer is different:
  • T(t) = T(0)/(1 + C t)^0.33
with E scaling instead like T.

Which answer is best? Experiments appear to favor 1 with its quicker decay, also shown in the above picture from Comlab.

The difference between 1 and 2 may be the difference between a blackbody in the form of a solid or a gas, where the blackbody nature/spectrum of a gas may be questioned.

1 kommentar:

  1. The lab report looks like an undergraduate lab report and deals with radiation and convection heat loss.