Green World Trust

[Pajaholic]

I've seen many claims for the "albedo effect" that as ice melts the underlying water that's exposed absorbs more energy than did the ice, hence speeding up the melt. Now I've seen lots of figures for the albedo of ice and snow from about 0.85 for new snow to about 0.5 for old white ice with the majority seeming to be in the region of 0.6 to 0.8. However, the pieces that say so much about the albedo effect never seem to give values for the albedo of the water the melting of the ice uncoveres. This intrigued me so I metaphorically rolled up my sleeves, grabbed my metaphoric spade, and started digging.

First, it's my understanding that the albedo of a body is the fraction of irradiation falling on that body that is reflected. So an albedo value of 0.85 means that 85% of the energy falling on the body is reflected, which implies that 15% is either transmitted or absorbed. Also, the albedo of snow and ice changes with conditions. One paper says that albedo increases under overcast conditions because there is less infra-red for the ice to absorb. Unless otherwise specified, albedo usually relates to broadband radiation (i.e. from above UV to below IR).

It took some searching to uncover anything about the albedo of water, but I did find a couple of references. One piece told me that albedo depends on the angle of incidence, that it's about 0.02 for energy arriving normal to the surface, and >0.9 for oblique angles. The other from the European Centre for Medium-range Weather Forecasts says that diffuse albedo for water is 0.06 and that direct albedo is given by the cosine of the zenith angle. We can quickly demonstrate that this is at best an approximation by considering a zenith angle of 90° - i.e. directly overhead. Cos(90°) = 0. So if the albedo was actually the cosine of the zenith angle, there'd be no reflection. Yet if we stare downwards into calm water we see a reflection of ourselves. Now that reflection is visible light (which water does not readily absorb), but it's still energy. That said, I suspect the cosine of the zenith angle is a good enough approximation to derive some reasonable values.

Direct albedo for water is Cosine of zenith angle. So at midday at the equinox it's:
latitude 50° ~0.77, 60° ~0.87, 70° ~0.94, 80° ~0.98, 90° ~1.00

Since the tropics are at 23.5°, and sin(x) = cos(90°-x), the direct albedo at midday on midsummer's day will be given by sin(latitude - 23.5°). For example, for the seventieth parallel (i.e the approximate start of the Arctic ice sheet) direct albedo will be sin(46.5°), which is approx 0.73. Note that we've only considered midday here and at all other times albedo will be greater because the sun is lower.

Difuse albedo is specified in the reference as 0.06, which is equivalent to direct albedo from a zenith angle of 86.56°, so I suspect that's taken normal to the surface - i.e. measured with the sun directly overhead. Real world, there'd be a strong bias towards the zenith angle (evidenced by soft shadows under overcast conditions) and so albedo must be greater than 0.06 under almost all conditions. I've calculated the diffuse albedo for even irradiation from all directions and inclinations at approx 0.64, so the actual albedo must be between this value and that for direct irradiation at the zenith angle.

Now the Arctic ice sheet extends from about the seventieth parallel. At that latitude my calculations suggest that the albedo for water will be greater than that for the ice that melted to uncover it for most of the time and as you get further North the difference is more pronounced.

However, I might have missed something so comments are more than welcome.

p/.

Calculation of albedo of water for even irradiation
If the direct albedo of water is the cosine of the zenith angle then the reflected irradiation arriving at some angle x is given by cos(x) and the total reflected radiation from all angles from 0 to pi/2 radians (0 to 90°) in one plane by the integral from 0 to pi/2 of cos(x).dx (=1) for incident radiation of pi/2.

Now this is for one quadrant of one plane. So if we rotate rectangular areas of height 1 that represent the total incident and reflected radiation about an axis, we get cylinders that represent the total incident and reflected radiation from all directions and angles.

Total incident radiation for one quadrant of one plane can be represented by a rectangle of base pi/2 and height 1, giving a cylinder volume of 2pi * pi/2 = pi^2.

The total reflected radiation for one quadrant of one plane can be represented by a rectangle of side and height 1, giving a cylinder volume of 2pi.

The total albedo for even irradiation is thus the reflected radiation divided by the incident radiation, which is 2pi/pi^2 = 2/pi ~ 0.64

[Richard111]

I have a book from the library: "Environmental Science" by A.R.W. & J.M. Jackson. I just looked up "albedo" in the index. Not mentioned! Sheesh!

Interesting post. Thanks. Kinda stuff I haven't thought about much.

Do you have information on IR penetration of water? I have read on other blogs that it is only about 2 microns. Barely through the surface film. I did some scuba diving in my distant youth and remember the sea was quite cold only a few meters down even if visibilty was good. Light penetration must be quite good remembering the amount of plant and fish life I could see way below me.

[Pajaholic]

Absorption is logarithmic with depth and depends on the absorbance at the particular frequency. I did see a formula for this on the 'net while researching albedo, but I have no reference. In the meantime, and if you have the time, you might find this from Google Books interesting since the title is "Light Absorption in Sea Water". It's only a preview, but the chapter you can access might be relevant.

p/.

[Lucy Skywalker]

Pajaholic I have some simple thoughts. I think of the photos of the Earth from space, you can get them up to date from the side panel at Watts Up, complete with land, clouds, sea, and misting-out at the edges. I saw a beautiful, beautiful one with the aurora corona but can't think where for now. My thought is, the seas look black or nearly so, and the edges of the globe show how far the whitening effect kicks in. I think that the flat picture is a good representation of the albedo factors in the right proportion though the mathematical proof of this eludes me. It just looks that way. So we have a picture of mainly near-black seas with a little peripheral whitening, either from cloud, haze, or sea-at-an-angle, and perhaps it doesn't matter which it is, or maybe it does.

Just my pennyworth.

I'll probably be off this forum until Tuesday.

[Pajaholic]

Any photo taken from space can only detect what is reflected back towards the camera. Reflections off white ice are scattered, so reflecting to a wide angle, whereas reflections off water more-closely follow the "mirror principle" - i.e. the angle of reflection is equal to the angle of incidence. Due both to the curvature of the planet, and to the positions of both the light source (usually the sun) and camera, the majority of energy reflecting off water is usually directed away from the camera. So it's unsurprising that water looks blacker.

However, if you were to take a photo of the planet with the sun lighting the planet from behind, I suspect you'd see evidence of large reflection off water where the zenith angle is equal to the camera angle.

p/.

[Richard111]

Thanks for the book link Pajaholic. It is indeed interesting. On page 54 there is a subtitle Fundamental Infrared Absorption Bands and it points out the IR band for water is 2.3µm to 8µm which is MUCH LESS THAN THE MAIN CO2 BAND OF 15µm.

So that makes it quite clear to me that CO2 "back radiation" has NO EFFECT ON AT LEAST 70% OF THE PLANET'S SURFACE. Add in all the worlds lakes and rivers and wetlands and wet areas that have just been rained on... just how much effect is this 0.0385ppm level of CO2 really having? Looking out my own windows I say absolutely non at all.

[hqmonaro]

The earths poles tilt towards the sun in summer, giving 24 hour daylight. The angle is changed significantly during those times. And during those times, the albedo of water is still going to be less than ice or snow.

[Richard111]

[Pajaholic]

[Pajaholic]

[Richard111]

I have read jorge's post and I don't think there is any conflict.

I have no argument against the properties of CO2, only in the interpretation of the effects. In this instance albedo of water is not relevant so I suppose I am OT. But, I still claim CO2 backradiation, which will be mostly in the vertical plane, does not add heat, or delay heat loss from the water surface. I accept that this radiation can increase the the transition phase for water from liquid to gas. This does not add heat energy, but assists in the COOLING of the water surface as vaporisation proceeds. The latent heat in the water vapour is moved to altitude by convection. At the dew point temperature the vapour changes state and releases that heat into the atmosphere and also reduces the air density resulting in an accellarated lapse rate of temperature upwards. I doubt much of this heat gets back to the surface but I can claim to have observed, many times, how night time temperatures on cloudy nights do not reduce at anywhere near the rate that occurs on clear sky nights.

Are there any experiments on the clear sky rate of cooling on any spot on the planet that shows a steadily reducing rate of cooling as CO2 in the air above increases?

[Pajaholic]

[hqmonaro]

[Pajaholic]

[hqmonaro]