‘Cheaper’….desalination eats a lot of power. This is causing a huge problem over a minor inconvenience.

The massic heat capacity of water is 4184 J⋅kg⁻¹⋅K⁻¹. To heat one 1 Liter (1 kg) of water from 30ºC to 100ºC it would take 4184×(100-30) = 2.929e5 J. We want 4 liters however, so we multiply that by 4 and get 2.929e5 J × 4 = 1,172e6 J To then turn that heated water into vapor it would require some more energy. The vaporization enthalpy of water is 4,066e4 J⋅mol⁻¹, and has a molar density of 1,80153e-2 kg⋅mol⁻¹(so 4 liters (4 kg) of water in moles would be 4 / 1,80153e-2 = 2,22033e2 mol), which means that to vaporize the four liters of water we would need 2,22033e2 × 4,066e4 = 9,028e6 J (I think I might have made a mistake here somewhere, because I don't think it would only need 8 times more energy to completely vaporize the water, compared to the amount of energy required to heat it, but I can't find the problem). So the total energy to heat and vaporize 30 ºC water would be 9,028e6 + 1,172e6 = 1.020e7 J

Let's take a 55x40x23 cm suitcase. And let's assume a solar irradiance of 1000 W⋅m⁻² (which is what this site says is a normal solar irradiance to be expected on a clear day on the equator). Let's assume three faces are exposed to the sun and all equally so (three faces receive 1000 W⋅m⁻² while the other three receive none, which would not happen since on a rectangular cuboid, like a suitcase, you can't have all three faces facing directly towards the sun). The box would be receiving (0.55×0.40+0.40×0.23+0.55×0.23)×1000 = 438.5 W, which means that over one hour (3600 s), it would receive 438.5×(3600) = 1,5786e6 J, which is less than the required 1.020e7 J (by almost an order of magnitude), so it wouldn't be possible to heat and vaporize 4 liters of water in an hour.

What am I missing?