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Earth's Equilibrium Temperature Video Lecture - Heat Transfer - Mechanical
Text Transcript from Video
okay welcome dialect online in our next
example and heat transfer to radiation
now we're going to talk about the
Earth's equilibrium temperature now of
course we have to oversimplify the model
because the earth does have an
atmosphere and the atmosphere does tend
to retain heat but if we ignore that
aspect of it what would the temperature
of the world be of the earth be if we
simply look at it from a simple
perspective of radiation yurts shining
onto the earth the earth receiving the
energy from the Sun and then the energy
being radiated out in all directions at
all times into the universe what would
temporary or it be if that was the case
again
we can say that the DQ DT is equal to
the emissivity and for the earth that's
probably about 0.9 Sigma times the
surface area times temperature to the
fourth power and now let's see we can
then go ahead and calculate how much
energy the earth receives from the Sun
now of course you can see that the
portions of the earth that they're
pointing directly to the Sun they
receive a whole lot more energy than
portions that are here on the side where
they just come and get a little bit of
the energy of the Sun but we can kind of
average it out we can assume that if we
were to cut the earth open like this and
we had all the energy received on that
flat surface of the cross section of the
earth that would be the same amount as
would be distributed over a semi sphere
so what we can say then is that the
total energy energy received from the
Sun is equal to the energy per unit area
times the area of the Sun all right so
maybe we'll just write it like this
energy per unit area times the total
area where the light shines if we were
to cut the earth in half and we have
this flat surface all right the energy
that we receive per unit area that's
about thirteen hundred and sixty-one
watts per square meter I know that in a
previous example the number that we got
was slightly different from that but
that's because there's some other
factors we have to consider at this rate
this is about the energy that we measure
out in space just above the Earth's
atmosphere that's the amount of energy
that we receive from the Sun
meter and if we then multiply that times
the cross-section of the earth the would
be PI R squared then we can get the
total energy received on the earth and
so this is equal to let's see here
thirteen hundred and sixty-one watts per
square meter times pi and times the
radius of the earth which is six
thousand three hundred and seventy eight
kilometers or six million seven three
hundred and seventy eight million meters
so don't forget the meters that would be
down meters and we have to square that
so that's the radius of the earth in
meters so let's see what we get
that would be 1361 times pi times six
three seven eight zero zero zero squared
equals and so the total amount of energy
received in the Sun per second is one
point seven four times ten to the
seventeenth watts
that's joules per second all right
now we've want to find the equilibrium
temperature we can go ahead and take
that number and plug it back into the DQ
DT so this goes back in here that's the
acuity and then we could solve for the
temperature of the earth would be all
right but now the area of course is no
longer going to be the cross-sectional
area of the earth it's not going to be
the total serves they're the earth
because the earth re radiates in all
directions not away from the Sun to the
side towards the Sun all directions all
right so we write in this equation we
could say that temperature is equal to
the fourth root of the DQ DT that's the
amount of energy received by the earth
divided by e Sigma a and so that is
equal to the fourth root of this number
right here one point seven four times 10
to the 17th 17 watts divided by 0.9
Sigma is five point six seven times ten
to the minus eight watts per meter
squared Kelvin to the fourth power and
then of course now we need to have the
surface area of the earth so what to
serve as air of a sphere it is four pi
times the radius squared so it would be
six three seven eight zero zero zero
meters squared there and that will give
us the equilibrium temperature of the
earth of course pushing aside all of the
factors such as atmospheres and clouds
and convection heat transfer and all
that solutes it's a lot more complicated
than that but from a simple perspective
what would this be so take that number
divided by point nine divided by five
point six seven
e to the eighth minus divided by four
divided by PI and divided by six three
seven eight zero zero zero squared
equals and then take the fourth root of
that and it's 286 that would be 286
Kelvin now how much is that in
centigrade degrees and how much is that
in Fahrenheit degrees now of course if
we subtract 273 from that we'll get 13
degrees centigrade 13 centigrade degrees
and then if we convert that to foreign
high degrees okay so then what we have
to do is we have to divide that by nine
no divided by five times nine and then
add 32 to that plus 32 and that gives us
55 degrees Fahrenheit which by the way
is the average temperature of the world
but again there's all kinds of factors
we have to take into play turns out that
we don't receive this kind of energy
from the earth actually we receive about
half of that so where was I so this is
the energy that was that is received by
the cloud tops and by the upper portion
of the atmosphere much of it is
reflected in back into space without
being absorbed by the top of the clouds
by the eyes by the oceans and so forth
somebody's being reflected or not
allowed through because of the
atmosphere and only a portion that gets
through so the real numbers would be
very different but this is kind of
interesting when you think about it if
we take all the energy received from the
Sun to be 1361 watts per square meter we
spread that over the earth and then we
understand how much
the temperature the world how much the
earth with them radiates back into space
it turns out the approximate number we
get is very very close to the average
temperature of the world so you can see
that there's kind of eight and
equilibrium status that they will take
on based upon how much energy receives
and I'm at energy that radiates back
into space
example and heat transfer to radiation
now we're going to talk about the
Earth's equilibrium temperature now of
course we have to oversimplify the model
because the earth does have an
atmosphere and the atmosphere does tend
to retain heat but if we ignore that
aspect of it what would the temperature
of the world be of the earth be if we
simply look at it from a simple
perspective of radiation yurts shining
onto the earth the earth receiving the
energy from the Sun and then the energy
being radiated out in all directions at
all times into the universe what would
temporary or it be if that was the case
again
we can say that the DQ DT is equal to
the emissivity and for the earth that's
probably about 0.9 Sigma times the
surface area times temperature to the
fourth power and now let's see we can
then go ahead and calculate how much
energy the earth receives from the Sun
now of course you can see that the
portions of the earth that they're
pointing directly to the Sun they
receive a whole lot more energy than
portions that are here on the side where
they just come and get a little bit of
the energy of the Sun but we can kind of
average it out we can assume that if we
were to cut the earth open like this and
we had all the energy received on that
flat surface of the cross section of the
earth that would be the same amount as
would be distributed over a semi sphere
so what we can say then is that the
total energy energy received from the
Sun is equal to the energy per unit area
times the area of the Sun all right so
maybe we'll just write it like this
energy per unit area times the total
area where the light shines if we were
to cut the earth in half and we have
this flat surface all right the energy
that we receive per unit area that's
about thirteen hundred and sixty-one
watts per square meter I know that in a
previous example the number that we got
was slightly different from that but
that's because there's some other
factors we have to consider at this rate
this is about the energy that we measure
out in space just above the Earth's
atmosphere that's the amount of energy
that we receive from the Sun
meter and if we then multiply that times
the cross-section of the earth the would
be PI R squared then we can get the
total energy received on the earth and
so this is equal to let's see here
thirteen hundred and sixty-one watts per
square meter times pi and times the
radius of the earth which is six
thousand three hundred and seventy eight
kilometers or six million seven three
hundred and seventy eight million meters
so don't forget the meters that would be
down meters and we have to square that
so that's the radius of the earth in
meters so let's see what we get
that would be 1361 times pi times six
three seven eight zero zero zero squared
equals and so the total amount of energy
received in the Sun per second is one
point seven four times ten to the
seventeenth watts
that's joules per second all right
now we've want to find the equilibrium
temperature we can go ahead and take
that number and plug it back into the DQ
DT so this goes back in here that's the
acuity and then we could solve for the
temperature of the earth would be all
right but now the area of course is no
longer going to be the cross-sectional
area of the earth it's not going to be
the total serves they're the earth
because the earth re radiates in all
directions not away from the Sun to the
side towards the Sun all directions all
right so we write in this equation we
could say that temperature is equal to
the fourth root of the DQ DT that's the
amount of energy received by the earth
divided by e Sigma a and so that is
equal to the fourth root of this number
right here one point seven four times 10
to the 17th 17 watts divided by 0.9
Sigma is five point six seven times ten
to the minus eight watts per meter
squared Kelvin to the fourth power and
then of course now we need to have the
surface area of the earth so what to
serve as air of a sphere it is four pi
times the radius squared so it would be
six three seven eight zero zero zero
meters squared there and that will give
us the equilibrium temperature of the
earth of course pushing aside all of the
factors such as atmospheres and clouds
and convection heat transfer and all
that solutes it's a lot more complicated
than that but from a simple perspective
what would this be so take that number
divided by point nine divided by five
point six seven
e to the eighth minus divided by four
divided by PI and divided by six three
seven eight zero zero zero squared
equals and then take the fourth root of
that and it's 286 that would be 286
Kelvin now how much is that in
centigrade degrees and how much is that
in Fahrenheit degrees now of course if
we subtract 273 from that we'll get 13
degrees centigrade 13 centigrade degrees
and then if we convert that to foreign
high degrees okay so then what we have
to do is we have to divide that by nine
no divided by five times nine and then
add 32 to that plus 32 and that gives us
55 degrees Fahrenheit which by the way
is the average temperature of the world
but again there's all kinds of factors
we have to take into play turns out that
we don't receive this kind of energy
from the earth actually we receive about
half of that so where was I so this is
the energy that was that is received by
the cloud tops and by the upper portion
of the atmosphere much of it is
reflected in back into space without
being absorbed by the top of the clouds
by the eyes by the oceans and so forth
somebody's being reflected or not
allowed through because of the
atmosphere and only a portion that gets
through so the real numbers would be
very different but this is kind of
interesting when you think about it if
we take all the energy received from the
Sun to be 1361 watts per square meter we
spread that over the earth and then we
understand how much
the temperature the world how much the
earth with them radiates back into space
it turns out the approximate number we
get is very very close to the average
temperature of the world so you can see
that there's kind of eight and
equilibrium status that they will take
on based upon how much energy receives
and I'm at energy that radiates back
into space
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Apr 19, 2026 Last updated
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