10 sq meters of ice is floating in the Arctic Ocean in mid-Sept - 12 hours of darkness, 12 hours of diffuse sunlight at 50 watts/sq meter.

Emissivity of ice = .97
temperature of the top of the ice = -20 C (253 K)
Albedo of sea ice = .62

How much radiant energy is absorbed by the ice in the 12 hours of sunshine?
How much radiant energy is lost by the ice in 24 hours?

Absorbed energy = 10 m^2 * (1 - 0.62)*50W/m^2 * 4.32*10^4 s = ___ J

Radiated energy = 10 m^2 * (sigma) *(0.97)(253^4)*8.64*10^4 s = _____ J
where sigma is the Stefan-Boltzmann constant, 0.5669*10^-7 W/m^2*K^4

The absorption of infrared energy from the sky has not been accounted for. It varies with could cover.

"Emissivity" refers to an average value in the infrared region where the ice radiates away most of its anergy. "Albedo" is an average reflectance for visible sunlight.

Thank you for the quick answer!

So, absorbed energy (over the 12 hours of daylight) is (1-albedo)*(available energy at the surface) ... that is, what radiation is not reflected from the surface is absorbed into the ice .. That's clear.

"The absorption of infrared energy from the sky has not been accounted for. It varies with could cover."

Wouldn't the 50 watts/sq meter diffuse radiation be that infrared radiation coming back from the clouds you mentioned? The Stefan-Boltzman equation
sigma*emissivity*(temperature)^4 does assume the ice is radiating into clear (black) space, right? Or should I make some assumption about temperature of the cloudy skies we are radiating up towards?

On-line on other sites, I keep finding various mixes of the S-B equation: some with no second temperature^4 value at all (as you used it above), some with a second temperature^4 value subtracted (apparently for the "sky" temperature) but no emissivity value, and still others with a third term (1-emmissivity).

But the reason for the various differences in the same fundamental S-B equation are not given.

I know in the real world we are not dealing with a perfect "black body surrounded by a perfect vacuum" but isn't that difference between an gray body and a black body exactly what the emissivity factor is intended to specifically correct for?

By the way, I got 82.1 x 10^6 watts for the absorbed energy (50 watts/m^2 over 12 hours) into the ice, and 194.6 x 10^6 watts for the radiated energy (lost to space) over 24 hours.

Net? In September, at the time of minimum sea ice extents, under these conditions, the ice cools off every day.

To find the amount of radiant energy absorbed by the ice in the 12 hours of sunshine, we can use the following formula:

Energy Absorbed = Incoming Solar Radiation * Absorptivity * Area * Time

1. Incoming Solar Radiation: The sunlight in the Arctic has a power of 50 watts/square meter.
2. Absorptivity: This is the same as emissivity, so for ice, it is 0.97.
3. Area: The area of the ice is given as 10 square meters.
4. Time: In this case, it is 12 hours.

Now let's calculate the amount of radiant energy absorbed:

Energy Absorbed = (50 * 0.97 * 10 * 12) Joules

Simplifying the equation:

Energy Absorbed = 5820 Joules

Therefore, the ice absorbs 5820 Joules of radiant energy in the 12 hours of sunshine.

To find the amount of radiant energy lost by the ice in 24 hours, we need to consider both the absorbed energy and the energy lost through radiation.

Energy Lost = Energy Absorbed + Energy Lost through Radiation

The formula to calculate the energy lost through radiation is:

Energy Lost through Radiation = Emissive Power * Emissivity * Area * Time

1. Emissive Power: This is the power emitted per square meter by the ice. To calculate this, we can use the Stefan-Boltzmann Law:

Emissive Power = σ * (Temperature^4)

where σ (sigma) is the Stefan-Boltzmann constant (5.67 × 10^-8 J/(m^2·K^4)) and Temperature is the temperature of the ice in Kelvin. In this case, it is -20°C, which is 253 K.

2. Emissivity: As mentioned, the emissivity of ice is 0.97.
3. Area: The area of the ice is given as 10 square meters.
4. Time: In this case, it is 24 hours.

First, let's calculate the emissive power:

Emissive Power = 5.67 × 10^-8 * (253^4) Joules/second

Now let's calculate the energy lost through radiation:

Energy Lost through Radiation = (Emissive Power * Emissivity * Area * Time)

Energy Lost through Radiation = (5.67 × 10^-8 * (253^4) * 0.97 * 10 * 24) Joules

Simplifying the equation:

Energy Lost through Radiation = 2.16 x 10^7 Joules

Therefore, the ice loses 2.16 x 10^7 Joules of radiant energy in 24 hours.

To find the total energy lost by the ice, we need to add the absorbed energy and the energy lost through radiation:

Total Energy Lost = Energy Absorbed + Energy Lost through Radiation

Total Energy Lost = 5820 Joules + 2.16 x 10^7 Joules

Total Energy Lost = 2.16 x 10^7 Joules + 5820 Joules

Total Energy Lost = 2.16 x 10^7 Joules

Therefore, the total energy lost by the ice in 24 hours is 2.16 x 10^7 Joules.