I'll answer the question and THEN answer a far longer unasked but vital portion of the question which you need to answer to be able to perform the task.
As laptop2d says, what you need is "resistance" - but this may not be a resistor in the normal sense. If you have a PV panel of known performance and can guarantee constant full sun then a properly sized resistor is useful. But if sun levels vary and clouds scud and the sun moves in the sky, as all happens, then any one resistor will not do anything like the best possible job.
A PV panel rated at 12V Vmp (Voltage at maximum power) and 10Wmp will produce current Imp in full sun of
I= W/V = 10/12 = 833 mA
and require a resistive load of
R = V^2/P = 12^2/10 - 14.4 Ohms.
A standard 15 Ohm resistor rated at say 20W would do well. Or say 2 x 4.7 Ohm 5W which are readily available.
If obtaining resistors is difficult then "resistance wire" from warious sources will do the job. This is typically "Nichome" (NiCr) wire and can be found in toaters, fishtamk heaters, electric heaters/radiators and more.
A toaster element is typically rated at 1000 W at 230V or 110 VAC so at 12V a suitable length of the same wire operated at full power would be rated at 12/230 x 1000 = 52 Watts or 12/110 x 100 = 109 Watts. By using longer lengths of wire so it does not (literally) glow in the dark, lower wattages can be obtained. If you can obtain a heater intended for low voltage operation (say a fish tank heater) then it may have better characteristics.
Heating wire in electric blankets is lower wattage per length (as 1000 Watts in a bed tends to make the user rapidly feel and/or be incinerated) so may be more suitable. This may have an insulating plastic coating making it suited for underwater use.
But, for shortish periods, standard resistors with a paint on insulation of choice on the live parts will probably be OK.
Resistors can be operated with almost all of their wiring out of water or can be clamped to a mostly immersed metal plate so they are above water level.
So - that's the easy part. BUT - is 10 Watts enough?
The answer is maybe yes, but it's a highly qualified maybe.
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Is 10 Watts enough?
10 Watts is a very low level of power for desalination.
As others noted, use of a thermal solar heater would be far cheaper per energy input.
1 Joule = 1 Watt.second
So your 10W panel outputs 10J/second in full sun or 36,000 j/hr in full sun
Water requires 4.2 Joules per cc per degree K (or C) temperature rise.
To heat 1cc of water from say 20C ambient to 100C (NOT boiling) requires
4.2J x (100-20) = 436 Joules.
We don't know where you are located - and it matters greatly - but I'll assume Northern hemisphere and not too too far North. So you may get 3 to 4 SSH (sunshine hours - hours of equivalent full sun) per day. 1 sun = 1000 W/m^2 and is the insolation level that performance for most PV panels is specified at.
In a day a 10W panel will provide W x s/hr x SSH = 10 x 3600 x 3 say = 108,000 Joule.
From above, raising 1cc of water from 20C to 100C (NOT boiling it) takes 436 J. So in a day your 10W panel could raise 108,000/436 = 247cc of water from 20C to 100C.
100C = "Boiling point" BUT if you are going to boil the water you need to add the "latent heat of vaporisation" to convert liquid water to gaseous water (steam). The LHoV of water at 1 atmosphere pressure = an additional 2258 J/g !!!. So to boil 1cc of water starting at 20C requires 436J to heat it to 100C and 2258J to then boil it = 2694J - say 2700J.
So your available 108,000J will boil 108,000/2700 = 40CC of water, starting at 20C. You can easily recover the 436J by using a counterflow heat exchanger ((essentially 2 pieces of Copper pipe in close thermal contact) giving 108,000/2258 = 48CC / day - not a very useful gain.
Worse, recovery of the LHoV is not possible without "clever tricks" as the source has to be hotter than the destination. This can be achieved by pressurising the source (boiling side) or lowering the pressure on the "sink" (incoming side) but this adds nasty complexity.
All is not lost.
If you look at a psychometric chart you'll note the rise of kg water/ kg of air as temperature increases. Most such charts are for HVAC applicatuions and do not go to temperatures near boiling point.
So - dealing with water at higher temperatures but below boiling is more liable to be successful.
This graph shows how much water is contained in air at varying degrees of saturation at various temperatures. The Y axis is in % water/air which is (probably) convertible to kg of water per kg of dry air under the same conditions.
Assume you can heat water and produce air at about the same temperature above the water surface. As an example, at 35C and 100% rh 1 kg of dry air will "support" about 55 grams of water, whereas at 25C it will contain about 30 grams. By cooling saturated air from 35C to 25C you can obtain about 25 grams of water. 1 kg of air occupies about 0.83 cubic metres or 830 litres of air.
As you use a larger differential you will get increasing amounts of water per litre of air handled. There are many sources of equations to derive these curves available on the internet - so you don't need to rely on graphs or wonder how they were derived.
To heat water from 25C to 35C takes 42 J/cc or water. You can recover about 30/55th = 55% of that by cooling the 35C air based on the above discussion - probably more if your RH falls below 100% in the cooler (as is likely).
Every 1cc heated recovers say 0.6cc.
Every cc heated takes 42 J
You have 108,000 J available.
That allows 108,000/42 x 0.6 =~ 1500cc/day with 3 SSH.
Getting there.
By examining the water gains obtained with differing Thot and Tcold (and some playing) you can achieve a cycle that is most efficient for your purpose.
The "counterflow" heat exchanger - A MAJOR "tool":
The above guesstimation assumes total energy loss on cooling.
If some or most of the energy could be reused when the wet air is cooled significant gains could be had.
A 'tool' which does this superbly is the counterflow heat exchanger. As the name suggests - the flow of hot but cooling and cool but heating fkuids are in opposite physical directions. This is an essential and amazingly effective aspect of the design.
Many leads to counterflow heat exchangers here and here
This immensely simple device is one of the wonders of the engineering world and far less appreciated than it should be.
In the diagram below if can be seen that hot fluid enters at left and leaves at right at a cooler temperature, while coolfluid enters at right and leaves at left at a higher temperature.
From here
This "counterflow" means that the differential temperature across the exchanger 'wall" is low and relatively constant. Cold is opposite cold and hot is opposite hot.
If warm air is passed through a counterflow heat exchanger, most of the heating energy can be recovered in the incoming air or heating the source water. The air system can be "closed loop" with air picking up vapour from the source and cooling and condensing at a location where liquid water flows into the outlet.
Air volume:
To get the above throughput a relatively immense amount of air must be used.
From above, you can expect about 25 grams of water per kg of dry air.
Or about maybe 30 g per cubic meter. To get say 1500g you need to heat and cool 1500/30 = 50 cubic metres = 50,000 litres of air.
To do this in say 3 hours (longer is available) you need a flow of 50000/3/3600 = about 5 litres/second. This flow rate has to pick up moisture from the water and then be cooled and then recycled. Not a trivial task.
An easier method:
If you use thermal rather than PV energy input it is easy to get a vastly larger energy input.
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MANY references
Wikipedias - Psychrometrics
Psychrometric calculator
Lots of references to Psychrometric charts and many charts and related discussions
Best Answer
If your supply voltage source is 25 volts then you need to use this to calculate the resistance needed.
\$Power = \dfrac{volts^2}{resistance}\$
Resistance = \$25^2 / 21.25 = 29.4\$ ohms
The error you made was to assume that your power supply will always supply 2 amps irrespective. In reality if the voltage is specified then hanging a load on the output will take precisely the current needed as defined by ohms law i.e. 0.85 amps in your case.