[kragen]

it's easy to calculate how big the effect is with units(1)

    You have: stefanboltzmann tempC(20)**4
    You want: W/m**2
     * 418.76592
     / 0.0023879689
    You have: stefanboltzmann tempC(0)**4
    You want: W/m**2
     * 315.65782
     / 0.0031679874

so in the temperature range of interest you're radiating 300–400 watts/m², derated for your surface emissivity at the relevant infrared wavelengths. as i understand it, the temperature will drop until the heat emitted through radiation is counterbalanced through heat that seeps in through the insulation and through the stagnant, stratified air above your pan of water, and, probably more importantly, through greenhouse-effect radiation from the warm atmosphere and any opaque objects you have unfortunately included in your field of view

tens or hundreds of watts of cooling is quite significant indeed when we're talking about making some ice or cooling some food