How long will it take 1/2 a gram of liquid mercury to evaporate
completely at 68 degrees Fahrenheit? How long will it take at 70
degrees Fahrenheit?

max...

It would take 8928.5714285714285714285714285714 hours for 
.5 grams of mercury to evaporate completely at 68F degrees,
assuming an evaporative surface area of 1 square centimeter.

This is based on the volatility of mercury, which is the 
measure of how quickly it evaporates, and is 0.056 mg/h.cm²,
or .056 milligrams per hour per square cm, at 68F or 20C:

"Given that the volatility of mercury at a temperature of
 20°C is 0.056 mg/h.cm²"
From a document titled 'Code of Practice Mercury Housekeeping',
page 13, on the United Nations Environment Programme - Chemicals:
http://www.chem.unep.ch/mercury/2003-ngo-sub/Eurochlor/Code%20of%20practice%20Hg%20housekeeping%20(Env%20Prot%2011).doc

The chart just below this statement, in the link above, 
indicates how mercury evaporates more quickly in proportion
to the increase in surface area exposed, so .5 grams in a
single globule won't evaporate nearly as fast as the same
amount dropped onto the floor and scattered into multiple
tiny globules. It illustrates how 1000 tiny droplets, vs 
one droplet of equal volume, will increase both the 
evaporative surface area and the rate of evaporation by
a factor of 10.


Now 68F = 20C, and 70F = 21.1111111C, which is a very small
increase in terms of the volatility of mercury, so this 
increase in temperature would take only slightly less time.
I am unable to determine the exact difference since I was
unable to find a formula for the volatility of mercury at
the temperature of 21C. Such formulae are typically given
a standard format based on normal room temperature. Out of
8928 hours, I would guess it would take just a few hours
less to evaporate at 70F.


The change in the volatility of mercury is nicely illustrated
in the charts shown on page 3 of this PDF file, titled
'Release of Mercury From Broken Fluorescent Bulbs' from the
website of the State of New Jersey:
http://www.state.nj.us/dep/dsr/research/mercury-bulbs.pdf

As you can see, there is no significant difference in
the volatility of mercury between 40F and 60F. At 85F,
however, you can see that the evaporation occurs almost
twice as quickly as it does from 40-60F.


I hope this satisfies your needs in asking the question,
as I sincerely doubt that a more accurate answer can be
provided.


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Additional information may be found from an exploration of
the links resulting from the Google searches outlined below.

Searches done, via Google:

mercury evaporation
://www.google.com/search?q=mercury+evaporation

"mercury evaporates"
://www.google.com/search?q=%22mercury+evaporates%22

mercury "rate of evaporation"
://www.google.com/search?q=mercury+%22rate+of+evaporation%22

"volatility of mercury" temperature
://www.google.com/search?q=%22volatility+of+mercury%22+temperature

very helpful

The mercury at 70 degrees would evaporate in 20/21.1 of the time of
the evaporation at 68 degrees, i.e., 8459 hours. Just for your
interest, perhaps, the danger of breaking neon lights is that the
little drop of mercury in the tube (which emits uv light when a charge
is run through the light, and that uv light is absorbed by the
phospholuminescent interior of the glass) is totally evaporated, and
mercury causes neurological damage when ingested/inhaled.

One minor point. The surface area is assumed to be constant whereas
the evaporation will cause it to deminish over time. This means that
rather than being a linear function, the evaporation will be an
exponential. Even so taking the 8kHrs as the half life of the oringal
sample, the danger is still obvious. Ingestion via fish contaminated
with methyl mercury presents more of a danger than inhalation and
should be avoided especially in young children.

I don't think the 20/21.1 ratio is accurate.  If this were true, there
would be an error at 0 degrees C.

Good point taiga. It would all have to be converted to kelvin. Also,
good point kiningersoll-- as the drop gets smaller, it evaporates
faster relative to its mass (because of surface area/volume ratio). So
even my temp ratio is only an approximation.