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).