[Note: Please do not try this at home. Consider trying this at your friend's house instead.]
How much energy would be released from the water?
 |
| A visual representation of the main concept of this blog post,
added because rigorous and extensive studies conducted on Mr. Melby have
prooved that he is more likely to read a blog post if it contains pictures. Image courtesy of Kelly Chen. |
The first thing we need to specify is the thickness of the ice. The
NHL
uses 0.75" thick ice at 16 degrees Fahrenheit (or, as their website
says, "approximately 3/4" of an inch thick"), but their ice rinks sit on
top of concrete, not water. Other websites, such as the
Minnesota DNR,
recommend 4" of clear ice (or 8" of white ice, which contains air
bubbles that make it considerably weaker) over a lake for an activity
like ice fishing. Using this information, I decided that freezing the
top 10 cm of the ocean would create a layer of ice sufficient for ice
skating. Since ice is less dense than water, 10 cm of water would
actually create slightly more than 10 cm of ice. If we assume that the
ice created is pure and clear (i.e. no air bubbles, salt, etc.), we can
calculate the thickness of ice that would be formed. According to the
86th edition of the CRC Handbook of Chemistry and Physics, the density
of water at 0 degrees Celsius and atmospheric pressure is 0.9998 g/cm^3,
while that of ice under the same conditions is 0.9167 g/cm^3. This
means that the volume of ice is 0.9998 g/cm^3 / 0.9167 g/cm^3 = 1.091
times that of water, so 10 cm water * 1.091 = 10.91 cm, or 4.294" ice. This should be (more or less) enough to make a decent ice rink.
Now
that we've decided how thick our ice needs to be, we can figure out the
total volume of seawater that we need to freeze to create it. If we
know the radius of the Earth, we can use the formula for the volume of a
sphere to calculate the volume of this "shell" of water. However,
there's a problem:
since the Earth is not perfectly round, its exact radius depends on where you measure it. According to
NASA,
the Earth's equatorial radius is 6378.1 km, while its polar radius is
only 6356.8 km. Rather than use either of these values, I decided to use
NASA's value for the Earth's
volumetric mean radius,
which is 6371.0 km. (Of course, this assumes that the radius of the
Earth's oceans is approximately equal to that of its landmasses.) If the
outer radius of our shell of ice is 6371.0 km, then its inner radius is
6371.0 km - 10 cm = 6370.9999 km. We can find the volume of the ice
shell by subtracting the volume of the inner sphere from that of the
outer sphere, so since the volume of a sphere is 4/3 * pi * radius^3,
the volume of the shell is 4/3 * pi * outer radius^3 - 4/3 * pi * inner
radius^3 = 4/3 * pi * (6371.0 km)^3 - 4/3 * pi * (6370.9999 km)^3 = 51
006.4 km^3. However, we must compensate for the fact that the Earth is
not completely covered in water. According to the
USGS Water Science School,
approximately 71% of the Earth's surface is covered in water, so the
actual volume of our ice shell would be 51 006.4 km^3 * 0.71 = 36
214.544 km^3, or 3.621E19 cm^3.
Now that we know the
volume of seawater we need to freeze, we can find the total mass of the
seawater by multiplying volume by density.
The ocean's density varies with temperature and salinity, but it has an average density of 1.027 g/cm^3, so the seawater's total mass is 3.621E19 cm^3 * 1.027 g/cm^3 = 3.719E19 g seawater.
The exact mechanics of phase changes in solutions are not well understood at this time, so let's simplify the problem: we can get a pretty good approximation of freezing seawater by ignoring all of the salt in it and assuming that only the water molecules in the solution actually undergo the phase change. To find the mass of water in this amount of seawater, we need to know the salinity of the seawater itself. Since
seawater is around 3.5% NaCl
(the mass of the other ions, algae, fish poop, etc. in it is
negligible, so here, I am assuming that seawater contains only water and
NaCl) by mass, it is 100% - 3.5% = 96.5% water by mass. Therefore, the
mass of water that we must freeze is 3.719E19 g * 0.965 = 3.589E19 g
water.
The surface temperature of the ocean varies with latitude, but as its average surface temperature is 17 degrees Celsius,
I will assume that all of our water is at this temperature. Now, in
order to freeze it, we must first lower its temperature to its freezing
point and then allow it to undergo the phase change itself. The freezing
point of pure water under normal conditions is 0 degrees Celsius, but
the addition of salt (or any other solute) produces a phenomenon known as
freezing-point depression, so the freezing point of seawater would actually be slightly lower than that of pure water. As it turns out,
seawater that is around 3.5% salt by mass actually freezes at -1.94 degrees Celsius.
The
total heat released by the water as it freezes is equal to the amount
released as we lower the temperature of the water to -1.94 degrees
Celsius plus the amount released as the water freezes. We can use two
equations to model this scenario: Q = m * C * delta-T represents the
change in heat of a system as the temperature changes, while Q = n *
delta-H_fus represents the change in heat of a system as a substance
melts or freezes. (In these equations, Q represents heat flow, m
represents the mass of the substance undergoing a Wha change in
temperature, C represents the specific heat of that substance, delta-T
represents the change in temperature (i.e. final temperature - initial
temperature) that occurs, n represents the number of moles of the
substance undergoing a temperature change, and delta-H_fus represents
the heat of fusion of that particular substance. However, as the value
of delta-H_fus used in this calculation is given in J/g, not J/mol, n is
replaced by m here.)
Although the specific heat of a substance changes slightly as temperature changes, I will be using
C = 4.184 J/(g*degree Celsius) here, as this is the value listed for water at 17.00 degrees Celsius (although C changes slightly with temperature, this inaccuracy will be negligible for our calculations). Our change in temperature here is -1.94 degrees Celsius - 17 degrees
Celsius = -18.94 degrees Celsius. The
heat of fusion of water at 0 degrees Celsius is 333 J/g; however, as this describes the
amount of energy absorbed when water melts and we want the amount of
energy released when it freezes, the value of delta-H_fus that we will
use is -333 J/g (i.e. a value of equal magnitude but opposite sign).
Thus, the
enthalpy (i.e. heat) change produced by lowering the
temperature of our 3.589E19 g of water is m * C * delta-T = 3.589E19 g *
4.184 J/(g*degree Celsius) * -18.94 degrees Celsius = -2.844E21 J,
while that produced by actually freezing it is m * delta-H_fus =
3.5898E19 g * -333 J/g = -1.195E22 J. Our total enthalpy change is
simply the sum of these two values, or -2.844E21 J + -1.195E22 J =
-1.480E22 J. Thus, turning the oceans into a giant ice rink would
release 14 800 000 000 000 000 000 000, or almost
fifteen sextillion,
joules of energy from the water!
How much energy is fifteen sextillion joules, really?
To put that huge number into perspective (somewhat), freezing the surface of the ocean would release as much energy
- as is stored chemically in 4.932E17 g, or 5.414E17 cm^3, of unsalted butter (at 7.17 kcal/g and 0.911 g/cm^3). That is enough butter to fill Lake Tahoe (volume = 122 160 280 acre*ft = 1.507E17 cm^3) 3.6 times over!
- as would be needed to meet the 2009 electrical consumption of New York City for the next 68 500 years! (In 2009, NYC consumed 60 billion kWh, or 2.16E17 J, of electricity.)
- as would completely reacting 1.712E8 kg uranium-235. (1 kg of U-235 releases approximately 24 000 000 kWh, or 8.64E13 J, of heat.) As a point of comparison, Little
Boy, the atomic bomb dropped on Hiroshima during WWII, contained only
64 kg of U-235, less than 1 kg of which actually underwent nuclear
fission. The amount of energy released by our oceans would actually be equivalent to the amount that would be released by detonating almost 100 million Hiroshima and Nagasaki atomic bombs-- combined. (The Hiroshima bomb is estimated to have released 15 kilotons of TNT (kt), and the Nagasaki bomb, 21 kt.)
- as it would take to turn 3.589E19 g ice at -1.94 degrees Celsius into water at 17 degrees Celsius. Yes, that's a
very clever comparison.
What would releasing all that energy do?
Long
story short, we'd probably all die very quickly thereafter.
(Thereafter, hereafter... Meh, same thing, really, in this case.)
But to elaborate on our impending doom, let's assume that the phase change is near-instantaneous.
All of the heat that we just released from the water has to go somewhere-- into the surroundings. This includes both the rest of the ocean underneath the ice rink and the atmosphere above it.
A warmer ocean would negatively impact aquatic ecosystems, change global weather patterns, and result in higher sea levels, among other things. As you can imagine, this would exacerbate many of the issues already caused by global warming, such as the
development of more powerful storms,
destruction of cities and farmland due to flooding, loss of habitat, erosion,
coral bleaching (from loss of algae),
increased spread of invasive marine species and diseases, extinction, and polar ice shelf collapses. Well, that seems like a pretty bleak outcome.
What about the atmosphere?
The
specific heat of air, at just over 1 J/(g*degree Celsius), is quite a bit smaller than that of water. This means that adding the same amount of energy to both will increase the temperature of the air more. Regardless, releasing so much energy from the surface of the oceans would definitely increase the temperature of the atmosphere by a pretty significant amount, a change that would also be quite disastrous to the planet. For example, the
rate of photosynthesis increases with temperature until a certain point, after which it drops down rapidly to zero. So, if we were to heat up our atmosphere a little bit too much, almost every
biome on Earth would break down as the enzymes required for photosynthesis became denatured. Not surprisingly
[
citation needed], this would have some rather serious [
citation needed] implications not only for our own species, but also for pretty much all others on the planet.
However, what we haven't considered yet is that if we assume that no (or very little) heat escapes from the outer edge of the atmosphere into space, we are essentially considering an
isolated system. This means that there would be no difference in total amount of energy before and after the phase change, so once the oceans return to
thermal equilibrium, a process that begins to occur immediately, we would end up right where we started. This makes it rather difficult to judge the effects of heating up the rest of the ocean, as well as the atmosphere, because both temperatures start to return to normal fairly quickly (although, to be fair, transferring so much energy back into the ice may take a little while). However, it is probable enough that at least some of the many serious consequences of pulling such a stunt would actually occur that most people (a possible exclusion to this statement can be found
here) would decide not to do so.
But if you
were to try this and we didn't all die off as a result of it first... well, then that just might be a
hack cool* enough to get you into
MIT. And I do hear that people are looking for ambitious risk-takers [
citation needed] these days...
*Pun unintended but noted.
