# 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?

• The assumption that heat losses are primarily radiative isn't necessarily true: see also Newton's Cooling Law: en.wikipedia.org/wiki/Newton%27s_law_of_cooling
– Gert
Commented Mar 25, 2016 at 16:53
• An object at high T and in vacuum will lose heat radiatively only. But an object at much lower temperature and surrounded by a fluid like air or water will lose heat mainly convectively.
– Gert
Commented Mar 25, 2016 at 17:01

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 $T(t)$.

• This is probably the answer the OP needs, but it's worthwhile to point out that it assumes that the objects are in thermal equilibrium internally, that is, the surface is the same temperature as the interior. Generally that condition will not be met, although it's probably a good approximation for small temperature differences between the object and its surroundings. Commented Mar 25, 2016 at 19:03
• @garyp~ that's correct. With a spatial temperature gradient $\nabla T$ inside the object the calculations become also much harder.
– Gert
Commented Mar 25, 2016 at 22:06

...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.

• Is there a way to mathematically relate the rates of cooling? Commented Mar 25, 2016 at 16:37