Hint: The enthalpy of fusion is 334Joules/gram. The energy of an object at 4C, 277.15K is 5.67e-8*277.15^4=334.5Wm-2, a Watt per second is equal to a Joule per sec.
Why 21.1C? There are approximately 4.2Joules per gram required to raise the temperature of water 1 degree centigrade. 21.1-(-1.9)=23C 23C times 4.2=96.6Joules per gram. 96.6Joules per gram plus 334=430.6 Joules per gram, balanced by 334+4.2*(21.1-0)=422.6 Joules per gram. (430.6+422.6)/2=426.1Joules per gram for fresh water. The equivalent energy of an object emitting 426.1Joules/second is 294.5K degrees or 21.35C. Why 21.35C and not 21.1C? An object at a temperature of 21.1C, 294.25K would emit 425.06Joules per second if it were a perfect black body. Water emits 425.06/426.1=0.99756 or 99.756% of a perfect black body. Plus or minus a schoosh, determine the more exact changes in specific enthalpy per degree and you should arrive at the emissivity of salt water.
Why 425.06Wm-2 and not 390+/- the average surface energy? The atmospheric window allows approximately 35.4Wm-2 to leave the surface without impacting the atmosphere. 425.06-35.4=389.7 Wm-2. From the heat of fusion, 334Joules per gram plus 23.52, the enthalpy required to change water at 4C to -1.9C, equals 357.52. Add the 35.4 joules per second energy not interactive and 392.92 Wm-2. Again use the more exact values for properties of salt water and fresh water, and you will arrive at a more precise estimate of the surface energy in equilibrium.
CO2 and other radiant gases appear to impact the 35.4Wm-2 atmospheric window which varies to maintain the ice, liquid ocean and moist air enthalpy balance. A very interesting system of feed backs.
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