Figuring out stuff is pretty easy. Explaining it can be the real bitch. It is easier some times to let people find their own starting point to figure things out for themselves. Look at the boxes. Let 32C be constant and -1.9C be constant. How much energy would it take for both to remain constant? Where would excess energy go?
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.