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Figure 7.15. Thermal
models of the lithospheric
plates beneath oceans
and continents. The
dashed line is the plate
thickness predicted by the
PSM plate model; k
(values of 2.5 and 3.3) is
the conductivity in
Wm −1 ◦ C −1 . (After Sclater
et al . (1981b).)
O ld
continent
Oceanic lithosphere
0
2.5
2.5
50
3.3
3.3
100
Thermal boundary layer
150
200
50
0
50
100
150
200
Age (Ma)
mantle and outer core, however, where conduction is not the primary method of
heat transfer, the methods and estimates of the previous sections are not appropri-
ate. In the mantle and the outer core, where convection is believed to be occurring,
heat is transported as hot material moves; thus, the rate of heat transfer is much
greater than that by conduction alone, and as a result the temperature gradient and
temperatures are much lower. In the interior of a vigorously convecting fluid, the
mean temperature gradient is approximately adiabatic. Hence, the temperature
gradient in the mantle is approximately adiabatic.
To estimate the adiabatic temperature gradient in the mantle, we need to use
some thermodynamics. Consider adiabatic expansion ,anexpansion in which
entropy is constant for the system (the system can be imagined to be in a sealed
and perfectly insulating rubber bag). Imagine that a rock unit that is initially at
depth z and temperature T is suddenly raised up to depth z . Assume that the rock
unit is a closed system, and let us consider the change in temperature that the unit
undergoes. When it reaches its new position z ,itishotter than the surrounding
rocks; but, because it was previously at a higher pressure, it expands and, in so
doing, cools. If the temperature to which it cools as a result of this expansion
is the temperature of the surrounding rocks, then the temperature gradient in
the rock pile is adiabatic . Thus, an adiabatic gradient is essentially the tempera-
ture analogue of the self-compression density model discussed in Section 8.1.2.
Temperature gradients in a convecting system are close to adiabatic.
To determine the adiabatic gradient, we need to determine the rate of change
of temperature T (in K not C) with pressure P at constant entropy S . Using the
reciprocal theorem (a mathematical trick), we can write this as
T
P
S =−
T
S
S
P
(7.86)
P
T
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