Does a cooling object lose mass as it radiates? Assume we have a red hot cannonball in space. It starts off with mass M at 1000K. Later it has cooled by radiation to 100K. Has the mass decreased?
 A: MSalters already said "yes". I would like to expand on that by computing the change.
Let's take a 10 kg cannon ball, made of lead. Heat capacity of 0.16 J/g/K means that in dropping from 1000 K to 100 K it has lost $10000\cdot 900 \cdot 0.16 \approx 1.4 MJ$. This corresponds (by $E=mc^2$) to a mass of $1.6 \cdot 10^{-11} kg$ or one part in $6\cdot 10^{11}$.
I cannot think of an experiment that will allow you to measure that mass change on an object in outer space.
UPDATE if you think of temperature as "a bunch of atoms moving", I was wondering whether the relativistic mass increment would be sufficient to explain this mass change.
The velocity of atoms in a solid is hard to compute - so I'm going to make the cannonball out of helium atoms (just because I can) in a thin shell. The mean kinetic energy is $\frac32 kT$, so mean velocity $v = \sqrt{\frac{3kT}{m}}\approx 2500 m/s$. When things cool down to 100 K, velocity will drop by $\sqrt{10}$ to about 800 m/s.
Now at 2500 m/s the relativistic factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \approx 1 + \frac{v^2}{2c^2}$. It is encouraging that this scales with $v^2$, just as $T$ scales with $v^2$. Writing this all for one atom:
$$\Delta m = m (\gamma - 1) = m \frac{v^2}{2c^2}$$
Now putting $\frac12 m v^2 = \frac32 kT$, we get
$$\Delta m = \frac{3kT}{2c^2}\\
\Delta m c^2 = \frac32kT$$
The change in mass really does scale with temperature! So even though I was using the average velocity of the atoms, it seems that this is sufficient to explain a real (if hard to measure) change in mass... relativity works. I love it when that happens.
A: Of course, it does, since: 
$$\frac{\partial E}{\partial t}  = \frac{\partial }{\partial t} \left(m \cdot c^2
 \right) $$ Very little, though
