H Jh kT

Solving for the ice thickness, we find h = KT~ Tg) (6.43)

This determines the ice thickness once Tg is known. The physical content of this statement is that the ice thickness grows until it is just thick enough to let through the amount of heat delivered to the bottom of the ice. Ice can exist in equilibrium if Tg < Tf, and the ice thickness approaches zero as Tg warms to the freezing point. Increasing the heat flux or decreasing the thermal conductivity would also thin the ice. The fact that it is the harmonic mean of the thermal conductivity that appears in thiss equation has important consequences. The harmonic mean gives greatest weight to regions of small conductivity, with the consequence that even a thin layer of very small conductivity can drive the harmonic mean to small values, and hence require thin ice. In particular, a relatively thin blanket of snow can hold in the heat diffusing through the ice, and cause the ice to thin dramatically - all other things being equal. The following exercise gives some feel for the numbers.

Exercise 6.9.1 (a) The mean geothermal heat flux on Earth is about 0.03 W/m2, and it was not much greater back in the Neoproterozoic. Snowball Earth simulations indicate tropical ice surface temperatures on the order of 230K for a globally glaciated planet. How thick do you expect the tropical ice to be if the entire layer has the thermal conductivity of ice? How would the thickness change if you added a one meter layer of snow at the top?

(b) In a situation like the present Earth with a great deal of open water, ocean currents can deliver heat to the bottom of an ice shelf or sea ice layer at rate much greater than the geothermal heat flux. Suppose that ocean currents transport a mean flux of 2W/m2 to the bottom of a polar ice layer which has a surface temperature of 250K. How thick is the ice in equilibrium?

(c) The atmosphere of Europa is so tenuous that it has essentially no radiative effect, so the temperature is determined by a balance between absorbed solar radiation and blackbody emission. Suppose that near the equator the annual mean absorbed solar radiation is 5 W/m2. The internal heat flux for Europa is not well constrained. Compute the equilibrium ice thickness assuming a heat flux of 0.01, 0.1 and 1 W/m2.

The second example in the exercise shows that a rather small amount of heat delivered to ice by oceanic heat fluxes can be very effective in melting back sea ice. This is true because essentially all the heat so delivered can be used in melting. In contrast, if one delivered the equivalent of 1

W/m2 to high latitude regions by heat transport in the atmosphere, a great deal of the heat would be lost by radiation to space, and only a small portion would actually be usable for melting ice.

Note that although the insulating properties of a snow layer thin the ice if the ice surface temeperature is held constant, this effect is offset by the fact that snow has a considerably higher albedo than ice, which reduces the surface temperature. Moreover, the low thermal conductivity and low density of snow allow it to cool very rapidly at night, particularly when a stable inversion forms. The daytime warming tends to be not so extreme, owing to stronger turbulent heat fluxes in a neutral or unstable surface layer. This process tends to reduce the daily-mean snow surface temperature, which again has a thickening effect on the ice.

Now let's bring in the effects of mass loss at the surface. In equilibrium, mass loss from the top must be balanced by freezing at the base. The latent heat of fusion adds to the flux delivered to the base of the ice. Hence, if all the mass sublimated from the surface is carried away and precipitated elsewhere, Fi is replaced by Fi + (Lf /Lsub)FL, where Lf is the latent heat of fusion and Lsub is the latent heat of sublimation. This makes the ice thinner than it was our previous estimate. In conditions cold enough to form ice we are generally in the weak evaporation limit, so Fl is small; moreover, for water Lf /Lsub = .118, which brings down the additional flux even more. The net flux delivered to the base is on the order of 1 W/m2 or less in typical conditions, and does not significantly increase the ice surface temperature. Thus, we can get the ice thickness including sublimation by simply replacing Fi in Eq. 6.43 with the modified basal heat flux. The latent heat flux FL can be estimated using the results of Section 6.8 in the weak evaporation limit. For Fl = 1W/m2, which is typical of the cold Snowball Earth tropics, the basal flux increases from .03W/m2 to .148W/m2, which thins the ice by a factor of 5. Clearly the effect of sublimation on ice thickness is very significant, and can allow the tropical ice on Snowball Earth to be much thinner than it otherwise might have been, though even with sublimation taken into account the ice is over 300 m thick when the mean surface temperature is 250K. With stronger sublimation, as would happen as CO2 increases and the ice surface warms towards the freezing point, the effect is even more pronounced.

If some regions are experience net ablation of ice through sublimation, others must be experiencing net accumulation, since the vapor that sublimates must ultimate precipitate out somewhere else. What happens to ice thickness in regions of net accumulation? To a point, accumulation at the surface can be balanced by melting at the base. However, the only supply of heat available to melt ice at the base is the geothermal heat flux, and a flux of .03 W/m2 can melt sustain a melt of only 3 mm of ice per year. If the accumulation exceeds this tiny rate, then if no other process intervenes the ice thickness increases until it freezes to the bottom. In reality, the generation of regions of thick ice would drive a flow of ice into regions where it can ablate by sublimation, thickening tropical ice and thinning polar ice. Things that flow are beyond the scope of this book.

We'll now consider one last variation on the theme of ice thickness. In the preceding calculations, it was assumed that all solar radiation that was not reflected was absorbed at the surface of the ice. In reality, some radiation will penetrate into the ice and be absorbed in the interior. If the penetration is significant, this can have a powerful effect on thinning the ice, because the low thermal conductivity of ice means that heat buried in the ice has a hard time getting out, and therefore accumulates until a considerable degree of warming has been achieved. To model this process, we modify the steady state thermal diffusion equation to allow for internal heating, which we represent as the vertical gradient of the downward solar flux Fq. The equation becomes

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