Cooling of a solid vs hollow sphere

Question

We have a solid and a hollow sphere made of same material, and of same dimensions. We have to compare the temperatures of the sphere after a long time of cooling.

So initially I can safely say that cooling by radiation will take place or: $$H=\epsilon A\sigma (\Delta T^4)$$

So initial rate of cooling will be same. However after inital conditions, the condcution will aslo take place inside the solid sphere, and here i start to have difficulty writing the equation for heat loss.

For the conduction, we have: $$H=\frac{KAdT}{dx}$$ Solving for a solid sphere case I get: $$H=KA\Delta T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So we have conduction which is bringing in heat from beneath and heat leaving from the surface by radiation, so, how can I write net equation for the heat loss by the sphere and compare it quantitatively with hollow sphere? Answer is that solid sphere would stay warmer than the hollow one.

Answer 1

(Note that I don't promise a rigorous proof, but this exposition may provide ideas.)

For both spheres, the conductive heat equation applies with axisymmetry:

$$\frac{\partial T(r,t)}{\partial t}=\frac{\alpha}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T(r,t)}{\partial r}\right)=\alpha\left(\frac{2}{r}\frac{\partial T(r,t)}{\partial r}+\frac{\partial^2T(r,t)}{\partial^2 r}\right),$$

with thermal diffusivity $\alpha$. The initial and boundary conditions are

$$T(r,0)=T_0;$$

$$\frac{dT(r_\text{inner},t)}{dr}=0;$$

$$-k\frac{dT(r_\text{outer},t)}{dr}=\epsilon\sigma[T(r_\text{outer},t)^4-T_\infty^4],$$

with thermal conductivity $k$, inner radius $r_\text{inner}$ (equal to 0 for the solid sphere), outer radius $r_\text{outer}$, emissivity $\epsilon$, Stefan–Boltzmann constant $\sigma$, and enclosure temperature $T_\infty$.

(The boundary conditions come from realizing that no outward heat flux can exist from the center/interior surface and from setting the conductive heat flux just inside the outer surface equal to the radiative transfer away from that surface.)

My aim is to avoid a numerical solution but rather to consider scaling arguments, for example, to obtain qualitative results that might build intuition. So consider the finite-difference version of the heat equation above:

$$\frac{\Delta T_t}{\Delta t}=\alpha\left[\frac{2}{r}\frac{\Delta T_r}{\Delta r}+\left(\frac{\Delta T_r}{\Delta r}\right)^2\right],$$

where $r=r_\text{outer}$ and $\Delta r=r_\text{outer}-r_\text{inner}$ and where $\Delta T_t$ and $\Delta T_r$ refer to temporal and spatial differences, respectively.

Consider the first moments of cooling at $r=r_\text{inner}$ (the center and inner surface of the solid and hollow sphere, respectively). The surface temperature $T(r_\text{outer},t)$ must be the same for both spheres, as the diffusive conduction process hasn't had time to "know" that the center is hollow or not. Thus, $\Delta T_r$ is constant, as are $\alpha$ and $r=r_\text{outer}$, and $\Delta r$ is smaller for the hollow sphere, so $\frac{\Delta T_t}{\Delta t}$ must be larger for the hollow sphere, corresponding to more rapid cooling. In a hand-wavy way, I'll conclude that the hollow sphere—which contains less mass and no heat source—must subsequently remain cooler as the two spheres asymptotically approach $T_\infty$.