How Much Heat is Needed to Vaporize Away all the Oceans on Planet Earth

Some Traumatic Thoughts

Running from Emergency Medical Care to Astronomy

How Much Heat Energy is needed to vaporize away all the oceans on Earth?

Facts, Calculations, Results & Figures

An Accident I Saw

Just this morning when I was driving along Jalan Batu Caves for a meeting, I saw a motorcyclist skidded on a sandy patch just as he was negotiating a bend at the traffic lights to join Karak Highway.

I felt ethically and morally duty bound to stop to help. I have neither gloves nor water in my car, and I am one who is very obsessed with washing my hands afterwards touching any drop of blood smeared on my hands after attending a bleeding patient. It is not just my fear of HIV or hepatitis B infections when handling infected blood, but my fear of even common skin infections such as Staphylococcal to Streptococcal infections. Of course it was silly of me as the risk of contacting these infections by touching a patient and not having gloves or water to wash my hands afterwards is very remote.

Infections caused by Streptococcus bacteria for instance can lead to streptococcal sore throat, also called strep throat. Then we also think of the risk of tetanus infection causing lockjaw, not to ourselves as doctors and health-care providers, but to the accident victim from the manure-rich soil particularly if the wounds are deep and penetrating. Of course this can easily be managed with an anti-tetanus serum (ATS) injection afterwards.

A Strep throat acquired in childhood for instance may lead to rheumatic fever and rheumatic heart disease later in life. It begins with a history of strep throat from streptococcal infection. The infection leads to bacterial endocarditis, a dangerous infection of the heart's lining or valves. The heart valves, especially the mitral valves are affected. It causes the blood to regurgitate from the ventricles back to the atrium at each ventricular contraction as the heart valves can neither fully open nor close. But that’s another story. Moreover, it is highly unlikely that I will ever get rheumatic mitral valve incompetence just attending to a bleeding patient. There is no connection at all I admit, or even there is, the probability statistically speaking is very remote indeed, perhaps a chance in a million. It is just my obsessive thoughts about getting infections from touching blood that separates my mind from reality.

But risk of all types of infections is always there. The pathogens are always in the air, water, soil, and even in your hands, food and clothes. Any health care provider working in a hospital is at risk. This includes even the patient, not just the doctor, the nurse or the paramedics. Infections that are acquired while a patient is in a hospital are referred to as nosocomial infections; a term derived from 'nosos' the Greek word for 'disease'. Nosocomial infections are diseases that we, as physicians and heath care professionals, give to our clients. Hospitals and clinics are places where sick people go with the expectation that they will get better. Unfortunately, there is a risk that clients may become infected because of their visits to these places.

After examining the motorcyclist, I found he had some lacerations on his arms and legs. There was no fracture as far as I could assess clinically. He is aware of the surroundings, respond to my questions. In short his LOC (Level of Consciousness) is a full 15 score on the GCS (Glasgow Coma Scale), and there was no evidence of neurological deficit, which would have been indicative of neuro-spinal injuries. His airways were clear, breathing normally, and his haemo (circulatory) dynamics were not compromised. He was not in shock After examining and assessing him through a primary and secondary survey, I was of the opinion it was not much of a life-threatening situation for him.

He had just some moderate lacerations and tear, moderate bleeding, which could well managed with pressure dressings, and later the lacerations closed with 6-8 sutures, and a prophylactic ATS injections. These could be done at any small private clinic, and there was no necessity of calling an ambulance to spare the ambulance from more urgent and emergency needs elsewhere. I could have easily driven him in my car to any nearby clinic (not necessary to an appropriate state-of-art hospital) to get some decent dressings for his wounds, if not for his bike, which he had to leave behind. So I proposed he pick up his bike and ride along, and that I shall follow him behind in my car just in case …?.

But to me, it was only a very minor event.. As I drove behind him, a far, far, greater fear ran through my mind. It was a thought how precious water is to our lives on this planet. Even little drops of water to wash my hands from blood and infected biological fluids suddenly became so obsessively urgent to me at that moment. What, I asked, as it ran through my mind, if the entire Earth is depleted of water. Could the oceans boil off, or could all the water in the oceans seep beneath the ocean floors if the floors opened up into a vast chasm generated by a colossal tectonic drift? What happens if the Greenhouse Effect comes to pass, and the ocean waters boiled off as superheated steam enveloping this entire Earth with a mantle I thought?

My mind transfixed from a very minor medical emergency in the street (a minor duty) to a much more fearful thoughts of grotesque dimension in astronomy which is my major interest of all the sciences, except in human physiology, pharmacology, human nutrition, and medical research.

Here what I fantasize. Read on, and follow my (science fiction) logic below:


Introduction to an Astronomical Nightmare

(Some basic data and information needed
for this Nightmare):

Radius of the Sun = 696 000 km
Volume of Sun = 4/3  r 3 = 1.4123 x 10 18 km 3

Equatorial Radius of the Earth = 6 378.140 km
Volume of Earth = 4/3  r 3 = 1.0868 x 10 12 km 3

Sun / Earth Volume Ratio = 1.3 million: 1
Hence the Sun is 1.3 million times bigger than the Earth in volume

Mass of Sun is 1.9891 x 1033 grams (1.9891 x 10 30) kg. This makes it 330,000 times more massive than the Earth. The Sun destroys itself at a rate of 5 x 10 9 kg sec –1 to 3.827 x 10 26 watts (3.827 x 10 26 joules of energy per second)

The Sun can maintain this current output of energy for about 5,000 million years more

The source of this solar energy is the proton-proton cycle in which hydrogen nuclei are converted to helium nuclei. Today, after more than 45000 million years of fusion in the core the concentration by mass of H 2 has been reduced from 75 % to about 35 %. Fusion is accompanied by a mass loss, which is converted, into energy

In the process of generating this vast amount of energy (4 x 10 26 watts), the heat and light is spread out into space. A tiny part of this heat is intercepted by Earth from an average distance (semi major axis) of 149.6 million km

In so doing, Earth receives 135.3  2.0 milliwatts / per cm (1.94  0.03 calories / cm 2 / minute. To put it in another way in the SI System, this is equivalent to 1353 watts per sq. metre. To express it another way, it is 1353 joules per second per square metre (1353 J s –1 m -1). This is called the Solar Constant.

Since the Radius of Earth is 6 378.140 km (6378 140 metres), and since the surface area of a sphere is 4r2, the surface area is 5.112 x 10 14 m –2 (approximately).

This means that, for every square metre of the Earth’s surface, 1353 watts x 5.112 x 10 14 m –2 = 6.917 x 10 17 watts (joules per second) will fall on it. This is spread out evenly day and night as the Earth rotates. The poles may be much colder than the equator for 6 months a year, but for the remaining 6 months of the year, it will receive the Sun’s energy continuously even at “night” Thus nearly about the same amount of energy is distributed over the Earth’s surface evenly over a long time frame.


The specific heat (also called specific heat capacity) is the amount of heat required to change a unit mass (or unit quantity, such as mole) of a substance by one degree in temperature

The specific heat capacity (abbreviated C, also called specific heat) of a substance is defined as the amount of heat energy (measured in Joules) required to raise the temperature of one kilogram of the substance by one Kelvin (K). The SI unit for specific heat capacity is the joule per kilogram Kelvin. Specific heat capacity is therefore heat capacity per unit mass.

The Kelvin (K) scale is a thermodynamic temperature scale, in which the lower fixed point is absolute zero, and the higher fixed point is the triple point of water at exactly 273.15 K. The melting point of ice based on the triple point is 273.15 K. Temperatures (t) on the Celsius scale can be converted to temperature T on the Kelvin scale: T/k = t/0C + 273.15. However, for practical purpose to simplify our calculations we shall still use the Celsius scale which most of us can understand better. Morever, we are not dealing with nano-physics of the world of atoms here.

The Specific Latent Heat of Vaporization is the amount of heat required to convert unit mass of a liquid into the vapour without a change in temperature

For water at its normal boiling point of 100 ºC, the latent specific latent heat of vaporization is 2260 kJ.kg-1. This means that to convert 1 kg of water at 100 ºC to 1 kg of steam at 100 º C, the water must absorb 2260 kJ of heat. Conversely, when 1 kg of steam at 100 º C condenses, it gives out 2260-kilo joules

Specific Latent Heat of ice = 3.4 x 10 5 J kg -1
Specific Heat Capacity of water (Cw) = 4.2 x 10 3 J kg –1 0 C -1
Latent Heat of Evaporation of water = 2.26 x 10 6 J kg -1
Heat needed to increase temperature of melted ice from 0 0 C to  0 C
= mil + miCw ( - 0)

There are approximately 400 glaciers and icebergs with a combined weight of (2.2067 x 10 19) kg

Density of ice = 917 kg/m3
Density of Liquid water = 1000 kg m-3

1 cubic km = 1000 m x 1000 m x 1000 m = 1 x 10 9 cubic metres
Since Density = Mass / Volume,
Therefore total mass of ice on planet Earth = density of ice x volume of ice = 917 kg x 24,064,000 cubic km x 10 9 (2.4064 x 1016 cubic metres) x 917 kg = 2.2067 x 1019 kg

(The above are the basic knowledge needed for my scientific nightmare. Now let me argue).

Average Temperature of All the Oceans:

The average temperature for all ocean waters is 3.51°C and its average salinity is 34.72 parts per thousand. For the ocean surrounding Antarctica (south of 55°), the average temperature is 0.71° and the average salinity is 34.65 parts per thousand. Of the major ocean regions, the North Atlantic is the warmest and saltiest (averages: 5.08°, 35.09 parts per thousand)

Source: Penguin

The Cold Dark Ocean Floors:

Sunlight cannot penetrate below a depth of about 660 feet, around the start of what's known as the bathyal zone (it ends where the water temperature drops to 4 degrees Celsius -- at about 6600 feet). Some fish and crustaceans at these depths are blind; other animals -- as many as half of the creatures in the deep oceans -- have become bioluminescent, producing their own light in specialized organs called photophores.

Without sunlight, there is no photosynthesis, and without phytoplankton to kick start the food web, animal life is sparse. Because of the scarcity of food in the deep sea, many fish have evolved bizarre adaptations to help them get what they can.

The greatest ocean depth has been sounded in the Challenger Deep of the Marianas, a distance of 10,294 m (35,798 ft) below sea level in the Pacific Ocean. It is located 338 km (210 miles) SW of Guam. It is the deepest at 10,294 metres (35,798 ft) known depression on the earth's surface. God only knows what lurks inside there.

Even the height of Mount Everest is only 8850 metres (29035 feet) high, which means the entire Mt Everest would be submerged into the Mariana Trench if it was placed there. We are unsure what are the temperatures of waters conceal in some of these awesomely deep trenches. Some of the ocean floors have vents and abyss where hot water may sprout out from underground volcanic activities. The hot water may dilute the remaining relatively cold masses of surrounding water. Then the hot and cold water may circulate around deep in the ocean floors. So the picture is very complicated. This makes calculations exceedingly complicated. Even if we do know all the information available, probably we may need a supercomputer to execute the equations containing all those ever-changing variables. But I think this is not necessary as we are looking at this planet as a whole, and not as a intricate changing fabric of physical variables. So we will use some oceanographic data that tells us that the average temperature of all the ocean waters on Earth is 3.51°C. We will adopt this figure for our calculations, and this estimate is not going to deviate very much from truth

Amount of Water on Earth

There is a lot of water on our planet. There are about 1 354 728 800 cubic kilometer (km3) of it. This water is spread over different kinds of places as given in table below:

Total amount of water in cubic kilometers (km3) (%)

Seas and Oceans 1 321 920 000 (97.57)
Ice-caps and glaciers 24 064 000 (1.77)
Deep groundwater 4 216 000 (0.31)
Shallow groundwater 4 216 000 (0.31)
Inclination and capillary water 68 000 (0.005)
Rivers 1 360 (0.0001)
Salt or brackish lakes 108 800 (0.008)
Fresh water lakes 122 400 (0.009)
Water-vapour in the air 12 240 (0.0009)

Total amount of solid water (ice) + liquid water = 1 354 716 560 km 3 This excludes all the water vapor in the air including all the clouds as they cannot be included in the calculation since they have already been “vaporized”

1 cubic km (km 3) = 1000 m x 1000m x 1000 m = 1 x 10 9 cubic metres (10 9 m 3)

We also need to realize that water fills ¾ of the Earth surface. So we can expect that ¾ of the heat delivered to Earth falls and warm up the oceans, the remaining heats up the dry land. So we may think we should not include this part of heat into the calculations? Are we right? Wrong! All the heat is now trapped, and the entire Earth is a heat trap. It is now like an oven in which a cake is being baked. It is now the total heat distributed and available everywhere, and not just some parts going into the oceans. The scenario then we will be entirely different from what is now where Earth receives the balance between what it receives, and what escapes back into space. This heat balance just equates so that Earth is not overheated at the moment. This heat balance also allows us to enjoy the cooler sea breeze in the day, and the warmer land air being discharged into the sea at night due to convection air currents. But this wind-heat dynamics over land and sea will no longer work when everything on Earth-land and oceans are all heated up share the same temperature. There may not be a difference in heat gradients by then. So the heat from the land through this heat-transfer mechanism is also heating up the oceans and seas.

The heat dynamics would be different unlike now. If it would still be the same as it is now, I would have taken that into consideration, and divided that up into the calculations. But I have not, because I cannot foresee there would be heat exchange between land and sea by then. The oceans will be actively boiling be then, and superheated steam will be everywhere – land sea, atmosphere, and some diffusing even into interplanetary space. The whole Earth is just heated up uniformly like an oven. There is just no temperature difference at all.

Stage One of Calculation (Ice on Earth):

Density = Mass x Volume
Since density of ice is 917 kg per cubic metre (917 kg m 3)
Therefore the total mass of ice on planet Earth = Density x Volume = 917 kg x (2.4064 x 10 16) m 3 = 2.207 x 10 19 kg
Hence, to melt 2.207 x 1019 kg of ice into water from 0 0 C – 0 0 C (same temperature), requires (3.4 x 105) x (2.207 x 1019) = 7.5 x 10 24 Joules

Stage Two (Changing All the Ice into Liquid Water):

To heat up 2.207 x 1019 kg of melted ice from 0 0 C to 100 0 C, requires;
Specific heat capacity of water (Cw) x weight of water in kg

= 4.2 x 103 J kg-1 x 2.207 x 1019 = 9.3 x 10 22 Joules

Stage Three (Changing All the Liquid Water into Gaseous Water):

To boil off 2.207 x 10 19 kg of water at 100 0 C completely into steam, requires:

Specific Latent Heat of Vaporization (of water) x Weight of water = 2.26 x 10 6 J kg-1 x 2.207 x 10 19 = 4.98 x 10 25 Joules

Therefore, to boil off 2.207 x 10 19 kg of ice-caps + glaciers + permanent snow completely into steam requires: (7.5 x 10 24) + (9.3 x 10 22) + (4.98 x 10 25)

= 5.74 x 10 25 Joules

Liquid Water (The Oceans, Seas, Rivers and Lakes):

The Average Temperature of all the oceans in the world is 3.51 0 C (see information earlier). Hence to raise this temperature to just boiling point (100 0 C) requires a temperature difference of 100 – 3.51 = 96.49 0 C

There are all in 1 330 652 560 cubic km (km 3) of liquid water on this planet. This is an estimated value. So we will round this up to 1330652560 x 10 9 = 1.33 x 10 18 cubic metres. (m 3). It includes all the oceans, seas, lakes, underground water, rivers, etc. Since the seas and the oceans make up the most waters (97.57 %) and they are at an average temperature of 3.51 0 C, the minor sources are assumed to be at the same temperature. This excludes all the ice. The calculation for ice has to be done separately, but the water vapour cannot be included in the calculation as it is already in the “gaseous” form and they are not going to be condensed again as we are going to heat up the entire planet after this.

One cubic metre of water = 1000 kg at 0 0 C, but we assume this is also true at 3.51 0 C. The difference is so insignificant that we can ignore it as we are dealing with astronomical figures.

Hence to bring the temperature of all the liquid waters (1.33 x 10 18 cubic metres) from 3.51 to 100 degrees Celsius requires:

(.1.33 x 10 18) x (4.2 x 10 3) J kg –1 x 96.49 0 C x 1000 kg =

5.39 x 10 26 Joules


Now to Change All the Liquid Water into Steam:

Now we need to change all that water (1.33 x 10 21 kg) at 100 0 C completely into steam

(1.33 x 10 21) kg x (2.26 x 10 6) J kg –1 =

3 x 10 27 Jolues

Finally, to vaporized all the masses of ice caps, glaciers, icebergs, permanent snow, plus all the liquid water on Earth completely into steam requires:

(5.74 x 10 25) + (5.39 x 10 26) + (3 x 10 27) =

3.6 x 10 27 Joules

This means it is 3100, 000 000, 000 000, 000 000, 000 000 (31000 million, million, million, million joules of heat energy from the Sun). In the American English language this Nightmare Figure is called

3.6 Octillion Joules

Just How Much is Lost?

The Sun destroys itself at a rate of (5 x 10 9) kg per second. In the process, it generates 3.827 x 10 26 joules of heat and light per second. We have seen it requires 3.6 x 10 27 joules of energy to vaporize all the oceans, seas, ice, lakes, river, and all the waters from Planet Earth (see calculation above). In order to supply this amount of energy to boil off all the waters, including all the glaciers and all the permanent ice on Earth, the Sun will have to destroy (5 x 10 9)  (3.827 x 10 26) x (3.1 x 10 27) =

4.05 x 10 10 kg of matter

This will yield a stupendous 3.6 octillion joules.

The Second Nightmare!


Imagine 3.6 octillion joules of heat are needed to vaporize all the water on Earth until it is bone-dry to the ocean floors. And how long will that take?

(3.1 x 10 27)  (3.827 x 10 26) =

8 seconds

Eight seconds, that’s all it takes if Earth were to be thrown into the Sun. And 1.3 Million Earths can jolly well be thrown into the Sun – A Colossal Lake of Fire that Will Burn with A Searing Temperature of

15 million °C

This nuclear fire will burn if not throughout
All Eternity, but for at least another

5,000 Million Years More

Save Your Souls if you ever get inside!

The Sun’s heat falls off inversely as the square of its distance. Hence, if Earth were to stay put in her orbit at a respectable mean distance of 149.6 million km away around the Sun, then that will take:

(3.1 x 10 27) joules  (6.917 x 10 17) joules per second (watts) = 4 479 768 786 seconds

But one Solar Year = 365.25 = 60 x 60 x 24 x 365.25 = 31 557 600 seconds. Hence it will take 4 479 768 786  31 557 600 =

141.95 years to empty all the oceans

The Third Nightmare

This means, not a drop to drink, let alone a drop to wash my hands

Save my OCD (Obsessive Compulsive Disorders)

This obsessive No Water thought is astronomically far, far too much for me to bear.

Now anyone cares for a drink? Get it now before 150 years

Beginning From

The Era of
Total Green House Effect

An Era When No Heat Can Escape
From Planet Earth


This is a Major TRAUMA psychologically for me with my OCD and washing hands obsession at the touch of every drop of blood!

But that chap who skidded off his motorbike this morning was relatively was just a minor drop of blood in the ocean compare to this psychological trauma. It would be easier to handle a much bigger traumas and MVA (motor vehicle accidents) in our streets. But not this one when all the oceans all boiled off.


A Warning Comment

Not So Easy As That!

Having said all that, the scenario of what will actually happen, and the time it will for all the water to disappear from the surface of Earth is not a picture as easy to paint from some simple calculations as that. The picture is going to be far more complex. The time taken, and the amount of heat needed may differ as much as 10 fold or more (or less) ? Many things can happen which no one can predict. The only way to know is to wait until it happens, and then take measurements. That is the only actual practical way to monitor the progress. Using updated measurements every now and then is the only approach to evaluate and reevaluate the status again and again, and only after much observation can we forecast and project the trend theoretically using statistical analysis. There is no easy way out. That’s the only way to tell. We can’t assume that 150 years is all that is needed to dry up all the oceans after a total green house effect has taken place? It is not as simple as that.

However, the laws of nature and physics have to be obeyed no matter what the scenario. They will not change, and they will never change even if Earth came to an end. The amount of heat requires to melt ice, the amount of heat needed to raise a certain amount of pure water from a certain temperature to another temperature, and the amount of heat needed to boil off boiling water completely into steam under a certain a pressure are all governed by physical laws. These values are all fixed, and they can never change. It is on these physical laws that we are basing our calculations. That is provided other confronting factors such as changing pressures; water-vapour equilibrium, concentration of the solute, etc don’t interfere, and may alter the initial physical constants. So our estimates are based on simple straight forward physic that other factors within the system don’t change or interfere. If they don’t, then those results holds true, accurate, and predictable. But they probably will not be as far as we can see.

Samples of many of the things that can happen to greatly alter the calculations are given below:

The Current Status of the Atmosphere:

An ocean of air envelops the surface of the Earth. It is estimated that this mantle of air that surrounds the Earth weighs some 5000 million, million (5 x 10 15) tons which exerts an atmospheric pressure (force / surface area) of 1013 mill bars or 760 mm of mercury Since air can be compressed, it is not surprising to find that most of the mass of air is concentrated at the lower altitudes. A quarter of the mass of the atmosphere lies below 2,000 metres, half of the mass below 5,000 metres, and around 75 % or three quarter at an altitude of around the height of Mt Everest, which is 8850 metres (29035 feet).

Despite the mass of the atmosphere, it is actually just only 0.3 % of the mass of the water in the ocean, so we can imagine what happens if all the water and ice on the surface of Earth were to evaporate and mixed up with the air above.

We can of course easily calculate out the additional volume being added One kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and converted into steam and maintained at 100 0 C, then for every kilogram of it, it will occupy a space of 1.673 cubic metres, and exert a pressure of 101.3 kPa (1.013 bars). But that’s under ideal lab conditions.

Even using the present parameters of just 0.0009 % of all the water available on Earth floating in the air, the pressure, density, composition and temperature varies differently at varying attitudes. What if all the waters and ice were to evaporate and mixed up with the air, plus the vastly increased in carbon dioxide levels from total greenhouse effect? What will these parameters be, and how would they affect and change the atmospheric scenarios? Even then, we are only assuming that the temperature will be maintain at 100 0 C and no higher. The Sun’s energy being trapped into the altered atmosphere is not going to stop there. The continuous penetration of solar energy into the altered atmosphere will continue to expand the super-heat steam-air-carbon dioxide mantle to a state where no scientist can predict.

What then will be the nett effect on the balance of temperatures, pressures, densities, layers of atmosphere, composition among others? Even right now it is impossible to give a definite upper limit of our atmosphere. For instance, the “lapse-rate” meaning the rate of which temperature decreases with increases altitude above sea level is 1.6 0 C for every 300 metres. There maybe temporary “inversion” or increase in temperature at low level. But generally the lapse rate does not vary very much. At about an altitude of about 11 km the lapse rate dropped to an almost constant value of – 55 0 C. We see these changing values even for our normal and “stabilized” atmosphere currently. What if the whole scenario is upset by entirely new dynamics, with much more heat plus all the oceans waters added in.

At the present moment we have the lowest layer (troposphere) where all the normal clouds and weather occurs, and where commercial jets fly. After this, comes the stratosphere where the ozone layer is, and where supersonic, hypersonic and probably surveillance, spy and military jets cruise. This stratosphere extends up to some 30 km up before coming to the ionosphere laying some 150 –250 km above the Earth’s surface. This is the layer of ionic discharge where radio signals are reflected back to Earth.

Above the ionosphere lies the exosphere, which lies with the border of interplanetary space. It lies between 720 upwards to some 2,500 km where it merges with the interplanetary medium. Hence it is not possible to give a definite answer to the upper boundary to the atmosphere. There is just no sharp border. The mass and density decrease until it merges with the density of interplanetary space.

So what would the picture like if all the waters on Earth just evaporate away? How far will the atmosphere be? As said earlier, one kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and changed into steam, then an additional space of 1.673 cubic metres is needed, and this will exert a pressure of 101.3 kPa (1.013 bars). This is provided it is maintained at 100 0 C. But it is not ?

Then again we are assuming that it is clean fresh water from the oceans all the time as it is being boiled off. We forget as the oceans are being evaporated off, and no rain can come down to replace the loss, the salt get gets more and more concentrated, and as it does so, boiling point also rises. This means more and more energy is needed as the ocean waters get more and more concentrated. This result is longer and longer time for the oceans to dry up since the Sun’s energy is constant per unit area. This affects the results of the simplistic calculations above. Of course in the initial stages, the melting glaciers will dilute the concentration of the salts. But this cannot last too long as the ice is only 1.77 % of all the waters on Earth. Soon that too will have to evaporate off.

Even if all the water is now boiled off, initially we may think it would be easy to use just straight logic on how much addition volume will be added to the air using the water-steam conversion data. We can even determine the addition thickness the atmosphere would be generated by the additional steam and even the new surface area over the steam-air mantle. Calculations like that are very simple and straightforward, but in reality, the problem is far from being as easy as that.

First of all the atmosphere is greatly going to continue to expand due to the continuous injection of heat from the Sun. The surface area of the new water-air mantle will continue to change, and so are all the parameters within. This cannot be worked out using straightforward mathematics. The problem is going to be very, very complex indeed, as it has to take into considerations other variables that cannot be handled by simplistic mathematical methods. Second, we do not even know how would all these superheated mixtures of gasses, water, including the heavier carbon dioxide have on the thermodynamics of all these gasses. Carbon dioxide is added only from just from the burning of fossil fuels, but much large qualities of them are also released from the oceans due decreased solubility by increasing ocean temperatures. We know the solubility of gasses in water decreases with increasing temperature, but what happens when additional vast quantities are released into the atmosphere, and also when the volume of water decreases as dissolved gasses in the oceans are also discharged into the atmosphere.

The vicious cycle will even be more disastrous and complex then we can think, let alone attempt to calculate. My imagination tells me nothing will be left over this hard and hot planet. Nothing volatile like superheated steam and highly energetic gases can remain. They will all escape the clutches of the gravity of Earth. The velocity of escape from Earth is currently about 10,000 meters per second. Any highly heated particle, whether air or water molecules, will exceed that speed and they will be thrown out into space. The Earth will probably be a dead world, devoid of any, not just more water, but air as well. It will be just like we see them in the rest of the other planets in our Solar System.

We may be wondering if we could then treat this Earth as a whole as if it was some kind of a black body emitting some electromagnetic radiation such as infrared. We might then apply the laws of thermodynamics of “black body radiation” to determine just exactly how much of this radiation heat managed to escape from the planet to be in thermal equilibrium with the surroundings Heat radiation enclosed in an isolated cavity such as within the walls of the atmospheric mantle with a temperature T may be used. We are thinking of a hypothetical Earth that emits such radiation in conformity with the classical thermodynamic system. But this works in our case where the atmosphere may undergo adiabatic expansion of the radiation. where T  V –1/3. From there we might even be tempted to work out Wein’s law, and Planck’s law of radiation. But can it apply? I am unsure, and I am not even going to try.

JB Lim
An Amateur Astronomer in Thought