I am so glad that you guys are starting to talk about "total loss water cooling systems." I think it is the answer for drag, and hill climbing cooling. Distilled or de-ionized water, perhaps with some corrosion inhibitor additive from boiler make up chemistry, and a surfactant (remember the drop of dawn dishsoap that I mentioned... one drop Vasili, one drop only!). Followed by a good spray with a water displacer like WD40 after the run. In a cycling road race, don't they hand water out to the riders? Some of that needs to then be dumped into the bikes cooling system. I like hollow frames for storing the coolant.
Lets throw some number out there:
Liquid Nitrogen 200 KJoules/Kg
R134a car freon 215 KJoules/Kg Note there is no real benefit to liquid nitrogen over R134a in the Heat of Vaporization race.
Ethyl Alcohol 855 KJoules/Kg Sure a lot better than the two above
Plain old Water 2260 KJoules/Kg ... makes me take notice!
1 lowly quart of water, or 1 Kilogram will absorb 2260 KJoules to vaporize!
Lets add to it the specific heat needed to raise the water from say 20 degC to 99 degC. Specific Heat of water, if I did the units right is 4.186 KJoules/Kg/degC Not too shabby either! So for 1 liter of water we add 331 KJoules
Our total ability for that one quart of water is 2260 + 331 = 2591 KJoules. Now for us who do not think in metric... oh the travesty... 1 Joule = 1 watt for one second. So that quart of water will take away 2,591,000 watt-seconds of energy.
Lets take that 20 minute ride to the top of the mountain. 20 X 60 = 1200 seconds
2,591,000 Watt-seconds can be harvested to turn our precious water to steam. If I want to do it all in that 1200 second ride to the top; I can magically absorb 2,159 watts each and every second in my quest for victory!! That's two kilo watts folks! At 80 % efficient motor; I can cool a motor that is running 10 KW input!
Now to hit submit, and then be

in 10 minutes when I learn of all the conversion errors I made from power to energy and back again...
PS: We know that none of this will be 100% efficient, some of the mist will be lost as water and not steam... so make it 50% and 5 KW input... that should do nicely!