Geology Reference
In-Depth Information
boundary layer at the bottom of the mantle, as will be discussed in the next
chapter.
We can quantify the rate at which heat is removed from the mantle by the
formation and subduction of plates. Referring to Figure 5.4(b) showing schematic
profiles of temperature through a thermal boundary layer, mantle that rises under
a mid-ocean ridge has the mantle temperature,
T
m
=
1300
◦
C, essentially right to
the surface. When this layer of mantle reaches a subduction zone about 100 Myr
later, it has cooled to a depth of about 100 km. Its temperature then varies from
T
0
=
0
◦
C at the surface to 1300
◦
C at 100 km depth. Its
average
temperature is then
about 650
◦
C. When it started at the ridge, its average temperature was 1300
◦
C.
Therefore its average temperature has dropped from 1300
◦
C to 650
◦
C, a
decrease
of 650
◦
C. Writing this out, the change in the average temperature,
T
,is
T
i
av
T
=
T
av
−
=
(
T
0
+
T
m
)
/
2
−
T
m
(6.1)
=
(
T
0
−
T
m
)
/
2
T
=−
T
m
/
2
,
where
T
av
is the initial average temperature at the ridge.
Referring back to Eq. (5.12) and the paragraph preceding it, we can write an
expression for the resulting change in the heat content of the lithosphere. A vertical
column of rock with a cross-sectional area of 1 m
2
and extending to a depth
d
has a
volume of
d
×
1 and a mass of
ρd
,where
ρ
is the density of the rock. Therefore
the change in heat content,
H
, of this rock column is
1
×
H
=
ρdC
P
T ,
(6.2)
where
C
P
is the specific heat of the rock.
Seafloor spreading creates an area of about 3 km
2
of new sea floor every year
[42]. Subduction removes a similar amount. Remembering that there are about 3
×
10
7
seconds in a year, and converting the area to square metres, the rate of removal
of sea floor by subduction is
S
0.1 m
2
/s.
Every square metre of sea floor that is removed by subduction has lost an amount
of heat
H
given by Eq. (6.2). Therefore the total rate of heat loss by cooling of
the oceanic lithosphere is
10
6
/
3
10
7
m
2
/s
=
3
×
×
=
Q
=
SH
=
SρdC
P
T
m
/
2
.
(6.3)
3300 kg/m
3
1000 J/kg
◦
C, this yields
Q
10
13
W
Using
ρ
=
and
C
P
=
=
2.1
×
=
21 TW.
