If it does , can heating an object cause change in it? How can the relation between change in temp and internal pressure be derived?
2 Answers
A solid, such as a sphere, is a statically indeterminate structure; it can (often does) contain internal stresses, and those can be pressure in the interior (or a variety of other conditions).
It is possible to make a structure that has negative pressure in the center, holding (in tension) to all the surface of that item. Prince Rupert drops are such a structure (easily made by cooling liquid glass) see a demonstration here that exhibit remarkable resistance to surface cracks such as are the starting point for normal fractures.
Tempered glass is a (slightly less spectacular) example of internal stresses created for a useful purpose.
In terms of creating high pressure inside a solid, an inclusion inside a diamond crystal can be laser-heated to create truly high pressures, but there are few practical applications for such stressed items (unless superballs' high bounce properties are 'practical').
- 10,853
Interesting question. The only pressure source in the center of solid sphere would be due to gravity force of a sphere. Using hydrostatic equilibrium equation for fluids : $$ {\frac {dP}{dr}}=-{\frac {GM(r)\rho (r)}{r^{2}}} $$ and integrating it, one can arrive at central pressure upper limit formula which is the same for calculating central pressure in stars: $$ P_{central} = \frac{5\,G \, m^2}{4 \, \pi \, R^4} $$
For Earth center this estimate gives pressure about $\approx 575~ \text{GPa}$ ($5.75 \times 10^{11} \text{Pa}$).
Btw, it's worth to note that for usual everyday objects this inner center pressure would be very small. For example if we take a $1 \text{kg}$ sphere with $1 \text{m}$ radius, it will have $ 10^{-11} \text{Pa}$ central pressure. So for all practical reasons this central pressure can be neglected for small objects with a mass $m \ll m_{_\text{Earth}}$. However for celestial objects this pressure becomes very important.
- 16,916