So as stated in the picture above, stress behavior in fluids and solids isn't the same. Why is it physically that way?
Why does the stress of fluid depend on rate of deformation unlike stress of solid that depends on deformation itself
continuum-mechanicsfluid dynamicssolid mechanicsstress-strain
Related Solutions
Your basic premise is wrong. Compressing things makes them heat up. With gasses this is even nicely predictable over a wide range. Look up something called the ideal gas law.
If you compress a gas enough then it will undergo a phase change and turn to liquid. This will release even more heat than just compressing the gas because the gaseous state of a substance has more energy than the liquid state. This is what the heat of vaporization is about, and why you have to heat a liquid to make it boil (change phase to its gaseous state).
Steam heat systems work on the reverse principle. The steam condenses on the inside of the pipe, thereby releasing a lot of heat, which heats the pipe, which heats the room. The condensed water returning to the boiler contains less energy than when it left the boiler as steam.
Ok, so figuring that stress is defined to be the force per unit area around a point, I'd like to know why it's defined that way, rather than say force per unit volume enclosing a point.
Here is the intuitive explanation.
Suppose you have a string S from which a weight W is hanging. We will suppose that the amount of weight is so much that the string is near the regime where it would snap, but is not quite there yet. We're going to think about what happens when we add another length of string S.
If you add it in parallel, so there are two strings holding W, then it seems obvious that you could then increase the mass to about 2W before the combined two-string system would be near-breaking.
By contrast, if you add it in series (W hangs from S hanging from S), then we know that the tension force is the same in both strings, and it seems likely that they'd simply both be near-breaking. The top string might even break, if the weight of the string underneath it is enough to put it past breaking.
This tells us that the material properties which concern us (like when a string breaks) respond to stresses, defined as forces divided by an area perpendicular to that force. The direction that lies alongside the force doesn't matter, qualitatively because it propagates the force rather than responding to it.
You can also examine this two-string thought-experiment with much smaller forces not near breaking, where Hooke's law should hold for lengthening, to find that if the original deforms by $\delta L = \ell$, the two-strings-in-parallel should deform by $\ell/2$ while the two-strings-in-series should deform by $2\ell,$ so that there seems to be a constant "stress / strain" relationship where the "strain" is defined as $\delta L / L.$ In other words you can look at a system with the normal spring-constant $F = k ~ \delta L$ relationship, but it is more helpful from a materials perspective to divide by the length of the "spring" $L$ and also its cross-sectional area $A$ to find $$\sigma = \frac FA = \frac {kL}{A} ~ \frac{\delta L}{L} = \lambda ~ \epsilon.$$ The elastic modulus $\lambda$ then also has units of pressure (force per unit area) and is more fundamental to the material (a material-property) that you're studying than the spring constant (a property of both the material and the setup) is.
Since the elastic modulus is a material-property, this hints that "stress" is the right definition of "force" and "strain" is the right definition of "displacement" at the more-microscopic level when we're peeking inside the substance.
In turn, it becomes very common in materials science to show the "stress/strain" curve of various materials. This starts out of course as a straight line through the origin with slope $\lambda$, but then as a substance deforms it will describe some sort of curve as added stress leads to further strain. So for example the plastic bags from the supermarket will curve upwards; they get stiffer as they stretch more.
Once you know that the stress is the right way to "microscopically" define force, elastic systems start to show off a similar problem: the simplest stretching of a beam consists in that beam not only lengthening but also narrowing. Microscopically, a little box inside the substance is not only feeling a force $+\sigma~dA$ on one side and a force $-\sigma~dA$ on the other side (so it is in force balance and provides tension), but it must also be feeling some forces on its other sides which "pinch" it smaller. So: the force is direction-dependent, and we therefore have not a stress vector (which we already didn't quite have -- the stress is in opposite directions on the top and bottom of the box), but a stress tensor: give me a direction and I'll give you the stress vector on a plane normal to that direction. (This also solves nicely the "stress on the top of the box is negative the stress on the bottom of the box" problem.)
Usually this simplifies a lot because there are eigenvectors of the stress tensor: directions where the stress points exactly normal to the plane it deforms. Those are called the "principal stresses". However there is no reason why they'd have to be orthogonal, especially, for example, in a crystal lattice which is not cubic.
Best Answer
Let's address a possible misconception: that if we switch from a solid to a fluid, the stress $\tau$ switches from being dependent on the deformation (only) to being dependent on the deformation rate (only). This is not correct.
Solids resist deformation and flow (deformation rate) and acceleration (rate of deformation rate) and so on. Consider ice, for example: Its shear modulus $G$, which quantifies its resistance to shear deformation $\gamma$, is several gigapascals. Its resistance to flow can be obtained from a deformation mechanism map:
Here, we can obtain the shear stress needed at any particular temperature to achieve a certain flow rate; this quantifies ice's viscosity $\mu$ or resistance to a shear deformation rate $\dot\gamma$.
In turn, ice's resistance to a time rate of change in its deformation rate ($\ddot\gamma$) is related to its density.
The point I'm making is that resistance generally exists for deformation and all of its time derivatives. You may have heard that the stress depends on the deformation (only), but this is a simplification.
Now, when we call something a fluid, we mean that it's bonded weakly enough that it rearranges quickly and nearly effortlessly under any load. Thus, we take its shear modulus—its resistance to shear $\gamma$—as zero. But the resistance to the time derivatives of the deformation remain. For further idealization, if the resistance to flow $\dot\gamma$ is negligible to us, then we might consider the fluid to be inviscid. But the resistance to acceleration $\ddot\gamma$ remains—the fluid isn't massless.
I hope this clarifies why it would inaccurate to say that stress in a solid and fluid depends (only) on the deformation and deformation rate, respectively. That's what simple models such as $\tau=G\gamma$ and $\tau=\mu\dot\gamma$ might imply, but that's only because these models are considering only static deformation and steady flow, respectively.