1) I do not understand if the temperature being measured is that of the liquid or of the entire system.
You have thermometer in liquid. I can't see why it will measure system's (Referring to the Latent heat of vaporization experiment we all must have witnessed in 10th grade or so)
2) Doesn't breaking the intermolecular bond automatically mean that the particles have become faster
I may not agree with you on this (see end)
Since evaporation has a cooling effect, when the boiling point is reached, the rate at which the liquid cools down because of evaporation becomes equal to the rate at which heat is added to the container, thus keeping the temperature of the liquid constant.
Now see, Evaporation is phase transition from liquid to vapors while Boiling is a phase transition from liquid to gases. More over. Evaporation may occur when the partial pressure of vapor of a substance is less than the equilibrium vapour pressure while Boiling, as opposed to evaporation, occurs below the surface. Boiling occurs when the equilibrium vapour pressure of the substance is greater than or equal to the environmental pressure. I can''t see how you are correlating them
If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same.
Note: The above explanation is borrowed from this Link
"... is there a way to show that you could build a perpetual motion machine from this?"
Yes. Focus the radiant heat from a thermal reservoir onto a spot that is hypothesized to be raised to a higher temperature through its concentration into a smaller area. Now connect heat engine - a Carnot engine - between the hot spot as the engine's heat intake and the original reservoir as the heat exhaust. Now the engine will run, outputting work. Your hypothesis means that you have an heat engine system spontaneously converting the heat in the thermal reservoir to work and there's your perpetual motion machine (of the so-called second kind).
Obligatory in any conversation of this kind is Randal Munroe's Fire From Moonlight article.
One way to understand all this is to note that optical systems are reversible, so that if light can pass from point A at the input to point B at the output, light can equally well go the other way. So if a hot body directs its radiant heat at another object through a lens system, the temperature of the latter will naturally begin to rise. That means that the second body will radiate back to towards the first body. If the second body became hotter than the first, it would be returning a higher heat power to the first along the reverse paths whence the incident heat came. Therefore, heat transfer will stop before the second body reaches the temperature of the first.
The second law of thermodynamics in optics is equivalent to the non-decreasing of étendue, which is the volume of a system of rays representing a light field in optical phase space and thus a measure of entropy. If étendue cannot be decreased, this means that density of rays in phase space cannot be increased; in turn this means that the divergence angles of a set of rays must increase if the area they pass through is shrunken down. This means that the light from any point on a hot body cannot be made brighter at the point where it reaches the target body.
This also is why a laser works differently if we try to reason as above. If energy reaches a body through a laser, the incident light paths taken have near to zero étendue - there's hardly any beam spreading at all. The second body will get hotter and hotter, but the radiant heat from the hot second body is all spread out in all directions (this is fundamental to blackbody radiation - there's no such thing as collimated blackbody radiation). So hardly any of the radiated light is accepted back along the extremely narrow range of paths back to the laser. Laser light is highly nonequilibrium light - it is the optical equivalent of thermodynamic work, rather than heat.
As well as by thermodynamic arguments, one can show that étendue is conserved very generally in passive optical systems using the Hamiltonian / symplectic geometry formulation of Fermat's principle. I discuss this in more detail in this answer here. Fermat's principle means that propagation through inhomogeneous mediums wherein the refractive index (whether the material be isotropic or otherwise) varies smoothly with position corresponds to Hamiltonian flows in optical phase space; mirrors, lenses and other "abrupt" transformations as well as smooth Hamiltonian flows can all be shown to impart symplectomorphisms on the state of the light in phase space, which means that they conserve certain differential forms, including the volume form. All these things mean that the volume of any system of rays in optical phase space is always conserved when the rays are transformed by these systems. This is the celebrated Liouville Theorem.
There is a clunkier but more perhaps accessible way to understand all this in optics. We linearize a system's behavior about any reference ray through the system, and write matrices that describe the linear transformation of all building block optical systems. It may seem that linearization involves approximation and thus something not generally true, but hold off with this thought - this is not the case. This is the Ray Transfer Matrix method and these linear transformations describe the action of the system on rays that are near to the reference (the "chief ray") ray of the light field in optical phase space. These matrices act on the state $X$ of a ray at the input plane of an optical subsystem:
$$X = \left(\begin{array}{c}x\\y\\n\,\gamma_x\\n\,\gamma_y\end{array}\right)\tag{1}$$
where $(x,\,y)$ is the position in the input plane of the ray, $(\gamma_x,\,\gamma_y)$ are the $x$ and $y$ components of the direction cosines of the ray's direction and $n$ is the refractive index at the input plane at the reference ray's position. The quantities $n\,\gamma_x$ and $n\,\gamma_y$ are the optical momentums conjugate (in the sense of Hamiltonian mechanics) to the positions $x$ and $y$; interestingly, they are indeed equivalent (modulo scaling by the constant $\hbar\,\omega/c$) to the $x$ and $y$ components of the photonic momentum $\hbar\,\vec{k}$, where $\vec{k}$ is the wavevector, but this fact is an aside. (1) describes our points in optical phase space.
Now we write down the matrices that represent the linearized action of every optical component we can think of; for example, a thin lens (representing the paraxial behavior of an optical surface) will impart the matrix:
$$\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\-\frac{1}{f}&0&1&0\\0&-\frac{1}{f}&0&1\end{array}\right)$$
If you study this matrix's action, you'll see that it transforms a collimated beam into one that converges to a point a distance $f$ from the input plane.
A key point to take heed of is that this matrix has a determinant of 1. If you go through the list of every possible passive optical component, you'll find that the matrices that describe their paraxial behavior all have unity determinant (they are unimodular). So they all multiply together to give a unimodular ray transfer matrix of the overall system built from these subsystems chained together.
This determinant is the Jacobian of the general, non-linearized, non approximate transformation that the system imparts on any system of rays. We can imagine recalculating a matrix from every neighborhood of every chief ray in an arbitrary, noninfinitessimal volume of rays in phase space. These matrices will all be unimodular, so what we've shown is the key idea:
The Jacobian $J(X)$ of the transformation wrought by any passive optical system is unity at all points $X$ in phase space.
This means that if we work out the volume $\int\mathrm{d}V$ of a system of rays in phase space, then the volume of their images $\int\,J(X)\,\mathrm{d}V$ will be exactly the same for any passive optical component. So we've shown the exact version of the law of conservation of étendue for optics without needing the full machinery of symplectic geometry and Hamiltonian mechanics.
Best Answer
There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law.
First method:
1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power incident on surface of earth about 1 kW, you get 600 W (consistent with the claimed 580 W from your question). If this is incident on an area of 1.14 cm$^2$, the power density is $\frac{580}{0.000114}\approx 5 MW/m^2$. A perfect black body of 1.14 cm$^2$, with this power incident on one side, and insulated perfectly from the other side, would be able to reach a temperature $T$ such that
$$\Phi = \sigma T^4$$
so:
$$T = \sqrt[4]{\frac{5 \cdot 10^6}{5.67\cdot 10^{-8}}} = 3000 K$$
(round numbers...)
However, if you tried to heat a disk (twice the area - no insulation on the back) the temperature reached would drop by $\sqrt[4]{2}$, to T = 2600 K (~2330°C). Note that the temperature doesn't drop all the way to 1500 K (~1230°C) - this is that 4th power in the Stefan-Boltzmann law shows its - ahem - power...
2) A second method would look at "how big The Sun looks" from the vantage point of the focus. When you are at the focal point of the mirrors, you "see" that it is as large as the satellite dish. That means that the heat flux increases, compared to the flux from The Sun, by the ratio of apparent areas. Which is actually the same thing as saying "you appear to be a lot closer to The Sun and can use the inverse square law to determine how much more power per unit area you experience".
Now The Sun looks like a disk that is 0.5° diameter; and with the dimensions given, your dish is equivalent to a disk with a diameter of 86 cm ($\sqrt{102\cdot73}=86.3$) and the focal distance is 138 cm (which I derived from the size of the focal spot, which is really an "image" of The Sun).
At a distance of 138 cm, a disk of 86 cm diameter "looks" 69x larger than The Sun - so it has an apparent area that is 4800x larger than The Sun - it therefore "feels the heat of 4800 suns". That is remarkably similar to the answer we got before, despite a different approach (but not really, if you look closely). So we will again get the same estimate for the temperature you can reach.
This second approach helps us understand that to get to higher temperatures, we need "The Sun to look even bigger" - that is, we need a larger dish or a smaller focal length. Making the focal length smaller will only work if the individual mirrors in the dish are small compared to the size of the focal spot - otherwise they will result in significant blurring of the focus and thus lower the power density. Simply increasing the size of the mirror array doesn't increase the power density - only the quality of the focus does. In principle the best you can do is create a giant 3D array of mirrors that make it look like The Sun is "everywhere" - a full $4\pi$ array would in principle give you the light of 50,000 suns (10x more than this mirror). Such a device would illuminate an object from all sides, and the temperature of that object would be (by the same equation as above) $\sqrt[4][10]$ x greater, or 5500 K (~5230°C). This is very close to the temperature of the surface of The Sun - and that is no surprise. If I had not used rounded values along the way (since there is a lot of estimating going on) I might have expected the answer to be 5776 K (~5500°C) - the surface temperature of The Sun, and the theoretical limit of such a device. So 5500 K (~5230°C) is "close enough for estimating".