Geology Reference
In-Depth Information
heat at constant pressure, which allows the material some thermal expansion as
it is warmed by one degree. Thermodynamicists also define the specific heat at
constant volume,
C
V
.) The temperature of the material in the thermal boundary
layer has decreased by an average of
T
m
/
2. Therefore, the change in heat content
per unit mass is
C
P
T
m
/
2. Density is mass per unit volume, so the change in heat
content per unit volume is
ρC
P
T
m
/
2. Now the heat loss in Eq. (5.11) is a heat flux,
i.e. the rate of heat loss
per unit area
of the top surface (the sea floor, if this is the
oceanic lithosphere). Therefore we need the change in heat content per unit area.
This will be the change in heat content of a column of rock of depth
d
and unit
cross-sectional area, which has a volume of
d
×
1
×
1. Thus we finally get the
amount of heat lost per unit area of the surface as
H
=
ρC
P
T
m
d/
2
.
(5.12)
If this amount of heat were lost at the rate given by Eq. (5.11), it would take a
time
ρC
P
T
m
d
2
/
2
KT
m
ρC
P
d
2
/
2
K,
t
s
=
H/q
1
=
=
(5.13a)
d
2
/
2
κ.
t
s
=
(5.13b)
There are several things to say about this series of expressions. Temperature cancels
out: the time does not depend on the temperature difference, only on the depth and
the material properties. The material properties have been combined in the last
expression into a quantity called the
thermal diffusivity
:
κ
=
K/ρC
P
.
(5.14)
A typical rock conductivity is about 3 W/m
◦
C. With a density of around 3000 kg/m
3
and a specific heat of 1000 J/kg
◦
C, we get
κ
10
−
6
m
2
/s. You can work through
the dimensions of these quantities to verify that
κ
has dimensions of length
2
/time,
which it must have for the dimensions in Eq. (5.13b) to balance.
How long would it take to cool the body to a depth of 10 m? Without prior
knowledge, you would not have much idea. It might be minutes or months. If we
use the value of
κ
already given, the answer is 5
=
10
7
s, which you might recall is
over a year. Well, this is an upper bound, but it indicates that it takes quite a long
time to cool rocks to a depth of 10 m. The foundations of houses in cold climates
are dug about 2 m into the ground so they extend below the level that freezes in
winter. This is why basements are common in cold climates.
How long would it take to cool the body to a depth of 20 m? Because of the
d
2
in Eq. (5.13b), it takes four times as long: 2
×
10
8
s or over 6 years. Why does
doubling the depth cause the time to increase by a factor of 4? If you go through
the earlier logic for the layer of depth 2
d
in Figure 5.4, you find that the rate of heat
×
