Since
some
materials that
are
to be tested
are
difficult to
shape into
a
sphere,
the
investigation has
also to
be
carried
out for the
cube,
which
can
easily
be
produced.
For the
cooling
of
a
point
of
a
cube from
a
distance
of a/4, where
a
is the
length
of the
edge,
we
have
3
712
K
t
=
C'-e_I
^
»cZ.
Thus
we
have again the constancy of the ratio of
two
temperatures
separated
by
time 4z.
K
is determined as before.
In this
way one
finds the coefficients of thermal
conduction
of
rock
species.
Granite
a
0.525
b 0.474
Gneiss
a
0.577
b 0.483
Limestone
a
0.451
b 0.404
Sandstone
a
0.427
b
0.181
Glass
a
0.125
b
0.108
a
and b
are
randomly picked,
non-extreme samples. The
conductivity of
silicates
depends
to
a
great extent
on
the quartz
contained in them,
which is
a
very
good conductor. In the
case
of sandstone, the clay
content
plays
a
great
role.
In
general,
the thermal
conductivity
of substances
depends only
on
the
material
properties
of the components and
not
on
the
physical
properties
of the
individual
[samples].
The earth in its totality
can
be considered
as a
silicate sphere
which radiates heat into the low-temperature universe. The insulating
effect of the atmosphere
can
be
neglected
as
small. Hence,
assuming
that
we
apply
the
case
of
our
sphere
to the
earth, i.e.,
if
we
also
neglect
the heat
produced
(by
contraction and chemical
processes),
we
can
state the minimum time
necessary
to cool the earth
by
one
percent.
39
3ir2
IC
-r .4z
t
4ct
pc
K
4.iTpc
4z.