How is rate of cooling of a body related to its Volume? I was wondering how two bodies made of the same material and
having the same Surface Area but different Volumes, heated to the same temperature, would cool. Would they cool at the same rate? I know that Stefan's Law states that $E=\sigma A T^4$ so theoretically they should radiate the same amount of energy per unit time and hence cool at the same rate. But doesn't this violate the conservation of energy? How could two bodies having different thermal energies(due to different volumes and masses) have the same rate of cooling? 
 A: 1. Case of radiative losses (only):
$$\frac{dQ}{dt}=\sigma AT^4$$
$$dQ=mc_pdT$$
$c_p$ is the specific heat capacity of the object. So the cooling rate $\frac{dT}{dt}$ is:
$$\frac{dT}{dt}=\frac{\sigma A}{mc_p}T^4$$
2. Case of convective losses (only) (with $h$ the heat transfer coefficient and $T_a$ the ambient temperature):
$$\frac{dQ}{dt}=hA(T-T_a)$$
Again, with $dQ=mc_pdT$:
$$\frac{dT}{dt}=\frac{h A}{mc_p}(T-T_a)$$
3. Conclusion:
In both cases:
$$\frac{dT}{dt} \propto \frac{A}{m}$$
So in both cases, all other thing being equal, objects with larger surface area will cool down faster, objects with larger mass will cool down slower.
Note that both expressions for $\frac{dT}{dt}$ can easily be integrated to find expressions for $T(t)$.
A: 
...they should radiate the same amount of energy per unit time and
  hence cool at the same rate.

The energy content is more related to the mass of the object than the volume.   (and at the end, you mention different masses as well as different volumes).
Because they begin at the same temperature and the same surface area, they radiate the same amount of energy initially.  But that energy loss changes their temperature as $Q = mc\Delta T$.  If you assume them to have the same mass, then they will cool at the same rate.  If they have different masses, then the smaller mass object will cool more rapidly.
