I believe the reason is due to the solution trapping water molecules in a cage around it. The reason water has a high specific heat is because it can rotate freely around its center of mass, there is a large number of degrees of freedom that can randomly vibrate and rotate in the pure water. When you have molecules in solution, they trap several water molecules close to them in a lowest-energy stiff configuration, and these molecules are like a tiny rigid body where thermal motion is not possible, because the quantum of oscillation frequency is higher than kT. This reduces the specific heat by an amount directly proportional to the solute.
This is probably strongest with salt, since the charged ionic solutes will produce a very strong cage. I would expect the effect with alcohols to be weaker, sugars weaker still, since I think the charged groups are less charged in these in order.
As a physics problem in a textbook, you could get somewhat close. Both the water and the steel have a heat capacity that relates the amount of temperature rise that would accompany an input of energy. If you make a few assumptions, you can relate the two.
The problem with a real-world application is that those assumptions may be far from valid. The two biggest ones I see are whether or not the energy input to both substances is similar. Any part of the sunlight that reflects or goes through the substance does not contribute to heating. So if the two materials behave differently (glass jar? shiny metal?), then the energy input could be quite a bit different.
The other factor is the energy loss from the material. As uninsulated materials heat up, they release greater amounts of heat energy. That allows the temperature to drop and makes the calculations harder. As an example, a simple stovetop can boil a pan of water, but would never be able to melt steel. Instead all the energy it puts into a hot piece of metal would just be radiated away and the temperature would stop climbing. Your standard solar oven isn't very fast and they only have to reach a delta of a couple hundred K. Making a solar forge would be much more difficult, but if you get enough power, you can almost overwhelm the energy loss. A MIT Solar dish is able to heat things to very high temperatures.
Now, if you want to ignore those effects (say you have very, very good insulation so that you can ignore cooling for a bit), then you can just take the heat capacity equation and fill in some numbers. See for example Hyperphysics Specific Heat
$$Q = mC \Delta T$$
$Q$ is total energy, but here you want to relate the power of the beam over a period of time. So that can be changed to
$$P \text{ }t = mC\Delta T $$
You don't say how much steel can be melted in a minute, so we can't plug in the numbers for the mass ($m$) (I hope it's not too much), but the rest of it you have. The change in temperature for the steel will be around $1000 K$, $t$ is less than $60s$ and $C$ for iron is about $450 \frac{J}{kg \text{ }K}$. That gets you the (minimum) power of the forge when you solve for $P$.
Turn it around and solve for $t$ with the jar of water, but there the water heat capacity is $4182 \frac{J}{kg \text{ }K}$ and the $\Delta T$ will be about $75K$ (assuming it starts at around room temperature).
Best Answer
It's safe to approximate that all of the microwave's energy is deposited in the water. To some extent you will also heat up the mug and the walls of the microwave, but we will neglect that in favor of a bigger effect.
If the water reached 100 ºC in the microwave before the heating cycle ended, the heat after that was "wasted" by creating water vapor.
Suppose what you want is water at 80 ºC. You pour some room-temperature water into a cup and microwave it for 60 s. After 40 s its temperature reaches 100 ºC; after this its temperature stops changing and vapor begins forming. At the end of the 60 s you remove the boiling water and add more room-temperature water. Because the heat of vaporization for water is quite large compared to the heat capacity for the liquid, very little of the boiling water actually left the cup as steam in the microwave. Essentially the final third of the microwave's heat is wasted.
Now suppose instead you add enough water that it would take 75 seconds to boil, but only put it in the microwave for 60 s. Voilà: all of the microwave's heat usefully raises the temperature of the water.