Dear Paul,
Thank you very much for your comment of my earlier article at:
http://scientificlogic.blogspot.com/search?q=how+much+heat+to+boil++off+oceans
I appreciate it very much. You have asked a very good
question, the answer to which I was from the very beginning tried to avoid
because I really do not have the answers.
My article was actually meant to be a very simple one for
the sake of scientific interest for ordinary readers. It was never meant to go
into the intricacies of various unknown reactions and inter reactions such as
whether or not the water will condense back to Earth when it boils off high up
into troposphere, stratosphere, mesosphere, the exosphere and beyond into the
magnetosphere, solar wind towards the Sun and outwards into the Solar System.
That was never my intention.
Mine article was just to ask a simple question as to how much
heat is needed to convert ice into water, and from water into steam till they
all boil off. I wrote it mainly for school science students and ordinary
readers just to generate an interest in popular science.
I naturally assume we will be dealing with water and ice when
brought to heat under ordinary atmospheric conditions, say in a lab or in the
kitchen when we boil water, or that some ice from our refrigerator.
If that was our intention, then using simple basic equations in
physics most people learn in school, then the question how much heat would be
needed if we were to boil off all the water on planet Earth will still hold
true and remains very valid.
But since you have asked, I have elaborate the article with a
little more additions which I tried to avoid as it was just meant for lay
readers. My aim is to make the complexities of Science, Mathematics and
Technology reachable to the lay masses in a popular way. It is so different
from publishing a research paper in a journal.
So let me stretch my initial article just by a little bit more.
Hopefully, it will answer your question. Many thanks again fort your
comments.
We know that in order to thaw a kilogram of pure ice under
ordinary conditions, the heat of fusion (melting or freezing) of ice is always
334 kJ /kg. Thus if we decide to melt 4.3 trillion (4.3 x 10 12)
tonnes or 4.3 x 10 15 kg of ice from Greenland, Antarctica, the
glaciers and ice caps then the heat energy required would be:
E (Energy) = 334 x (4.3 x 10 12) x 1000 = 1.436 x 10 18
kilojoules. (1 tonne = 1000 kg). This equation does not
change under ordinary physical conditions.
Similarly, the amount of heat energy required to change
temperature of one gram of liquid water by one degree Celsius (at 15 0
C) will require 1 calorie. This is equivalent to 4.184 joule. Thus if we were
to change 4.3 trillion (4.3 x 1012) tons or 4.3 x 1015 kg
of melted ice from 00 C to its boiling point at 1000
C under atmospheric pressure, then the amount of heat required will
be:
4.184 J x (4.3 x 1015) tonnes x 1000 kg x 1000 gm =
1.8 x 10 22 joules or 18,000,000,000 terajoules (TJ)
A terajoule (TJ) is equal to one trillion (1012)
joules.
In the same way, the amount of heat required to boil
off 4.3 x 1018 kg of pure water at 100 0
C under standard sea-level pressure = (4.3 x 1518) kg x
2260 = 9.72 x 10 21 Joules since it requires 2260 KJ of heat to boil
off 1 kg of water at 100 0 C under STP, This is the latent heat of
evaporation of water.
You might ask where then can Earth get the energy to boil off
all the waters from its surface? Is this possible?
There are only two possible sources.
·
Geothermal energy
·
The Energy of the Sun
We shall discuss these two possibilities shortly. But first
allow me to list out the content of this follow-up article.
Content of this Article:
1. The Hydrosphere
2. How much water
3. Global Ice Mass
4. How much heat ice to water
5. Temperature of Oceans
6. How much heat needed to boil water?
7. How much heat required from water to steam?
8. The enthalpy of vaporization x mass of water:
9. Geothermal Energy
10. The Sun
11. Mass of Sun
12. The Sun, a Main Sequence Star:
13. Energy Output of Sun
14. The Death of the Sun
15. Second life:
16. Geothermal Energy
17. Total Energy of the Sun
18. What happens if the Sun becomes a supernova?
19. The Milky Way Galaxy
20. What Happens if Sun Explodes
21. Supernova
22. Volume of Earth-Sun Sphere
23. Light time
24. Solar Energy Received:
25. Solar Energy:
26. Distance of Sun:
27. Theoretical Calculations vs Satellite Measurements:
28. The Total Energy of the Sun:
29. Calculated vs. Estimated Energy Output:
30. Gravitation Force
31. The Volume of the Oceans:
32. Vaporized Water and Waterless World: The Pull
33. Geosynchronous Satellite
34. Height of Satellite:
35. Speed of a Geostationary Satellite:
36. Why does a geostationary satellite not fall down?
37. The Separation of the Waters: A Cloud of Waters:
38. Clouds float, Rain falls:
39. Water Condensing:
40. Making it Easier:
41. A Narration Summary from the Calculations:
42. The Searing Scenario:
43. Scaling it down:
44. How much water then?
45. A Mini Supernova:
46. Gravity acts only…?
47. Neither would there be a merry-go-round:
48. A Barren Earth:
49. The Creation of Earth:
50. Chemical marriage of hydrogen and oxygen:
51. The Waters Above and Below:
52. Torrential Rains for A Millennium:
53. The Birth of the Oceans:
54. The Earth and the Oceans How much hydrogen and oxygen:
55. The Chemical Calculations:
56. How shall approach this problem?
57. Mass of Hydrogen atom: In Deep Space:
58. Background Radiation:
59. How Much?
60. What a Horrendous Volume
61. How Big was The Gas Cloud?
62. Light Time
63. Bigger or Smaller
64. How far between Sun to Pluto and back?
65. Genesis
66. The following equivalents were used in the calculations for the above article:
The Hydrosphere:
The hydrosphere described in physical geography is the combined
mass of water found on, under, and over the surface of a planet.
Scientists estimate total mass of the Earth's hydrosphere is
about 1.4 × 1018 metric tons (1.4 x 10 21 kg) or about
0.023% of the Earth's total mass. The Earth’s atmosphere contains about
20 × 1012 tonnes of water. Water has a density of one tonne of water
per cubic metre. Some 75% of the Earth's surface is covered by water.
This represents an area of about 361 million square kilometers, or 139.5
million square miles.
Since the total amount of water on Earth is estimated at 1.4 x
10 21 kg, let us assume there may be another 10 % of it unaccounted
for, hidden somewhere on the planet. This put the total amount to be about 1.4
x 10 21 kg + 10 % hidden amount = 1.54 X 10 21 kg
How much water:
The total amount of ice + liquid water + water vapour + 10% extra
allowance for hidden and unaccounted water on planet Earth is estimated to be:
1.4 x 10 21 kg + 10/100 (1.4 x 10 21) =
1.54 x 10 21 kg
Mass of Earth = 5.974 x 1024 kilograms
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 1024 kilograms - 1.54 x 10 21 kg
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 1024 kilograms - 1.54 x 10 21 kg
= 5.9726 x 10 24 kg
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 1024
kilograms - 1.54 x 10 21 kg
= 5.97 x 10 24 kg
Global Ice Mass:
It has been estimated that total global ice mass loss from
Greenland, Antarctica, glaciers and ice caps was about 4.3 trillion tons since
the turn of the century a NASA study found.
That amount has raised the global sea level by 0.5 inches (12
millimeters). How much ice is that? It would cover the entire United States 1.5
feet deep, scientists said.
How much heat: ice
to water?
Hence the amount of heat required to raise 1.54 X 10 21
kg of water from an average ocean temperature of 17 0 C to 100 0
C as given by:
Q =cm Δ t
(where, Q = amount of heat required, c =specific heat, m =
mass, Δ t = change in temperature)
= (100-17) x 4.184 (specific heat of water) x (1.54
x 10 21) kg x 1000 (joules into kilojoules) = 5.35 x 10 26
kilojoules (ok)
(The average temperature of all the ocean waters is 17 0 C).
Temperature of the
Oceans:
Water temperature in the deepest parts of the ocean averages
about 36°F (2°C). The average temperature of ocean surface waters = 17 degrees
Celsius.
The average water temperature worldwide was 62.6 degrees
Fahrenheit (17 C) according to the National Climatic Data Center.
Every summer the hottest places for water are usually the red
sea and the Persian Gulf. The Gulf of Mexico is pretty hot. However the water
at Abu Dabi in the UAE is usually the hottest. It can get up to 94/95 deg F (34
/ 35 deg C) degrees in summer. The coolest recorded ocean temperature was 15.17
degrees C (59.3 degrees F) in December 1909.
How much heat needed to
boil water?
A calorie is defined as an amount of heat required to change
temperature of one gram of liquid water by one degree Celsius.
Thus, the amount of heat energy required to change temperature
of one gram of liquid water by one degree Celsius (at 15 0 C) will
require 1 calorie. This is equivalent to 4.184 joule. Thus if we were to change
4.3 trillion (4.3 x 1012) tons or 4.3 x 1015 kg of melted
ice from 00 C to its boiling point at 1000
C under atmospheric pressure, then the amount of heat required will
be:
4.184 J x (4.3 x 1015) tonnes x 1000 kg x 1000
gm = 1.8 x 10 22 joules or 18,000,000,000 terajoules (TJ)
(A terajoule (TJ) is equal to one trillion (1012)
joules).
In the same way, the amount of heat required to boil
off 4.3 x 1018 kg of pure water at 100 0
C under standard sea-level pressure = (4.3 x 1518) kg x
2260 = 9.72 x 10 21 Joules since it requires 2260 KJ of heat to boil
off 1 kg of water. This is the latent heat of evaporation of water.
How much heat required from water to steam?
Now having reached all the waters to standard boiling point at
sea-level STP, the amount of heat required to boil them all off will require:
The enthalpy of vaporization x mass of water:
= (2257 kJ kg−1) x (1.54 x 10 21) =
3.48 x 1024 kilojoules or 3.48 x 10 27 joules
Hence the total heat requirements to change all the average
ocean and global waters estimated at 17 0 C into
ordinary steam at 100 0 C under normal sea-level pressure,
would be:
5.35 x 10 26 + 3.48 x 10 24 = 5.38 x 10 26
kilojoules (okay)
(heat of fusion of ice is 334 kJ/kg)
This is the latent heat of evaporation of water is 2257
kilojoules / kg, meaning, it requires 2257 kilojoules of heat to boil off 1 kg
of water at 100 0 C under STP,
In the same way, the amount of heat required to boil
off 4.3 x 1018 kg of pure water at 100 0
C under standard sea-level pressure = (4.3 x 10 18) kg x
2257 = 9.7 x 10 21 kilojoules (9.7051 x 10 24)
Joules since it requires 2257 KJ of heat to boil off 1 kg of water. This is the
enthalpy of vaporization of water, or the latent heat of evaporation of
water. This applies provided there is no change in the state of other
physical conditions such as changes in the pressure as the water is evaporated
off into the colder outer space re- condensation of the water back to Earth,
increase of heat to super heat the steam, etc.
However, if the total amount of water in the entire hydrosphere
estimated to be 1.54 x 10 21 kg were to be boiled off, then the
amount of heat required would be:
(1.54 x 10 21 kg) x (100-17) x
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 1024
kilograms - 1.54 x 10 21 kg = 5.97 x 10 24 kg
(Mass of Earth = 5.974 x 1024 kilograms)
(Mass of total waters = 1.54 x 10 21 kg)
The enthalpy of vaporization, sometimes called, the heat of vaporization or heat of evaporation, is the energy required to change a given quantity of a substance from a liquid into a its gaseous stage at a given pressure (often atmospheric pressure).
Normally it is measured at the normal boiling point of a substance example water into steam at 100 0 C.
The enthalpy of vaporization, sometimes called, the heat of vaporization or heat of evaporation, is the energy required to change a given quantity of a substance from a liquid into a its gaseous stage at a given pressure (often atmospheric pressure).
Normally it is measured at the normal boiling point of a substance example water into steam at 100 0 C.
Geothermal Energy:
Now let us answer the question where would it be possible to get
so much heat to boil off all the waters of the oceans and elsewhere hidden on
Earth? One nearest and tremendous source of heat naturally lies just below our
feet – in the Earth’s core.
The heat from the Earth’s core is extremely high. The
temperature of the interior of Earth is estimated between 5650 to
7,000 Kelvin.
The temperature of the interior of Earth is estimated to be
between 5650 and 7,000 Kelvin. Total heat loss from the earth’s
interior to the surface is estimated at 44.2 TW (4.42 × 1013 watts).
This is about 1/10 watt/square meter on average. This amount is about
1/10,000 of the energy from the Sun.
The total heat locked up in the interior of earth is estimated
at 1031 joules. But it only requires 1.8 x 10 22 Joules
to change 4.3 trillion (4.3 x 1012) tons or 4.3 x 1015 kg
of melted ice on Earth from 00 C to its boiling point at
1000 C under atmospheric pressure, and another 9.72 x 10 21
Joules to boil off a total of 4.3 x 1018 kg of water.
We have shown that the total heat energy required to boil off
all the ice + water on Earth is:
1.8 x 10 22 + 9.72 x 10 21 = 2.77 x 10
22 Joules
(2260 KJ of heat required to boil off 1 kg of water).
But the heat locked away as geothermal energy in the Earth’s
interior is 1031 Joules. This is a whooping 361 million times more
heat in the interior of Earth than it would be required to boil off every drop
of water from the Earth’s surface if the Earth were to crack open into many
parts (very, very unlikely), and all the waters on the Earth’s surface drains
right into the interior, or if all the geothermal heat were to suddenly leak
out onto oceans beds, seas, rivers, and lakes, and everywhere in the
hydrosphere.
Of course for the Earth to crack open to gush out all its
trapped heat, or for all the waters on the surface o=f Earth to gush into the
interior of Earth can never happen, but just for the sake of argument and
interest in imagination for this article, let us assume it happens.
If this possibility is just in a fringe of our imagination, let
us then consider the energy from the Sun.
The Energy of the Sun:
The Sun was formed from the collapse of part of a giant
molecular cloud of hydrogen and helium about 4.57 billion (4,570,000,000) years
ago. (1 billion is 109 or 100 million). The Sun pours out a
horrendous amount of energy each second, and it can continue to generate its
stupendous energy for another 5,000 million years. But before that, let’s
briefly talk something about the Sun.
Mass of Sun:
The mass of our Sun is 2×1030 kilograms. This is
330,000 times that of Earth.
The mass of the Sun (solar mass) makes up 99.86% of the total
mass of the Solar System. Hydrogen is the principal composition of the Sun. It
accounts for 75 % of the Sun's mass, the rest is mostly helium
As in most stars, our Sun is called a main-sequence star,
generating its energy through nuclear fusion by converting its hydrogen nuclei
into helium. In its core, the Sun fuses 620 million metric tons of
hydrogen into helium each second.
Such a reaction, known as a proton–proton chain occurs around a
whooping tune of 9.2×1037 times per second within the core of the
Sun. Four free protons of hydrogen nuclei) are being used in the conversion,
and about 3.7×1038 protons are being converted into alpha particles
of helium nuclei every second.
There are about 8.9×1056 free protons in the Sun, or
about 6.2×1011 kg of hydrogen nuclei are being converted each
second.
About 0.7% of the fused mass is released in the mass-energy
conversion, the rate of conversion is in the order of 4.26 million metric
tons per second, yielding 3.846×1026 W of energy. This is the
power of 9.192×1010 megatons of TNT each second.
The Sun is a Main
Sequence Star:
The Sun, like most stars in the Universe, is one the main
sequence stage of life.
Sun is middle-aged about 4.57 billion (4,570,000,000) years old
The Sun is about halfway through its main-sequence evolution. . At this
rate, the Sun has so far converted around 100 Earth-masses of matter into
energy.
The Sun will spend a total of 10 billion years as a
main-sequence star.
Energy Output of Sun:
Every second, 600 million tons of hydrogen is converted into
helium in the Sun’s core, generating 4 x 1027 Watts of energy, and
neutrinos. This process has been going on for 4.57 billion years since
its formation. But there isn’t an unlimited amount of hydrogen in the
core of the Sun. In fact, it’s only got another 7 billion years worth of fuel
left.
The Death of the Sun:
Approximately 6000 million years from now, the current chapter
of our Sun will come to a close. The Sun will run out of hydrogen fuel.
When this occurs, the helium in the conversion from hydrogen will
built up. This will make it unpredictable and unstable. It will then
collapse under its own gravitational pull. This immense pressure within the
Sun’s core as it collapses will cause the interior to heat up making it becomes
denser.
A Red Giant:
It will then begin to bloat up to become a red giant as part of
stellar evolution.
In the scenario that follows the red giant phase, extreme
thermal pulsations will cause the Sun to fling off its outer layers, forming a
planetary nebula. What remains after the outer layers are evicted is the
exceedingly hot stellar core. This will slowly cool and fade into a white dwarf
over billions of years. It will be a scenario of a low to medium mass
stars
The expanding red giant Sun will devour the orbits of Mercury
Venus, the Earth, and perhaps into the outer Solar System as well. The
intense heat of an expanding and dying red Sun will burn Earth and boil off all
the waters within. Nothing can survive then. The Sun will then end its life as
a main sequence star
Second life:
But that’s not the final end. When it enters its stage as
a red giant, there will be a built up heat and pressure in the heart of the red
giant. The red giant Sun will trigger off a another fusion, but this time it
will convert helium into carbon instead of hydrogen into helium.
It will continue this fusion for another 100 million years
until all helium fuel is exhausted. In its death breath, the envelope of helium
becomes unbalanced causing the Sun to pulse strongly. It will then blow off
several portions of its atmosphere into space; many of these pulses will reach
Earth and the outer planets taking all the other planets into the grave with
it.
Total Energy of
Sun:
The Sun is much farther away than what lies in the interior of
Earth below our feet. So we can expect to get much less energy. But is this so?
Let us have a look at some facts and figures.
If the Sun's current energy output is (3.839 × 10 26) Joules
per second, and if we assume that the Sun has a 10, 000 million (1 X 10 10)
year main sequence lifetime, then we can expect it to have a total
energy output equivalent to 3.839 × 10 26 J/s, multiply by (1 x 10 10)
x 31557600 = 1.2 x 10 44 Joules. This is a horrendously huge amounts
of energy beyond imagination.
(1 solar year = 86400 sec. a day x 365.25 solar days a year =
31,557,600 seconds).
Even then, how does this energy output compare with the energy
released by a supernova? Let us have a look.
Give or take a factor 10 the total energy output of a supernova is 10^46 J, one percent of which is visible light, and 99% neutrinos. The solar mass of a supernova is 10 – 100 times that of our Sun. So the total energy output of a supernova ought to be 10 – 100 times as great as that of the Sun should our Sun decides to become a supernova, namely between 1045 – 1046 Joules per second.
Once again, rest assured our Sun is not massive enough to become
a supernova. We need not worry.
If the Sun's current energy output is (3.839 × 10 26) Joules
per second, and if we assume that the Sun has a 10, 000 million (1 X 10 10)
year main sequence lifetime, then we can expect it to have a total
energy output equivalent to 3.839 × 10 26 J/s, multiply by (1 x 10 10)
x 31557600 = 1.2 x 10 44 Joules.
(1 solar year = 86400 sec. a day x 365.25 solar days a year =
31,557,600 seconds).
But what happens if the Sun
becomes a supernova?
Stars with a solar mass of ten or greater can explode into a
supernova as their inert iron cores collapse into an tremendously dense neutron
star or even a black hole.
The Sun does not have enough mass to qualify into a supernova.
Instead, in about 5 billion years just as its hydrogen fuel runs out, it will
enter a red giant phase. Its outer shell will expand as the last remaining
hydrogen fuel in its core is consumed.
The Giant Sun Expanding:
The red giant Sun will steadily expand until the core
temperature reaches around 100 million Kelvin, The core will begin
producing carbon. Intense thermal pulsations will cause the Sun to eject its
outer layers forming a planetary nebula leaving only an extremely hot core.
This will slowly cool and fade into a white dwarf over tens of billions of
years. This shall be the fate of our Sun. (OKAY)
In a supernova event should the Sun decides to explode (not
possible) then heat and light may arrive on Earth in 8 min 19 sec, but its
shock waves sweeping out an expanding shell of solar materials consisting of
hydrogen, helium, neutrinos and other elements traveling at 30, 000 km per
second ( 10 % the speed of light) may take about 83 minutes or more to arrive.
The Energy Outshines the Entire Galaxy:
But since the energy output of a supernova outshines the energy
output of an entire galaxy like our Milky Way Galaxy of 150,000 million stars,
then if we presume the Sun is going to behave like that then we can expect the
energy output of the Sun will suddenly increase by 200–400 thousand million
times.
Let us take a very modest figure of just 150 000 000 000 stars
for our Milk Way Galaxy, each star like our own Sun releasing 3.846 × 1026
J of energy per second, then we can expect a sudden release in the order 5.8 x
1037 J each second. However, once again allow me to remind my gentle
readers that our Sun is not a super massive star, and it can never become a
supernova.
Instead it will slowly outlive its life in another about 7
billion (7000 000 000) years before becoming a red giant star to engulf
the Mercury, Venus, Earth and most of the outer planets beyond. It is very
unlikely the Sun will explode like a supernova.
Such a Rare Event:
In fact a super nova is so rare that there were only about
supernovae recorded in the entire history of mankind, the last prominent one
was the Crab Nebula.
So rest assured, nothing will happen to our Sun. But just for
the sake of this article, let us assume it will go supernova. In that case the
Sun will suddenly release 5.8 x 1037 J of energy each second. With
such stupendous energy output, not just all the water on Earth, but the entire
Earth and all its elements within will be annihilated completely in a blink of
a second.
There is no question of asking how much heat energy will be
required to boil off all the waters on this planet.
The Milky Way Galaxy:
Our Milky Way is a barred spiral galaxy, estimated of between
100,000–120,000 light-years across. It is a normal galaxy the population is
between 200 to 400 billion stars (200 000 million – 400 000 million). Our Sun
is just one of the average stars in our Milky Way Galaxy.
The number of stars in smaller galaxies is about 100 billion
(100,000,000,000).
What Happens if Sun
Explodes?
If the Sun decides to explode into a supernova (not
possible) the energy releases would be 6.8 x 10 29 Joules.
This is 195 times more than is needed to boil off the entire
waters found and hidden on planet Earth
Supernova:
The Sun can never become a supernova. It will have to be a super
massive Sun with some 10 solar masses before it can explode into a supernova.
A supernova is an explosion of a massive super giant star. It
may shine with the brightness of 10 billion suns! The total energy output may be 1044
joules, as much as the total output of the sun during its 10 billion year
lifetime.
Should it explode, it can last only about 2 weeks during which
it will radiate as much energy as the Sun would emit over its entire life span
of 100 billion years. The explosion will throw off much or all of a
star's material to the tune of 10 % the speed of light, namely at
30,000 km/s sending off shock waves into the surrounding deep space.
This shock wave sweeps up an escalating shell of gas and dust called the
remnants of a supernova.
Even though no supernova has been seen in the Milky Way since
1604, supernovae remnants indicate that on average, a supernova event occurs
approximately only once every 50 years in our own Milky Way Galaxy. The
supernova seen in the year AD 1054 gave rise to the Crab Nebula, and the last
SN remnants seen in the year 1604 was the Kepler's supernova remnant.
The cloudy remnants of SN 1054 are now known as the Crab Nebula.
Supernova plays an important role by enriching the interstellar
medium with higher mass elements. Additionally, the expanding shells of shock
waves bursting out from supernova explosions can activate the birth of new
stars.
Volume of Earth-Sun Sphere:
Let us do some simple calculations before narrating further.
The Astronomical Unit is the distance between the Earth and the
Sun. As the oribit of the Earth round the Sun is an ellipse, the mean Earth-Sun
distance is 149,597,870.700 kilometres or 1.495979 x 10 11
metres. This is the Earth-Sun mean distance.
Since the volume (V) of a sphere is given by:
V = 4/3 πr3
Hence, the volume of the sphere in space enclosed by the Sun to
Earth’s distance as the radius would be:
1.402 x 10 34 cubic metres.
The surface area of such a sphere with a
radius of 1 A.U. will have a volume equals to:
A = π r2
= 7 x 1022 square metres
Area of Sphere of Earth:
The surface area of a sphere is given by :
A = π r2
The radius of the Earth varies between 6,353 and 6,384 km.
(3,947–3,968 miles). The mean radius is 6,371 km (6371,000 metres)
Applying this formula, surface area of Earth would be about 5.1
x 10 14 square metres if we ignore the tiny bumps of mountains and
hills, the dips of oceans and seas and other uneven terrain here and there all
over the skin thin Earth’s crust
Light time:
The time taken for light from the Sun to reach
Earth is:
Sun-Earth distance ÷ Speed of light
=1.495979 x 10 11 metres ÷ 299 792
458 m /s
= 499 seconds (8 min 19 sec)
Solar Energy Received:
The Earth surface area is = 5.1 x 10 14
square metres. This is (5.1 x 10 14) ÷ (7 x 1022) x 100 =
7.29 x 10-7 % or 0.0 000 007 % of the Sun-Earth spherical area.
Since the Sun emits 3.846 × 1026 watts (Joules) per
second (1 watt second = 1 J)
Hence each second the Earth theoretically should receive;
Energy output of Sun per second ÷ Surface area of Sun-Earth
sphere x Surface area of Earth ÷ ½ (sunlit surface)
= (3.846 x1026) ÷ (7 x 1022) x (5.1 x 10 14)
÷ 2 = 1.4 x 1018 Joules per second or (1.4 x 1018)
watt-second for the sunlit part at any time.
Solar Energy:
The energy output of Sun is 3.8×1026 joule per
second or 3.8×1026 watt second
(1 watt second = 1 joule)
Hence in its life span of 10 billion years (3.15576 x 1017 seconds),
the Sun will have expended (3.846 × 1026) x (3.15576 x 1017)
= 1.2 x 10 44 Joules of energy. If it decides to go supernova and
releases all these horrendous amount of energy in, say 15 days (1,296,000
seconds). Hence each second, the amount of energy emitted will be 9.26 x 10 37
J.
Distance of Sun:
In order for us to calculate out how much solar energy the Earth
receive at least in theory, we need to know how far the Sun is away from Earth,
and how much of energy is spread out into space over a sphere sustained
by the Earth-Sun distance.
The mean distance of the Sun from the Earth is at an average
distance of 149.6 million kilometers called an Astronomical Unit
(1 AU). This distance varies as the Earth moves from perihelion when
it is closest to the Sun at 147,098,291 km in January to aphelion when it is
farthest from the Sun at 152,098,233 km in July as Earth describes an
elliptical orbit round the Sun.
Over this distance, light from the Sun, takes 8 minutes and
19 seconds to reach Earth.
Thus if the energy from the Sun is spread out over a sphere of 7
x 1022 square metres sustained by the Sun-Earth distance, then Earth
will get part of the energy determined by the surface area of
Earth.
The surface area of Earth or a sphere is given by :
A = π r2
The radius of the Earth varies between 6,353 and 6,384 km.
(3,947–3,968 miles). The mean radius is 6,371 km (6371,000 metres)
Applying this formula, surface area of Earth would be about 5.1
x 10 14 square metres if we ignore the tiny bumps of mountains and
hills, the dips of oceans and seas and other uneven terrain here and there all
over the skin thin Earth’s crust
Since the surface area of Earth is 5.1 x 10 14 square
metres, Earth will receive:
Energy output / Total area of Sun-Earth sphere x Area of Earth
= 9.365 x 10 37 ÷ 7 x 1022 x 5.1 x
10 14 = 6.8 x 10 29 J (okay)
The Earth surface area is = 5.1 x 10 14
square metres. This is (5.1 x 10 14) ÷ (7 x 1022) x 100 =
7.29 x 10-7 % or 0.0 000 007 % of the Sun-Earth spherical area.
The surface area of a sphere can be calculated by:
A = ∏ r2
where r, is the mean radius of Earth at 6,371 km (6371,000
metres) between 6,353 and 6,384 km. (3,947–3,968 miles), the surface area
of Earth would be about 5.1 x 10 14 square metres if we ignore the
tiny bumps of mountains and hills, the dips of oceans and seas and other uneven
terrain here and there all over the skin thin Earth’s crust
The Earth receives 174 petawatts (PW) of solar radiation at
the upper atmosphere. Approximately 30% of this is reflected back to space
while the rest is absorbed by clouds, oceans and land masses
The solar constant embraces all types of solar radiation, not
just the visible light. It is measured by satellite to be roughly 1.361
kilowatts per square meter (kW/m²) at solar minimum (measurements done by
satellite)
Let us now calculate the solar energy Earth receives at maximum
solar output using first principle:
Energy output of Sun per second ÷ Surface area of Sun-Earth
sphere x Surface area of Earth ÷ ½ (sunlit surface)
= (3.846 x1026) ÷ (7 x 1022) x (5.1 x 10 14)
÷ 2 = 1.4 x 1018 Joules per second or (1.4 x 1018)
watt-second
= 1.4 x 10 18 watt-second (max solar output) ÷ 5.1 x
10 14 (surface area of Earth) ÷ 1000 (watts into kilowatts) x 0.7 (70
% absorbed, 30 % reflected back into space) = 1.92 KW / sq. metre
Theoretical Calculations
vs Satellite Measurements:
·
1.9 kilowatts per square metre (calculated out by first
principle at solar maximum)
·
1.361 kilowatts per square meter (kW/m²) as measured by
satellite at solar minimum
The Beauty of Mathematics:
Please note the very close match, one at maximum solar output
(determined by simple calculations), and the other (by actual satellite
measurements) at minimum solar output.
The above calculation using simple school-level mathematics
demonstrates the beauty of mathematic.
The Total Energy of the
Sun:
There energy output of Sun in its entire life of 10
billion years = (3.15576 x 1017) seconds x 3.846 × 1026 J
= 1.2 x 1044 Joules ok
Since the Sun emits 3.846 × 1026 watts (Joules) per
second (1 watt second = 1 J), its energy output in one second in its
15 days of it being a supernova will be 1.2 x 1044
Joules ÷ 1,296,000 seconds = (9.26 x 10 37) .
Another way of putting it, if there are 250 000 million stars in
our Milky Way, and if a supernova in it explodes, its very short life-span of
just a few days will outshines the energy output of an entire galaxy as claimed
by astronomers and astrophysicists and then the total energy output for the
entire galaxy =
(3.846 × 1026) x 2.5 x 10 11 = 9.615 x 10 37
Joules
Calculated vs. Estimated
Energy Output:
1. 9.26 x
10 37) J (calculated out here by first principle)
2. 9.615 x
10 37 J (estimated by the consensus of scientists elsewhere)
The Beauty of Mathematics Again:
We can see the remarkably similar results of total energy of our
Sun as calculated out above using first principles, and the total energy output
of all 250 000 million stars in a galaxy as scientists said ‘the energy output
of a supernova will outshines the total energy output of an entire galaxy’.
What a beautiful statement!
Gravitation Force:
Let us now talk on something else, but related.
The gravitation force of attraction between two bodies is
proportional the product of their masses and inversely proportional to the
square of their distances.
Let us take an example. Applying the law of universal
gravitation, the gravitation force (F) between the Earth and Moon is
proportional to the product of their masses (m1 and m2),
and inversely proportional to the square of the distance (d) between them:
F = G. m1m2 / d2
= 6.67300 × 10-11 x (5.974 x 1024) x
(7.36x1022) kg ÷ (384,400,000 metres)2
= 1.9856 x 10 20 Newton (ok)
Where, G (Gravitational Constant) = 6.67300 × 10-11 m3
kg-1 s-2 (Newton)
M1 = mass of Earth = 5.974 x 10 24 kg
M2 = mass of the Moon = 7.36x10 22 kg
d = average distance between Earth and elliptical orbit of Moon
= 384,400 kilometers (384,400,000 metres)
The Volume of the Oceans:
The mass of the oceans is in the region of 1.35×1018 metric
tons (1.35 x 10 21) kg. This is roughly 1/4400 of the total mass of
the Earth. In terms of coverage, the oceans cover up an area of 3.618×108 km2
with a mean depth of 3,682 m (3.682 km). An estimated volume of the oceans
is 1.332×109 km3.
Most of the water (97.5%) is salt water; the remaining 2.5% is
fresh water. Of the fresh water, about 68.7%, is in the form of ice.
Vaporized Water and Waterless World: The Pull
Let us now imagine the world is now completely devoid of water
due to say a sudden surge of heat from the Sun causing all the ice to melt off,
and all the water to suddenly vaporized away. But before we proceed, I have to
admit this can never happen even in the foreseeable future because our Sun is
not massive enough for it to suddenly explode into a supernova.
It has to be a super massive Sun or star with 10 times in mass
in order this to happen. But for the sake of this article, let us assume our Sun
can go bang like a supernova.
Let us assume all the waters on Earth have vaporized in a puff.
Let us imagine the heat is just sufficient to send all the waters into space,
say the same height as a geostationary satellite 35900 km above the
Earth’s surface.
What then will be the pull then between the waterless Earth and
all the water 35900 km (35900000 metres) above the surface of a waterless
Earth.
Let us calculate:
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 1024
kilograms - 1.54 x 10 21 kg = 5.97 x 10 24 kg (ok)
Mass of Earth = 5.974 x 1024 kilograms
Mass of total waters = 1.54 x 10 21 kg
Force of gravity between a waterless Earth and the mass of water
separated out from it by heat of boiling to a distance of say,
a geosynchronous orbits of around 37, 000 km (23,000 miles) away.
Applying the law of universal gravitation, the attractive force
(F) between the waterless Earth and all the waters separated from it to a
distance of 35,900 km (35,900,000 metres) away from Earth centre
will be proportional to the product of their masses (m1 and m2),
and inversely proportional to the square of the distance (d) between them:
F = G.m1m2 / r2
= 6.67300 × 10-11 x (5.9726 x 10 24 x 1.4
x 10 21) kg ÷ (3.59 x 107 + radius of Earth ) 2
= 6.67300 × 10-11 x (5.9726 x 10 24 x 1.4
x 10 21) kg ÷ (3.59 x 107 + 6.37 x 106 metres)
2
= 5.58 x 10 35 ÷ 1.79 x 10 15 = 3.12 x 10 20
Newton
But for the sake of our imagination, let us assume under almost
impossible situation, the Sun does for just a fraction of a second and suddenly
it goes back to normal stability again.
Force of gravity between a waterless Earth and the mass of water
separated out from it by heat of boiling to a distance of say,
a geosynchronous orbits of around 37, 000 km (23,000 miles) away.
Applying the law of universal gravitation again, the attractive
force (F) between the waterless Earth and all the waters separated from
it to a distance of 37,000 km (37,000,000 metres) away from Earth centre
will be proportional to the product of their masses (m1
and m2), and inversely proportional to the square of the
distance (r) between them:
F = G.m1m2 / r2
= 6.67300 × 10-11 x (5.9726 x 10 24 x 1.4
x 10 21) kg ÷ (37,000,000) 2
= 4.075765 x 10 20 Newton
Geosynchronous Satellite:
Now let us take the example of a satellite that is neither
pulled down by Earth’s gravity, nor fly off at a tangent into space. We
shall apply this knowledge later in this article. So kindly be patient, even
though we may think it has nothing to do with the ocean waters being boiled
away.
A geosynchronous satellite orbits the earth with an orbital
period of 24 hours. This synchronizes with the period of the earth's rotational
motion. Thus a geostationary satellite appears permanently fixed over the same
point on the Earth.
We can calculate the height of such a satellite as it orbits in
the same time as the rotation of Earth in 24 hours
In order to cut short a lengthy explanation and its calculation,
allow me to give the final equation. We can derive the results by first
principle for the motion of a satellite around Earth as:
R3 = (T2 x G x M) ÷ (4 x pi2)
R3 = (86400 s)2 x (6.673 x 10-11
N m2/kg2) x (5.98x1024 kg) ÷ (4 x (3.1415)2
= 2.979 x 10 24 ÷ 39.476
= 7.54 x 1022 m3
(where, T = orbital period (86400 sec), G = gravitational
constant (6.673 x 10-11 N m2/kg2),
M = Mass of Earth (5.98x1024 kg).
The approximate radius of orbit of the satellite from the centre
of Earth would be the cube root of R3 (7.54 x 1022)
metres.
R = 4.23 x 107 m (42,300 km) from the Earth’s centre
Height of Satellite:
Since R (4.23 x 10 7 m ) is the distance of the
satellite from centre of the Earth we need to
subtract the mean radius of Earth (6.37 x 106 m)
to find the height of the satellite above the surface of Earth
Thus, (4.23 x 10 7 m) – (6.37 x 106
m) = 35,930,000 m = 35930 km
Speed of a Geostationary Satellite:
The angular speed ω of the satellite can also be found by
dividing the angle travelled in one revolution (360° = 2Ï€ rad) by the orbital
period (the time it takes to make one full revolution). In the case of a
geostationary orbit, the orbital period is one sidereal day, or 86,164.09054
seconds). This would be:
ω = 2 π rad / 86164 = 7.2921 x 10-5 rad / sec
By multiplying this angular velocity by its radius of orbit
(4.23 x 10 7) its orbiting speed would be about 3084.6 metres / sec
or 11,104.6 km / hr.
The radius of orbit indicates the distance that the satellite is
from the center of the earth. Now that the radius of orbit has been found, the
height above the earth can be calculated.
Since the earth's surface is 6.37 x 106 m from its
center (that's the mean radius of the earth), the satellite must be 4.23 x 107
m – 6.37 x 106 = 35,930,000 metres (35,930 km) above the
Earth’s surface. More accurately it is 35,786 km above the Earth’s
equator.
Why does a geostationary satellite not fall down like rain?
Why does a satellite stays stationary up there not fall down due
to gravity? A satellite stays up there in a circular orbit because of its
centrifugal force outwards just balances against the pull of gravity
(centripetal force) that is directed inwards towards the centre of the circular
motion. If not for gravity the satellite will be flung off at a tangent towards
outer space.
It is these two forces that keep a satellite in a stable orbit
around the Earth. But eventually a satellite will still fall down when it loses
its speed and kinetic energy as it collides with myriads of micrometeorites in
space.
The Separation of the Waters: A Cloud of Waters:
What about if all the waters were separated from Earth and
remain in the same orbit height of a geosynchronous satellite? Would it
not all condense, and fall down back to Earth?
In fact we have calculated that even if all 5.974 x 10 24
kg of total amount of water on earth were to be boiled off to the
height of say, a geostationary satellite, the force of attraction between
the waters that have separated and the waterless Earth below will still have a
whooping force of:
3.12 x 10 20 Newton
This is even 1.57 times the force of attraction between the
Earth and the Moon at 1.9856 x 10 20 Newton.
Of course the Moon is much farther away at 384,400 kilometers
(384,400,000 metres), whereas the separated water was assumed at 4.23 x 107
m (42,300 km). This is over 9 times farther away, whereas the force of
gravitational attraction falls off inversely as the square of their distances,
namely, the further away the two objects, the greater the diminishing fall of
attraction. But the gravity is still there no matter how far the objects.
Clouds float, Rain falls:
A cloud floats high up, and does not fall down even by the
action of gravity. Why is this so?
A typical cloud droplet has a size of 10 micrometres, and it
would take 15 million cloud droplets to form an average raindrop. The terminal
velocity of a rain droplet is about 0.3 cm/s or about 10 metres per hour.
The fall of rain droplets from the cloud at 3500 m at this speed
would take 350 hours. However, when the droplets coalesce do they form
heavier rain drops, some 300 times larger The rain clouds will then fall
down quickly
The rain droplets moreover is being pushed upwards against
gravity by the warmer air below so much so they are actually rising up instead
of falling down. That does not mean clouds are so light that gravity
cannot act on it.
If the water droplets, no matter how light, will accelerate down
to Earth in the same rate of 9.81 m/s2 as lead or a
brick had there been no air resistance in a vacuum. Of course, a
droplet caught in a downdraft will descend at the rate of the downdraft plus
its terminal velocity.
So in the same vein it does not mean that if all the waters on
Earth have evaporated and stayed up there at the same height
of say, a geosynchronous satellite at 35,786 km above the Earth’s
equator, it cannot fall down like a satellite because of ‘zero’ gravity
up there.
This is not true at all. Gravity is universal. Its force, albeit
very weak is the only force that permeates across the entire Universe, except
that it falls off inversely as the square of its distance.
Water Condensing:
Of course, even if all the waters have been boiled off, and
there is no more heat left, the super heated steam, steam or water vapour will
be exposed to the frigid temperature of outer space. This will cause them to
condense and cause them to fall down to Earth.
There will be 5.974 x 10 24 kg of them up
there, and there will no air, no resistance in a vacuum to block them to
rapidly condense and fall down to Earth if there is no more heat left to keep
them apart from Earth. Eventually they will cool and fall down as torrential
rains for years until all the oceans, seas, lakes, rivers, and land are filled
up again with water.
Making it Easier:
In order to make it easier for us to visualize astronomical
dimensions, we need to scale down our Sun and Earth and their distance.
·
If the Earth’s mean radius of 6.37 x 106 m is shrunk
down to just 1.0 mm, its volume would be: V = 4/3 πr3 =
4.19 cubic mm
·
Similarly, the Sun with a radius of 6.955 x 10 8
metre will have to be a scaled down by:
1/(6.37 x 106) x 6.955 x 10 8 = 109.18 mm.
It volume would be 4/3 π (109.18)3 = 5451523.38 cubic mm = 5.45
litres.
·
The Earth and Sun will be separated by a distance of:
1/(6.37 x 106) x 149,597,870,700 = 23484.75 mm =
23.48 metres apart
·
Scale down speed of the supernova explosion: 23.48 metres / 4990
sec = 4.7 x 10 -3 metres or 0.47 cm per second.
A Narration Summary from the Calculations:
The Searing Scenario:
Let us now imagine we now have a scaled-down that is only Earth
4.19 cubic mm in volume. Imagine we now use the hottest chemical
available on Earth – the torch of an oxy-acetylene flame at 3000 0 C, a flame
that can easily cut through steel.
The flame volume from the jet of an oxyacetylene torch is just
about 5 cubic milliliter. Let us now direct this blast of flame over 4.19 cubic
mm of water. What happens? Obviously all the water will vapourize
instantly in a flicker of a second.
Scaling it down:
Let us now take the second scenario of an extremely searing
flame, one that represents the plasma remnants of a star or Sun that has
exploded into a supernova by scaling down the size of the Sun and Earth.
On such a scale, the Sun would be 5.45 litres in size, and the
Sun-Earth’s distance would be 23.48 metres (2348 cm) apart. It has
exploded into a supernova where the temperature is known to reach about 100
billion (1 x 10 12) kelvin (100 gigakelvin).
This is 6000 times hotter than the core temperature of the Sun,
or 333 million times hotter than an oxyacetylene flame at 3,000 0 C,
the hottest available chemical flame on Earth.
Since it has been measured the speed of ejection of stellar
materials from a supernova can reach a velocity 10 % the speed of light at
30,000 km/s (30,000,000 m /s) will take 4986.59569 sec = 83.10992817 min =
1.38516547 hours = 1 hr 23 mins
On such a scale, the speed of the supernova ejection will be
2348 cm / 4986.59 sec = 0.47 cm / sec
I leave this scenario what will happen to the entire Earth
to my readers to imagine. I am talking about the entire planet Earth, and not
just the oceans, and all the waters found on Earth.
How much water then?
We must not forget the Earth is like an apple, most of it mass
is in its interior with its very hot molten core and the Earth’s crust
containing all the water on Earth is just like a thin apple skin with some
drops of water on it. is just Remember even if we give it the most generous
estimate of 1.54 x 10 21 kg of all the visible, invisible and hidden
waters on planet Earth compared to the mass of Earth, it is just:
Mass of water / mass of Earth x 100 %
= 1.54 x 10 21 / 5.974 x 1024 kilograms x
100 = 0.026 %
Since the density of water is 1 kg per litre (1 gm /ml), the
total water would be 1.54 x 10 21 litres. But the volume of our
scaled down Earth is 4.19 cubic mm = 4.19 x 10 -3 ml. (1 ml =1000 cubic millimeter). Hence with 0.026 % water,
the actual amount of water is 1.09 x 10 -3 (0.001 ml.)
A Mini Supernova:
Now, I want you to imagine a horrendously searing ball of
nuclear fire 5.45 litres in volume representing the Sun suddenly releasing some
5.8 x 1037 Joules of energy each second, at an unspeakable
temperature of 100 billion (1000, 000,000,000) Kelvin, 333 million times hotter
than the interior of the Sun, inching its way at a rate of 0.47 cm / sec
towards a miniaturized Earth just 4.19 cubic mm in volume, and containing just
0.001 ml of water.
What happen to that Earth and all the waters then? Need I say
more? I leave this to the imagination of my gentle readers to ponder over
during coffee, dinner or bed time.
Gravity acts only…?
If we have a situation where there is just sufficient heat to
boil off every drop of water on Earth and send them hurling up to the height of
40,000 km where geosynchronous satellites are placed, and let us now
assume there is suddenly no more heat left to push them any
further, we can expect the pull of Earth’s gravity albeit weaker at this
height, to bring the waters down again. That 40,000 km it is far beyond
the where air is.
It is almost a vacuum up there. There will be no air
resistance to drag the entire mass of water down. With that kind of heat that
can boil off the entire ocean, seas, lakes, rivers glaciers and atmospheric
waters, even the entire atmosphere will be torn apart and pushed away.
Even if there is no more heat, both water and air would not stay
as a layer over the Earth because they will continue to fly away outwards as
nothing could stop them. They will obey Newton first law of motion which states
that every object will continues in its state of rest, or of uniform motion in
a straight line, unless acted upon by an external force.
High up at the altitude of over 35, -40,000 km, Earth’s
gravitation force may not be sufficient to pull the waters back to Earth if the
momentum is great enough. It will escape gravity completely.
Neither would there be a
merry-go-round:
But if it remains motionless, gravity will act no matter how
weak. But there is no reason it will stop moving outwards into space. There is
no external force to act against its motion. There is also no reason to
suppose the mass of water will go into orbit round the Earth like a
satellite.
There is nothing to push it round and round in orbit. If
it does the centripetal pull of gravity will just balance the opposite
centrifugal force outwards and the waters will stay put in orbit round the
Earth.
Why should this be so? Why should it orbit the Earth in
the first place? We would expect the exceedingly hot super heated steam to be
ejected straight outwards into space. It will probably go in puffs of
rings and shells like puffs of steam into the air.
A Barren Earth:
The Earth will be completely barren, of water. It would be
similar to the other planets in the Solar System. The Earth would remain very
hot then for any water to condense back once it was blown off into the reaches
of outer space. I suspect the heat radiation would keep pushing all those
waters further, and further away.
Once at a distance, the Earth’s gravity will rapidly weaken
inversely as the square of its distance. The gravity would no longer be
effective then.
The Creation of Earth:
As the formless mass of dark clouds accreted due to inter-atomic
and gravitation crush, the heat generated would be so immense that no
ordinary chemical reaction can take place. The heat itself will energize the
hydrogen and oxygen atoms to keep them apart to form water.
In other words, the hydrogen and oxygen atoms will be kept as
separate elements to prevent ordinary chemical combustion. There was
completely no water then. The Earth will also be molten, formless, and devoid
of water and life.
Chemical marriage of hydrogen and oxygen:
But as the Earth cools over the eons, the hydrogen and oxygen
enveloping Earth would allow them to combine in stages for the first time. But
they still cannot come down as rain as the Earth was still exceedingly hot. Any
attempt to come down as rain, the steam generated by the very hot Earth, would
instantly cause them to boil off once again on contact to rejoin the
superheated steam still enveloping Earth in space.
But over the millions of years, the heat would slowly be loss
due to continuous cycles of rains and instant boiling off, and remaining heat
radiated out into space. The Earth then began to cool. Each cyclic event will
require a stupendous amount of heat to boil and cool off. It is an energy
transfer from heat into mechanical energy of cyclic rains. There would be
cycles of heat losses then.
The Waters Above and
Below:
The water enveloping the Earth would exist as a separate
filament above the Earth, from what was below. By and by, they would then be
dragged down to Earth by gravity. Over the millions of years, the waters
existing as a separate filament of superheated steam would cool rapidly at a
frigid temperature of near zero Kelvin in space.
Torrential Rains for A Millennium:
As both Earth and water cooled, torrential rains will pour down
onto Earth for the first time. The rains would probably last for hundreds, if
not thousands of years to form the oceans for the earliest time. The
filament above will then be separated from the filament below.
The Birth of the Oceans:
It is my belief during the initial formation of Earth from a
large mass of interstellar clouds containing an abundance of hydrogen and
oxygen the Earth was exceedingly hot due to the accretion of a mass of the dark
clouds coming together. This dark mass of cloud would be void, and without
any form.
The Earth and the Oceans:
The dark formless mass of interstellar cloud without any
division that came and created Earth and eons later, water and the oceans were
formed. It was initially just one dark formless mass devoid of anything
living.
Although I do not have the scientific evidence to base my
hypothesis, it is my very personal and private belief based on my humble
understanding of astronomy and astrophysics.
How much hydrogen and oxygen:
The question that
just ran into my mind as I pen the last sentence above was, how much hydrogen
and oxygen were needed during the genesis of all the waters found in the
oceans, seas, lakes, rivers, ice-caps, glaciers, and throughout the entire
hydrosphere on Earth?
In order for us to
understand that, we need to know how much water was synthesized out of hydrogen
and oxygen from the passing gas cloud during the Creation of Earth? To answer
that, we must first understand basic chemistry with this equation:
2 H2 + O2
→ 2 H2O
In order to
calculate how much hydrogen and oxygen were needed to form water; we need to
know the mole ratio of hydrogen and oxygen. In chemistry this is technically
called the stoichiometric ratio which is the amount of one reactant with the amount
of another reactant in the reaction.
In our case the
reactants are hydrogen and oxygen as shown in the above chemical reactions. This
means, for every two moles of hydrogen, two moles of water are produced. The
mole ratio between H2 and H2O is 1 mol H2 ÷ 1
mol H2O.
The Chemical Calculations:
Now, we need to calculate
out the theoretical yield of water to form all the ocean waters, the seas,
lakes, rivers, the ice caps, the glaciers, water in the atmosphere, and any
hidden waters elsewhere in the entire hydrosphere on this planet Earth.
How shall approach this problem?
We shall do this:
Convert the molar mass of hydrogen into
moles
1.
Apply the mole ratio between hydrogen and
water
2.
Apply the molar mass of water to change
moles water to mass of water
This applies for all reactants in a
chemical reaction, not just hydrogen and oxygen to form water. In our case, it
is the synthesis of water.
Thus,
Wt. of water produced
= wt. of reactant x (1 mol reactant ÷ molar mass of reactant) x (mole ratio
product ÷ reactant) x (molar mass of product ÷ 1 mol product)
In order to
calculate the amount of oxygen required, we need to know the mole ratio of
oxygen to water. The equation tells us, that for every mole of oxygen used, 2
moles of water are produced. The mole ratio between oxygen and water is 1 mol O2
÷ 2 mol H2O.
Thus the amount of oxygen required to
generate the entire mass of water on this planet estimated at 1.54 x 10 21
kg (1.54 x 10 24 gm) would require
= 1.54 x 10 24 gm
H2O x (1 mol H2O ÷ 18 g) x (1 mol O2 ÷ 2 mol H2O)
x (32 g O2 ÷ 1 mol H2)
1.37 x 10 24 gm or 1.37 x 10 21 kg of oxygen
Hence, the mass of
hydrogen required would be:
= (1.54 x 10 21 kg)
– (1.37 x 10 21 kg) = 1.7 x 10 20 kg hydrogen
Mass of Hydrogen atom:
But the mass of a
hydrogen atom consists of:
A proton of mass
= 1.7 x 10 -27 kg and an electron of mass =
9.1 x 10 -31 kg. Hence, the total mass is = 1.7 x 10 – 27
kg.
But since there are 1.7 x 10 20 kg of hydrogen in all the oceans and the
total waters elsewhere on this planet, then the number of hydrogen atoms in all
of them would be
= 1.7 x 10 20 ÷
1.7 x 10 – 27 = 1 x 10 47
In Deep Space:
Now, let us now look
at the amount of hydrogen in deep space.
There is no absolute vacuum in space. There is always a molecule or two even in between the horrendously vast chasms of space between the galaxies.
Matter is very, very tenuously distributed throughout the Universe. The average density of gas in our Milky Way galaxy is about one atom per cubic centimeter. This is by far a much more better vacuum than scientists can achieve in a laboratory. But this can mean a lot of matter inside a galaxy.
Throughout this vast universe, the most abundant
element is hydrogen (90 %), followed by helium (9 %) with oxygen, silica and
the rest of the elements in the Mendeleev's Periodic Table at 1 %. In
between the galaxies (intergalactic space) the amount of hydrogen is even more tenuous,
probably with just a hydrogen atom per cubic metre of space. But there is
always something there no matter how feeble and thin.
In short, there is no absolute vacuum throughout
the entire universe. All the elements on
the Earth came from a massive gas cloud that collapsed to form the Sun and the
planets including Earth.
Background
Radiation:
Even the temperature in throughout the universe is not spared. In every
direction, there is a very low energy and very uniform radiation permeating the
entire Universe. Astronomers found there is a left over of near 3
degree Kelvin Background Radiation.
Sometimes, this is called Cosmic
Background Radiation, or the
Microwave
Background. These were
termed because this radiation is basically a black body with temperature
slightly less than 3 degrees Kelvin (about 2.76 K).
So we can see there are always some remnants of something, whether temperature
or matter left over from a Big Bang during the creation of the Universe. There
is no such thing as an absolute vacuum, or an absolute zero temperature
anywhere in the Universe.
How Much?
But how much matter, especially the hydrogen,
oxygen and the silicon are there in the gas cloud that created this Earth? We
can never know for sure because we were never there when the Sun, Earth and the
rest of the Solar System was created. But we can make a hypothetical guess without
going into the statistical complexity of explaining Null Hypothesis and
Probabilities.
Assumption:
Let us assume instead of just one hydrogen
atom per cubic metre as in intergalactic space, let us multiply that density conservatively
to just a million times more. Compare this with a calculation using Avagadro's constant that tells us there are
6.02 x 10 23 molecules in a mol of gas. To cut a calculation short
for my gentle readers, there would be 2.69 x 10 25 molecules of air per
cubic metre at STP.
Hence if we assume that
there were a million hydrogen atoms in each cubic metre of the gas cloud that
created the oceans on Earth, then it can be shown that the volume of gas cloud
needed to accommodate 1 x 10 47 number of hydrogen turns out
to be 10 47 / 10 6 =
10 41 cubic metres.
What a Horrendous Volume:
But the volume of
Earth and Sun is only 1.082 x 10 21 m3 and 1.409 x 10 27 m3 respectively.
This implies that the volume of gas cloud containing 1 atom of hydrogen per
cubic metre required to form the Earth’s oceans was approximately 1 x 10 20
times (100 quintillion or 1 followed by 20 zeros) and 1 x 10 14 (100
trillion or 1 followed by 14 zeros) times larger in volume than those of the
Earth and Sun respectively.
What a bizarre volume
of hydrogen and oxygen needed just to create the ocean and other waters on
Planet Earth.
How Big was The Gas
Cloud?
The next question we would like to ask is, how far would
such a ball of hydrogen gas stretch in order to create all the oceans and
waters on Planet Earth during Creation?
Let us calculate again.
Volume of a ball of Gas Cloud = 4 π r 3
÷ 3
Its radius would be:
3V = 4 π r 3
r3 = 3V ÷ 4 Ï€
r =
3√ (3V ÷ 4Ï€)
= 3√
( 3 x
10 41 ÷ 4 Ï€)
= 3√ 2.4 x 10 40 =
2.9 x 10 13 metres in radius
Hence diameter of a ball of gas cloud = 5.8
x 10 13 metres across
(Where, 10 41 cubic metres was
the volume of the amount of hydrogen needed at a density of 1 million atoms per cubic
metre)
Light Time:
An unexcited
hydrogen atom has a radius called the Bohr radius. This is a physical
constant, approximately equal to the distance between the proton and electron
in an unexcited hydrogen atom at ground state. It is named after physicist
Niels Bohr, and is called the Bohr model of an atom.
The radius of a hydrogen atom or the Bohr
radius has a value of 5.2917721092 × 10−11 m. That’s 0.000 000
000 05292 of a second, or 52.92 picosecond.
We know that the speed of light
in a vacuum is = 299 792 458 m / s or nearly 300 000 km per second.
Hence it takes 1.765 x 10-19 second (0.1765 attosecond) for light to cross the diameter of a hydrogen atom at ground state.
Since the volume (V) of a sphere is 4 π r3
÷ 3
Its radius (r) will be:
r = 3√
3V÷4 Ï€
Thus, a sphere with a volume of 1 cubic
metre will have a radius of:
r = 3√ 0.2387324146
= 0.6203504909 metres, or a diameter of 1.240700982
metres.
If we were to line up all the hydrogen
atoms side-by-side across a sphere of 1 cubic metre in volume, there will be
2.3 x 10 10 (23 billion) hydrogen atoms all in.
This means, it takes light 4.1 x 10 -9
seconds (4.1 nanosecond) to cross 2.3 x 10 10 atoms lying across the
diameter of a sphere one cubic metre in volume.
If we assume the Gas Cloud was a sphere
with a diameter of 5.8 x 10 13 metres across, containing 10 47 hydrogen atoms with a density of a million atoms per cubic
metre, it would take light travelling at
nearly 300 000 km per second, 53 hrs 44 min
27 seconds to cross.
Bigger or Smaller:
How large or small the Gas Cloud depends on
the numbers of hydrogen atoms per cubic metre.
It may be very much smaller if the hydrogen density was far greater and much
larger if it was more tenuous.
We do not know exactly, but as we work it
out, for a Cloud of this magnitude, there must be at least a million hydrogen
atoms per cubic metre.
But conventional chemistry tells us, there
must be at least 10 41 hydrogen atoms to create all the waters on
the hydrosphere of his planet
This would be the size of the Cloud upon
which all the waters on Earth was created.
How far between Sun to Pluto and back?
The distance of Pluto from the Sun is
between 30 to 49 AU (4.4–7.4 billion km ( 4400 000 000 to 7400 000 000 km or 4.4 x 10 12 – 7.4 x 10 12
metres).
This means the Gas Cloud that created all the
oceans and waters on Earth must have stretched up to nearly 8 times the
distance of Pluto from the Sun.
In other words, its diameter could enclose the
distance between the Sun and Pluto to and fro 4 times over.
What an horrendous distance of hydrogen needed
at 1 atom per cubic metre needed to create the oceans.
What an Almighty
Hand God has in His Creation!
Genesis:
Genesis:
And the earth was without form, and void; and darkness was upon
the face of the deep. And the Spirit of God moved upon the face of the waters.
(Genesis 1: 2). This verse speaks about the lifeless and molten formless Earth.
And God said, Let there be a firmament in the midst of the
waters, and let it divide the waters from the waters
And God made the firmament, and divided the waters which were
under the firmament from the waters which were above the firmament: and it was
so (Genesis 1: 6-7)
Verses 6 and 7 speak about how water was divided into two stages
on probably an exceedingly hot Earth, the very hot steam above, and the
partially cooler, and condensed water below on the surface of Earth.
Of course in geological time scale this took millions of years
in human time scale, but in God’s eternal time scale, it was ‘just the second
day’ to Him.
For a thousand years in your sight are like a day that has just
gone by, or like a watch in the night (2 Peter 3:8)
The following equivalents were used in the calculations for the
above article:
1. 1 solar
day: 60 x 60 x 24 = 86400 sec.
2. 1 Solar
Year: 365.25 solar days = 31,557,600 seconds
3. 15
solar days: 86400 x 15 = 1,296,000 seconds
4. 4.57
billion years = 4570,000,000 years = 1.44 x 10 17
seconds
5. 10
billion years (1010): 1010 x 31557600 = 3.15576 x
1017 seconds
6. 1 watt
hour = 3600 J
7.
1 kilowatt hour = 3.6×106 J (or 3.6 MJ)
8. 1 watt
second = 1 Joule
9. Speed
of light = 299 792 458 m / s
10. Age of
the Sun: 4.57 billion (4,570,000,000) years old
11. Lifespan
of our Sun: 10, 000 million (1 X 10 10) years (main
sequence lifetime)
12. Mass of
Sun: 1.98892 × 1030 (nearly two nonillion kilograms
13. Mass of
Earth: 5.974 x 1024 kilograms
14. Sun is
332929.4 times more massive than Earth
15. Mass of
Moon: 7.36x10 22 kg
16. Radius
of the Sun: 6.955 x 10 8 metre = 6.955 x 10 11 mm
(695,500 km)
17. Mean
Radius of Earth: 6.37 x 106 m (6.37 x 10 9 mm)
18. Sun:
Earth radius ratio =109:1
19. 1
Astronomical Unit (Earth-Sun distance) : 149,597,870,700 metres
20. Volume
(V) of a sphere is: V = 4/3 πr3
21. Volume
of Sun = 1.409 x 10 27 m3
22. Volume
of Earth = 1.082 x 10 21 m3
23. Sun
/Earth volume ratio: 1,302,218 : 1
24. Gravitational
Constant = 6.67300 × 10-11 m3 kg-1 s-2 (Newton)
25. 1000 mm
= 1 metre
26. 1 metre
= 1000mm
27. 1 litre
= 1 000 000 cubic millimeter
28. 1 metre
= 100 cm
29. 1 litre
= 1 000 000 cubic millimeter
30. 1 cubic
metre = 1 000 000 000 cubic millimeter
31. 1 cubic
metre = 1000 000 cc = 1000 litres
32. Mass of
ocean waters: 1.35×1018 metric tons (1.35 x 10 21)
kg
33. Mass of
water in hydrosphere: 1.4 × 1018 tonnes (1.4 x 10 21 kg)
34. Mass of
hidden water at 10 % of hydrosphere: 1.4 x 10 20 kg
35. Mass of
total water: 1.54 x 10 21 kg
36. Mass of
Earth: mass of total water : 3879:1
37. Mass of
total water on Earth: 1.54 x 10 21 kg
38. Mass of
total water: mass of earth = 2.57 x 10 -4 (0.025 %)
39. Zero
Kelvin = - 273.15 0 C
40. (1 x 10
12) Kelvin = (1 x 10 12) 0 C
41. (1 x 10
12) Kelvin = 333 million times hotter than an oxyacetylene flame at
3,000 0 C
42. Average
temperature of ocean waters: 17 0 C (National Climatic Data Center)
43. Heat of fusion of ice is 334
kJ/kg
44. Specific heat of water: 1
calorie/gram °C = 4.186 joule/gram °C
45. Total
geothermal energy locked up: 1031 joules
46. Energy
output of a Supernova =1.2 x 1044 Joules
47. Temperature of a
supernova is about 100 billion (1 x 10 12) kelvin (100 gigakelvin).
This is 6000 times the core temperature of the Sun
48. Energy
output of Sun = (3.839 × 10 26) Joules per second
49. Energy
Total of the Sun: 1.2 x 1044 Joules
50. Solar
Constant (calculated): 1.92 KW. m-2
51. Solar
Constant (measured by satellite): 1.361 KW. m-2
52. No. of
stars in Milky Way Galaxy: 200 000 million – 400 000 million (average no: 250
000 million)
53. Diameter
of Milky Way Galaxy: 100,000–120,000 light-years across
54. 1 tonne
= 1000 kg
55. One Newton (N) in
SI unit for force required to accelerate a mass of one kilogram at a rate of
one metre per second squared
56. N = kg. m /s2
57. Heat of fusion of ice is 334
kJ/kg
58. Enthalpy of vaporization
of water : 2257 kJ kg−1
59. 1 calorie = 4.1868 joule
1 J = 0.238846 cal
1 J = 0.238846 cal
60. Radius of the Sun: 6.955
x 10 8 metre = 6.955 x 10 11 mm (695,500 km)
61. Mean Radius of Earth:
6.37 x 106 m (6.37 x 10 9 mm)
62. 1 Astronomical Unit
(Earth-Sun distance) : 149,597,870,700 metres
63. Speed of light =
299 792 458 m / s
64. Sun-Earth travel
light-time = 499 seconds = 8.3167 min = 8 min. 19 se
65. Sun-Earth supernova
travel time = 4990 sec = 83.16746397 min = 1.3861244 hours =
1 hr 23 min 10 sec
Mathematics is the Science of Logic. Facts and Figures
do not lie.
Thank you for your patience in reading.
lim ju boo
Founder and First President
The Astronomical Society of Malaysia
Dated: 23rd July, 2012
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