# If Mars has a frozen ocean, how can it be used by settlers?

Some of what I've read theorises that Mars' oceans evaporated and blew away, but it seems to me that (potential) observed water flows and exposure of (potential) water ice under the surface dust by the Phoenix lander suggests the liklihood that a lot of that ocean is still there, aside from the polar ice caps.

So my question is, ignoring the innumerous hazards of going to and settling Mars, how could we harvest that ice and turn it into water to be used for drinking and fuel, with the least amount of cost and effort? Could there be an automated way of doing this so that robots could do it whilst humans stay on Earth for a while initially?

The Martian atmospheric pressure is around 600 Pascals. At that pressure the boiling point of water is around 0°C. The average temperature on the surface of Mars is -55°C. The melting point of water is also lower, but this usually changes by a much smaller fraction so we'll assume you want to get it to just below boiling. So you only need to increase the temperature by 55°C to be able to store it in liquid form (and then slowly raise its temperature and pressure thereafter). The specific heat capacity of water is around 2,000J/Kg.K. That means you need 2,000 Joules of energy to raise 1kg of water by 1°C. So to raise 1kg by 55°C you need: 2,000x55 = 110,000 Joules. On top of that you need to provide the energy for the enthalpy of fusion. for water this is 334,000J/kg, bringing your total value up to 444,000J/kg. Now you just need to know how much water you need per day.

Aside from that though there's one more useful calculation; the size of solar array required for melting 1kg. The solar flux at Mars is around 557W/m2. Assuming you get that for around half the day (if you have tracking panels) then you're looking at 557 * 12 * 3600 (Martian day is pretty much the same as Earth) = 24,062,400 Joules/m2/day. Solar panels efficiencies are pretty bad, around 14% is typical: 24,062,400 *0.14 = 3,368,736. That means you need 444,000/3,368,736 = 0.1318 m2/kg/day of solar array to melt ice.

This is achievable by robots, but giving an example is a little to broad to answer here.

I hope my thesis has this density of references.

• 1) even water below the boiling point will evaporate, just a bit slower (ig in the open), 2) no nead to transform sunlight into electricity to melt water, conversion into heat is more efficient than PV, 3) Melting water is another 3335,5 kJ/kg
– mart
Jun 3, 2014 at 11:16
• @mart For point (2), that depends on the albedo. If pure ice reflects 90% then going through the electricity would melt it faster. The alternative is to somehow darken the ice to increase its absorption, but that would need to be done continuously, so may be harder to automate. Not sure. Jun 3, 2014 at 14:07
• @Mart, very true on point 3 - my bad answer amended now. For point 1 I was making a worst case assumption for the amount of energy required. For point 2 changing the sunlight into electricity allows for easy storage/transport of the energy. Jun 3, 2014 at 14:22
• I do not hope your thesis has this density of references to Wikipedia ;-) Dec 17, 2015 at 22:20