5Magmas and Their Emplacement

A basaltic melt at $1200\,{}^{\circ}\mathrm{C}$ flows with a viscosity of about $70\,\mathrm{Pa}\,\mathrm{s}$: thick, but it pours. A dry rhyolitic melt at $800\,{}^{\circ}\mathrm{C}$ has a viscosity near $10^{11}\,\mathrm{Pa}\,\mathrm{s}$, and a block of it would take hours to sag visibly under its own weight. Between the two lie ten orders of magnitude, more than between water and the basalt, and most of the difference comes from two things: the silica that links the melt into a network, and the temperature, which is several hundred degrees lower in silicic magmas. A few per cent of dissolved water, which a rhyolitic magma can hold at depth, brings its viscosity down by a factor of $10^{5}$, and loses it again as the magma rises and the water comes out as bubbles. These numbers decide whether a magma reaches the surface quietly, explodes, or freezes kilometres down as a dyke, a sill or a pluton. This chapter treats a magma as a material: its density, viscosity and volatiles, the physics of its ascent through cracks and as blobs, the shapes it takes when it stops, and the heat it gives to the rocks around it.

You already know

Viscous flow (physics books’ Year 2 volume): in a Newtonian fluid of viscosity $\eta$ the shear stress is $\eta$ times the velocity gradient; between two parallel walls a pressure gradient drives a parabolic (plane Poiseuille) profile; a sphere moving slowly through a viscous fluid feels the Stokes drag $6\pi\eta a v$. Heat conduction (same volume): temperature obeys $\partial T/\partial t = \kappa\,\partial^2T/\partial x^2$, with diffusivity $\kappa = k/(\rho c)$, and a step of temperature spreads as an error function over a distance of order $\sqrt{\kappa t}$. Buoyancy (physics books’ Year 1 volume): a body immersed in a denser fluid feels a net upward force equal to the weight of fluid displaced minus its own. Chemical equilibrium (chemistry books): at equilibrium the activities of products and reactants satisfy the law of mass action with a constant $K$ that depends on temperature. From the Year 1 volume of this series: the lithostatic pressure $p = \rho g z$ and the geotherm; from the school volumes: mafic and felsic rocks, phenocrysts and porphyritic textures.

5.1 What a magma is

Definition 5.1 (Magma)

A magma is molten or partly molten rock beneath the surface: a silicate melt, the liquid made mostly of silicon, oxygen and metal cations, together with any crystals and gas bubbles it carries. A crystal mush is a magma so rich in crystals that they touch and form a framework: it cannot flow as a liquid, and its melt can only move through the pores between crystals.

The composition of the melt is given, as for rocks, in weight per cent of oxides. Three melts serve as references throughout the chapter, all of them studied in the laboratory: a tholeiitic basalt dredged from a mid-ocean ridge (about $51\,\%$ SiO$_2$, $12\,\%$ iron oxide, $7\,\%$ MgO), an andesite ($63\,\%$ SiO$_2$) and a rhyolite ($78\,\%$ SiO$_2$, $8\,\%$ Na$_2$O $+$ K$_2$O and almost no iron or magnesium). Their eruption temperatures differ as much as their compositions: basaltic magmas erupt at $1000\text{ to }1200\,{}^{\circ}\mathrm{C}$ (lava streaming through a tube on Kilauea has been measured near $1200\,{}^{\circ}\mathrm{C}$), andesitic magmas at $800\text{ to }1000\,{}^{\circ}\mathrm{C}$, rhyolitic magmas at $650\text{ to }800\,{}^{\circ}\mathrm{C}$. A rhyolitic magma is cooler because it is the residue of a long crystallisation or the first melt of crustal rocks, both of which happen at low temperature (Chapter 8).

The crystal content decides whether a magma flows. Field and geochronological studies of large granite bodies conclude that bulk magmatic flow needs a melt fraction of at least $30\text{ to }50\,\%$ by volume; below that the crystals are welded to their neighbours and the material deforms as a solid with melt-filled pores. Much of the magma stored in the crust is thought to sit in that state, as a mush that becomes mobile again when hot magma is added from below.

5.2 Density and viscosity

5.2.1 The density of a melt

A silicate melt has no crystal lattice, but its volume can still be built from its components. Experiments on many melts show that each oxide contributes a nearly fixed volume per mole, its partial molar volume, which changes slowly and linearly with temperature and pressure; the volume of the melt is the sum of these contributions. The table below gives the values used in this chapter.

Oxide	$\bar V_i$ ($\mathrm{cm}^{3}/\mathrm{mol}$)	$\partial\bar V_i/\partial T$ ($10^{-3}$ $\mathrm{cm}^{3}/(\mathrm{mol}\,\mathrm{K})$)	$\partial\bar V_i/\partial p$ ($10^{-4}$ $\mathrm{cm}^{3}/(\mathrm{mol}\,\mathrm{bar})$)
SiO$_2$	26.86	0	$-1.89$
Al$_2$O$_3$	37.42	2.62	$-2.26$
FeO	12.68	3.69	$-0.45$
MgO	12.02	3.27	$0.27$
CaO	16.90	3.74	$0.34$
Na$_2$O	29.65	7.68	$-2.40$
K$_2$O	47.28	12.08	$-6.75$
H$_2$O	22.9	9.5	$-3.20$

Proposition 5.2 (Density of a silicate melt)

Let a melt contain oxide components $i$ with mole fractions $X_i$, molar masses $M_i$ and partial molar volumes $\bar V_i(T, p) = \bar V_i^{\circ} + (\partial\bar V_i/\partial T)(T - T_{\mathrm{ref}}) + (\partial\bar V_i/\partial p)(p - p_{\mathrm{ref}})$. If the components mix ideally in volume,

$$\rho = \frac{\sum_i X_i M_i}{\sum_i X_i \bar V_i(T,p)}.$$

Water has a molar mass of only $18\,\mathrm{g}/\mathrm{mol}$ but a partial molar volume of $22.9\,\mathrm{cm}^{3}/\mathrm{mol}$, close to that of silica: it lowers the density of a melt strongly. Applied to the three reference melts, the proposition gives $2730\,\mathrm{kg}/\mathrm{m}^{3}$ for the basalt at $1200\,{}^{\circ}\mathrm{C}$, $2480\,\mathrm{kg}/\mathrm{m}^{3}$ for the andesite at $1000\,{}^{\circ}\mathrm{C}$ and $2340\,\mathrm{kg}/\mathrm{m}^{3}$ for the dry rhyolite at $850\,{}^{\circ}\mathrm{C}$; one per cent of dissolved water takes about $70\,\mathrm{kg}/\mathrm{m}^{3}$ off the basalt, as much as several hundred degrees of heating. A basaltic melt is therefore lighter than the mantle peridotite it comes from (about $3380\,\mathrm{kg}/\mathrm{m}^{3}$ below the crust in the reference Earth model) and a little lighter than the lower crust of that model ($2900\,\mathrm{kg}/\mathrm{m}^{3}$), while a rhyolitic melt is lighter even than its upper crust ($2600\,\mathrm{kg}/\mathrm{m}^{3}$). This difference drives magmas upward.

5.2.2 Viscosity

The viscosity of a silicate melt is set by the network of SiO$_4$ tetrahedra linked through shared oxygens: to flow, the melt must break and remake silicon–oxygen bonds. Silica and alumina build the network; the cations of magnesium, iron, calcium and the alkalis break it, and so does water, which turns bridging oxygens into hydroxyl groups. Laboratory measurements on natural melts cover more than fifteen orders of magnitude, from $0.1\,\mathrm{Pa}\,\mathrm{s}$ to $10^{14}\,\mathrm{Pa}\,\mathrm{s}$, and they do not follow a simple Arrhenius law: the viscosity rises faster than exponentially in $1/T$ as the melt cools towards its glass transition.

Proposition 5.3 (Vogel–Fulcher–Tammann law)

The viscosity of a silicate melt of fixed composition follows, to within a few tenths of a logarithmic unit,

$$\log_{10}\eta = A + \frac{B}{T - C},$$

with $\eta$ in pascal seconds and $T$ in kelvins. For natural melts $A$ is close to $-4.55$ whatever the composition; $B$ and $C$ (in kelvins) are linear combinations of the mole percentages of the oxides, with a few products of them, fitted to about $1800$ measurements.

The figure below shows the law for the three reference melts and for the rhyolite with dissolved water. At $1200\,{}^{\circ}\mathrm{C}$ the basalt has $\eta \approx 70\,\mathrm{Pa}\,\mathrm{s}$; the andesite at $1000\,{}^{\circ}\mathrm{C}$ is about $10^{6}\,\mathrm{Pa}\,\mathrm{s}$; the dry rhyolite at $800\,{}^{\circ}\mathrm{C}$ about $10^{10.9}$ $\mathrm{Pa}\,\mathrm{s}$. Four per cent of water brings the same rhyolite down to $10^{5.5}$ $\mathrm{Pa}\,\mathrm{s}$, six per cent to $10^{4.6}$ $\mathrm{Pa}\,\mathrm{s}$. An independent equation fitted only to hydrous granitic melts gives the same values to within half a logarithmic unit (Exercise 5.1). The effect of water is largest at low temperature, where a few tenths of a per cent change the viscosity by orders of magnitude.

Crystals make a magma stiffer still. A suspension of rigid particles resists shear more than its liquid alone, because the liquid must flow around the particles, and when the particles are crowded enough to touch they lock into a framework.

Proposition 5.4 (Einstein–Roscoe relation)

A suspension of a volume fraction $\phi$ of rigid spheres in a Newtonian liquid of viscosity $\eta_{\ell}$ has the viscosity

$$\eta = \eta_{\ell}\,(1 - R\phi)^{-2.5}, \qquad R = 1/\phi_m \approx 1.35,$$

where $\phi_m$ is the maximum packing fraction, at which the suspension jams. For small $\phi$ this reduces to Einstein’s $\eta = \eta_{\ell}(1 + 2.5\phi)$.

Twenty per cent of crystals double the viscosity, forty per cent multiply it by seven, sixty per cent by more than sixty (Exercise 5.5); in real magmas, where the crystals are elongated and of many sizes, the lock-up comes earlier, near the $50\text{ to }70\,\%$ of crystals at which the field evidence places the end of bulk flow.

5.3 Volatiles

Definition 5.5 (Magmatic volatiles)

A magmatic volatile is a component of a magma that is dissolved in the melt at depth and forms a separate gas phase at low pressure: mainly water, then carbon dioxide, then sulfur species. Volatile exsolution is the separation of a volatile from the melt into bubbles of gas. The volatile saturation pressure of a melt is the pressure below which its dissolved volatiles exceed their solubility, so that a gas phase must form.

Mafic magmas typically carry $1\text{ to }3\,\%$ of volatiles by mass, intermediate magmas $3\text{ to }4\,\%$, felsic magmas $4\text{ to }7\,\%$, and the exsolution of $4\text{ to }6\,\%$ of water drives most explosive eruptions. How much water a melt can hold depends above all on pressure. The figure below shows measurements on basaltic and rhyolitic melts held at a known water pressure until saturated, then quenched to glass and analysed: the solubility rises steeply at low pressure and more slowly at high pressure.

Proposition 5.6 (Square-root law for water)

Suppose water dissolves in a silicate melt mostly as hydroxyl groups, through the reaction $\mathrm{H_2O\,(gas)} + \mathrm{O^{2-}\,(melt)} \rightleftharpoons 2\,\mathrm{OH^-\,(melt)}$, with the oxygen of the melt in large excess. Then, at low water contents, the dissolved water is proportional to the square root of the water pressure: $w \approx s\sqrt{p}$, with $s$ a constant of the melt and the temperature.

Fitted by least squares to the measurements, $w = s\sqrt p$ gives $s = 0.34$ weight per cent per $\sqrt{\mathrm{MPa}}$ for the basalts and $s = 0.44$ for the rhyolites; every measurement lies within one per cent of water of its curve. The saturation pressure of a melt with $w$ per cent of water is then $p_s = (w/s)^2$: $129\,\mathrm{MPa}$ for a rhyolite with $5\,\%$ of water, about $5\,\mathrm{km}$ down in the crust, but only $2\,\mathrm{MPa}$ for a basalt with $0.5\,\%$, a hundred metres or less.

Carbon dioxide is far less soluble than water. It dissolves as molecules in rhyolites and as carbonate groups in basalts, more in alkaline melts than in tholeiites, and its solubility also rises with pressure. A magma carrying both gases therefore starts to exsolve deep, with a gas very rich in CO$_2$, and the gas becomes richer in water as the magma rises. Basalts erupted on the deep sea floor contain vesicles of more than $95\,\%$ CO$_2$, which required saturation with CO$_2$ even under $5\,\mathrm{km}$ of water. Estimates of the initial CO$_2$ content of the parental basalts of Kilauea range from $0.2\text{ to }0.6\,\%$, and melt inclusions trapped in their olivine crystals record saturation pressures up to $500\,\mathrm{MPa}$.

Exsolved gas takes up a large volume. One per cent of water in a magma at $900\,{}^{\circ}\mathrm{C}$ occupies, as an ideal gas, about $12\,\%$ of the volume at $100\,\mathrm{MPa}$, $57\,\%$ at $10\,\mathrm{MPa}$ and more than $99\,\%$ at atmospheric pressure (Exercise 5.7). The bubbles lower the density of the magma further and drive it upward faster; in a viscous magma they cannot escape, and Chapter 6 shows how they fragment it.

5.4 Ascent

A body of magma less dense than its surroundings feels a buoyancy force of $\Delta\rho\,g$ per unit volume, where $\Delta\rho$ is the density difference. It can rise in two ways: through a crack it opens ahead of itself, or as a hot, soft blob that pushes the surrounding rock aside.

5.4.1 Dykes

At depth the rock is under a large lithostatic pressure, but a crack filled with magma at a slightly higher pressure than the least principal stress can open and lengthen at its tip. Magma fills the crack as it grows, and the result is a thin sheet of magma, metres wide and kilometres long, that cuts its way upward: a dyke in the making. The magma then flows between two nearly parallel walls.

Proposition 5.7 (Flow of magma in a dyke)

A Newtonian magma of viscosity $\eta$ driven upward by a pressure gradient $G$ (for buoyant flow, $G = \Delta\rho\,g$) between two parallel walls a distance $w$ apart flows with the mean velocity

$$\bar u = \frac{G w^2}{12\,\eta},$$

and the volume flux per unit length of dyke is $q = \bar u w = G w^3/(12\eta)$.

The cube of the width makes wide dykes far more efficient than narrow ones. A basalt of $100\,\mathrm{Pa}\,\mathrm{s}$ in a dyke $1\,\mathrm{m}$ wide, with $\Delta\rho = 300\,\mathrm{kg}/\mathrm{m}^{3}$, rises at about $2.5\,\mathrm{m}/\mathrm{s}$ and crosses $10\,\mathrm{km}$ of crust in an hour; a rhyolite of $10^{6}\,\mathrm{Pa}\,\mathrm{s}$ in the same dyke would take more than a year, and would freeze on the way, so silicic magmas need much wider dykes (Exercise 5.6). The Reynolds number $\rho\bar u w/\eta$ of these flows is small enough for them to stay laminar. Ascent rates estimated for magmas during eruptions separate effusive eruptions, below about $0.1\,\mathrm{m}/\mathrm{s}$, from explosive ones above it, with a transitional range from $0.005\text{ to }0.25\,\mathrm{m}/\mathrm{s}$ where both styles occur; kimberlites, which come from depths beyond $150\,\mathrm{km}$, have been estimated at $0.1\text{ to }30\,\mathrm{m}/\mathrm{s}$, and a diffusion study of one of them gives $0.02\text{ to }0.23\,\mathrm{m}/\mathrm{s}$ averaged over the last $80\text{ to }110\,\mathrm{km}$, a journey of $5\text{ to }74$ days.

5.4.2 Diapirs

Definition 5.8 (Diapir, level of neutral buoyancy, stoping)

A diapir is a body of buoyant material, magma or partly molten rock, that rises through its surroundings by flowing them aside. The level of neutral buoyancy of a magma is the depth at which its density equals that of the surrounding rock: below it the magma is buoyant, above it it is denser than its surroundings. Magmatic stoping is the detachment of blocks of the wall or roof rock, which sink into the magma; blocks that stay enclosed in the solidified rock are xenoliths.

Proposition 5.9 (Stokes ascent)

A sphere of radius $a$ and density deficit $\Delta\rho$ rising slowly through a Newtonian medium of viscosity $\eta$ reaches the terminal velocity

$$v = \frac{2\,\Delta\rho\,g\,a^2}{9\,\eta}$$

if it is rigid, and $1.5$ times this if it is a fluid of negligible viscosity.

The rock around a diapir is not a fluid of $100\,\mathrm{Pa}\,\mathrm{s}$ but a hot solid, whose viscosity is many orders of magnitude higher than that of any melt: in the crust, a sphere of magma a kilometre across rises only millimetres a year, and loses its heat long before it has risen far (Exercise 5.9). Diapirs work where the surroundings are hot and soft, in the mantle and in the deep roots of mountain belts, and where the buoyant material is a partly molten mass kilometres across; the shallow crust is crossed by dykes.

5.5 Emplacement

Most magma never reaches the surface. It stops where it is no longer buoyant, or where it is blocked by a stronger layer, and spreads into a body whose shape records how it made room for itself.

Definition 5.10 (Intrusive bodies)

A pluton is any body of magma that solidified within the crust. A dyke is a sheet-like pluton that cuts across the layering of its host rock (or any sheet if the host has no layering); a sill is a sheet-like pluton parallel to the layering. A laccolith is a sill that has thickened by pushing up the overlying layers into a dome. A batholith is a large pluton, or an assembly of plutons, with an exposed area of more than $100\,\mathrm{km}^{2}$.

Where a magma stops depends first on its density. In a crust whose density increases with depth, a magma rises to its level of neutral buoyancy and spreads sideways there, as a sill or a lens; light sediments near the surface are especially effective traps for dense basaltic magma. A strong, stiff layer can also stop a dyke and divert it along its base. The figure below shows the principle.

A pluton must make room for itself. It can lift its roof, as a laccolith does; it can push its walls aside where they are soft; it can fill space opened by faulting; and it can stope its roof, block by block. Large plutons were long pictured as huge tanks of magma emplaced at once. Dating has changed that picture. The zircons of the Tuolumne Intrusive Suite, a granite batholith in California, show that it was emplaced between $95$ and $85\,\mathrm{Ma}$, oldest at the margins and youngest at the centre, and one of its units alone over more than $3\,\mathrm{Myr}$, between $92.8\,\mathrm{Ma}$ and $88.8\,\mathrm{Ma}$. A single body of magma $5\,\mathrm{km}$ thick and $20\,\mathrm{km}$ wide emplaced at $15\,\mathrm{km}$ depth would have cooled below $750\,{}^{\circ}\mathrm{C}$, where it is about half crystallised, in about $500\,000$ years. Such plutons must have been assembled incrementally, from many smaller intrusions such as sheets and dykes, over millions of years (Exercise 5.11).

5.6 Cooling and contact metamorphism

An intrusion loses its heat by conduction into its host rocks. The simplest model treats the intrusion and its host as one conducting medium of uniform diffusivity $\kappa$: at time zero a sheet of thickness $b$ is at the intrusion temperature $T_m$ and the host at $T_0$. The solution of the heat equation, with $x$ measured from the middle of the sheet, is

The figure below shows it for a sill $100\,\mathrm{m}$ thick.

Proposition 5.11 (Contact temperature)

In this model, as long as $\sqrt{\kappa t}$ is small compared with $b$, the temperature at the contact of the intrusion stays at the mean $(T_m + T_0)/2$ of the intrusion and host temperatures.

A basaltic dyke at $1150\,{}^{\circ}\mathrm{C}$ in rocks at $50\,{}^{\circ}\mathrm{C}$ thus heats its contact to $600\,{}^{\circ}\mathrm{C}$. The heat of crystallisation, typically $700\text{ to }1000\,\mathrm{MJ}/\mathrm{m}^{3}$, keeps the magma hot longer and raises this value: with $900\,\mathrm{MJ}/\mathrm{m}^{3}$ released between $1150$ and $950\,{}^{\circ}\mathrm{C}$ the same model gives $690\,{}^{\circ}\mathrm{C}$, while temperature-dependent conductivities and wet host rocks lower both estimates by 50 to $75\,{}^{\circ}\mathrm{C}$.

The time scale of cooling is the time heat needs to diffuse across the intrusion, of order $(b/2)^2/\kappa$: about $16$ days for a dyke $2\,\mathrm{m}$ wide, a century for a sill $100\,\mathrm{m}$ thick, a hundred thousand years for a pluton of a few kilometres. The latent heat lengthens it, and so does the need to solidify the magma rather than merely cool it.

Proposition 5.12 (Solidification of an intrusion)

Let magma at its solidification temperature $T_m$, with latent heat $L$ per unit volume, fill the space $y > 0$ beside a host rock at $T_0$ occupying $y < 0$, both of heat capacity $\rho c$ per unit volume and diffusivity $\kappa$. The solidification front advances as $y_s = 2\lambda\sqrt{\kappa t}$, where $\lambda$ is the root of

$$\frac{L\sqrt\pi}{\rho c\,(T_m - T_0)} = \frac{e^{-\lambda^2}}{\lambda\,(1 + \operatorname{erf}\lambda)},$$

and the contact stays at $T_0 + (T_m - T_0)/(1 + \operatorname{erf}\lambda)$. A sill of thickness $b$, solidifying from both walls, is solid after $t = b^2/(16\lambda^2\kappa)$.

With the latent heat of a basalt and a host at about $100\,{}^{\circ}\mathrm{C}$, $\lambda$ is about $0.65$: a sill $100\,\mathrm{m}$ thick solidifies in about seventy years, and its contact stays near $740\,{}^{\circ}\mathrm{C}$ rather than $625\,{}^{\circ}\mathrm{C}$ while it does. The weekend problem works through this case. Because the time grows as the square of the thickness, a sill ten times thicker takes a hundred times longer.

Definition 5.13 (Contact aureole and its rocks)

A chilled margin is the fine-grained or glassy rim of an intrusion, where the magma cooled quickly against colder host rock. A contact aureole is the zone of host rock around an intrusion that was metamorphosed by its heat. Hornfels is the hard, fine-grained, non-foliated rock that mudstones and volcanic rocks become in a contact aureole. Skarn is the rock formed where hot magmatic fluids react with carbonate host rock, depositing calcium–iron–magnesium silicates such as garnet and pyroxene, often with magnetite and metal sulfides.

Contact aureoles range from a few centimetres around small dykes and sills to several tens of metres around a large stock, and contact metamorphism spans about $300\,{}^{\circ}\mathrm{C}$ to more than $800\,{}^{\circ}\mathrm{C}$. A mudstone becomes a hornfels, a limestone a marble, a sandstone a quartzite. A magma emplaced deep in rocks that are already hot makes almost no aureole: the width depends on the size and temperature of the intrusion and on the temperature of its host, as the cooling model predicts (Exercise 5.12). The minerals that grow in an aureole, and what they say about temperature, belong to Chapter 10; the metals that skarns concentrate to Chapter 29.

In the field — Reading the contacts of a dyke

Where a dyke is exposed, a geologist first follows its contacts with the host rock. A margin finer-grained or glassier than the centre of the dyke shows that the magma was chilled against cold rock; a dyke $25\,\mathrm{cm}$ wide studied on the coast of British Columbia has chilled margins $2\,\mathrm{cm}$ thick. The host rock is examined next to the contact for baking: a change of colour, a hardening, new minerals that fade away from the dyke. The width of the baked zone and of the chilled margin, measured against the width of the dyke, indicates how hot the host was when the dyke arrived; a dyke with no chilled margin and no aureole was intruded into rocks that were already hot. Samples are taken across the dyke, from margin to centre, for the geochemistry of its melt and for the size of its crystals.

5.7 Exercises

Exercise 5.1 ★

For hydrous granitic melts, an equation fitted to measurements gives $\log_{10}\eta = [-3.545 + 0.833\ln w] + [9601 - 2368\ln w]/\{T - [195.7 + 32.25\ln w]\}$, with $\eta$ in $\mathrm{Pa}\,\mathrm{s}$, $w$ the water content in weight per cent and $T$ in kelvins. Compute $\log_{10}\eta$ at $800\,{}^{\circ}\mathrm{C}$ for $w = 1$ and $w = 4$, and the factor by which the three extra per cent of water reduce the viscosity.

Solution

Solution of Exercise 5.1.

$T = 1073\,\mathrm{K}$. For $w = 1$, $\ln w = 0$: $\log_{10}\eta = -3.545 + 9601/(1073 - 195.7) = -3.545 + 10.94 = 7.40$. For $w = 4$, $\ln 4 = 1.386$: $A = -3.545 + 1.155 = -2.390$, $B = 9601 - 3283 = 6318$, $C = 195.7 + 44.7 = 240.4$, and $\log_{10}\eta = -2.390 + 6318/832.6 = -2.390 + 7.588 = 5.20$. The viscosity falls by $10^{7.40 - 5.20} = 10^{2.20} \approx 160$.

Exercise 5.2 ★

With the square-root law, $s = 0.44$ for a rhyolite and $s = 0.34$ for a basalt (weight per cent per $\sqrt{\mathrm{MPa}}$), compute the saturation pressure of a rhyolite with $6\,\%$ of water and of a basalt with $0.5\,\%$, and the corresponding depths in a crust of density $2600\,\mathrm{kg}/\mathrm{m}^{3}$ ($g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$).

Solution

Solution of Exercise 5.2.

Rhyolite: $p_s = (6/0.44)^2 = 186\,\mathrm{MPa}$, at a depth of $186 \times 10^{6}/(2600\times9.81) = 7.3\,\mathrm{km}$. Basalt: $p_s = (0.5/0.34)^2 = 2.16\,\mathrm{MPa}$, at $2.16 \times 10^{6}/(2600\times9.81) = 85\,\mathrm{m}$. The rhyolite starts to form bubbles deep in the crust, the basalt only near the surface (if it carries no CO$_2$).

Exercise 5.3 ★

Compute the contact temperature of a basaltic dyke at $1150\,{}^{\circ}\mathrm{C}$ intruded into rocks at $50\,{}^{\circ}\mathrm{C}$, then into rocks at $400\,{}^{\circ}\mathrm{C}$. Which of the two dykes will have the wider chilled margin?

Solution

Solution of Exercise 5.3.

$(1150 + 50)/2 = 600\,{}^{\circ}\mathrm{C}$ and $(1150 + 400)/2 = 775\,{}^{\circ}\mathrm{C}$. The dyke in the cold rock is chilled more strongly at its contact, and has the wider chilled margin.

Exercise 5.4 ★

Name each body. (a) A sheet of dolerite $30\,\mathrm{m}$ thick lying between two beds of sandstone over $20\,\mathrm{km}$. (b) A vertical sheet of basalt $2\,\mathrm{m}$ wide cutting horizontal limestones. (c) A granite exposed over $2000\,\mathrm{km}^{2}$, made of a dozen units of different ages. (d) A lens of porphyry $1\,\mathrm{km}$ across, flat at its base, above which the beds are arched up. (e) A horizontal sheet of basalt in a granite.

Solution

Solution of Exercise 5.4.

(a) A sill, parallel to the bedding. (b) A dyke, cutting across the bedding. (c) A batholith, more than $100\,\mathrm{km}^{2}$ and assembled from many intrusions. (d) A laccolith, which has domed its roof. (e) A dyke: granite has no layering, so any sheet in it is a dyke, whatever its orientation.

Exercise 5.5 ★★

A basaltic melt has $\eta_{\ell} = 100\,\mathrm{Pa}\,\mathrm{s}$. With the Einstein–Roscoe relation ($R = 1.35$), compute the viscosity of the magma with $5\,\%$, $20\,\%$, $40\,\%$ and $60\,\%$ of crystals, and compare the first with Einstein’s dilute formula. At what crystal fraction does the relation predict jamming?

Solution

Solution of Exercise 5.5.

$(1 - 1.35\phi)^{-2.5}$: for $\phi = 0.05$, $0.9325^{-2.5} = 1.19$, so $119\,\mathrm{Pa}\,\mathrm{s}$, against $1 + 2.5\times0.05 = 1.125$, $113\,\mathrm{Pa}\,\mathrm{s}$, with Einstein’s formula (close, as it should be in the dilute limit). For $0.20$: $0.73^{-2.5} = 2.20$, $220\,\mathrm{Pa}\,\mathrm{s}$; for $0.40$: $0.46^{-2.5} = 6.97$, $697\,\mathrm{Pa}\,\mathrm{s}$; for $0.60$: $0.19^{-2.5} = 63.6$, about $6360\,\mathrm{Pa}\,\mathrm{s}$. Jamming at $\phi = 1/1.35 = 0.74$.

Exercise 5.6 ★★

Magma of density $300\,\mathrm{kg}/\mathrm{m}^{3}$ less than its walls rises through a dyke. (a) For a basalt ($\eta = 100\,\mathrm{Pa}\,\mathrm{s}$, $\rho = 2700\,\mathrm{kg}/\mathrm{m}^{3}$) in a dyke $1\,\mathrm{m}$ wide, compute the mean velocity, the flux per metre of dyke, the Reynolds number and the time to rise $10\,\mathrm{km}$. (b) Same time for a rhyolite of $10^{6}\,\mathrm{Pa}\,\mathrm{s}$ in a dyke $1\,\mathrm{m}$ wide, then $10\,\mathrm{m}$ wide. (c) A dyke $1\,\mathrm{m}$ wide loses its heat in a time of order $(w/2)^2/\kappa$, with $\kappa = 1 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$: which of the three ascents can succeed?

Solution

Solution of Exercise 5.6.

$G = 300\times9.81 = 2943\,\mathrm{Pa}/\mathrm{m}$. (a) $\bar u = 2943\times1^2/(12\times100) = 2.45\,\mathrm{m}/\mathrm{s}$; $q = 2.45\,\mathrm{m}^{2}/\mathrm{s}$; $Re = 2700\times2.45\times1/100 = 66$, laminar; $t = 10000/2.45 = 4080\,\mathrm{s}$, about $1.1\,\mathrm{h}$. (b) $\bar u = 2943/(12\times10^{6}) = 2.45 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$, $t = 4.08 \times 10^{7}\,\mathrm{s} = 472$ days; with $w = 10\,\mathrm{m}$, $\bar u$ is a hundred times larger, $0.0245\,\mathrm{m}/\mathrm{s}$, and $t = 4.08 \times 10^{5}\,\mathrm{s} = 4.7$ days. (c) The $1\,\mathrm{m}$ dyke cools in about $0.5^2/10^{-6} = 2.5 \times 10^{5}\,\mathrm{s}$, 2.9 days; the $10\,\mathrm{m}$ dyke in $5^2/10^{-6} = 2.5 \times 10^{7}\,\mathrm{s}$, 289 days. The basalt ($1.1\,\mathrm{h}$) and the rhyolite in the wide dyke (4.7 days) arrive before their dykes freeze; the rhyolite in the $1\,\mathrm{m}$ dyke (472 days) cannot.

Exercise 5.7 ★★

A magma of density $2400\,\mathrm{kg}/\mathrm{m}^{3}$ at $900\,{}^{\circ}\mathrm{C}$ releases $1\,\%$ of its mass as water vapour. Treating the vapour as an ideal gas ($R = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$, $18.015\,\mathrm{g}/\mathrm{mol}$), compute the volume fraction of gas in the magma at $100\,\mathrm{MPa}$, $10\,\mathrm{MPa}$ and $0.1\,\mathrm{MPa}$. Which of the three values is the least reliable, and why?

Solution

Solution of Exercise 5.7.

Per kilogram of magma, $n = 10/18.015 = 0.555\,\mathrm{mol}$ of water and $nRT = 0.555\times8.314\times1173 = 5414\,\mathrm{J}$; the remaining melt occupies $0.99/2400 = 4.13 \times 10^{-4}\,\mathrm{m}^{3}$. At $100\,\mathrm{MPa}$: $V = 5.41 \times 10^{-5}\,\mathrm{m}^{3}$, fraction $5.41/(5.41 + 41.3) = 0.12$. At $10\,\mathrm{MPa}$: $V = 5.41 \times 10^{-4}\,\mathrm{m}^{3}$, fraction $0.57$. At $0.1\,\mathrm{MPa}$: $V = 0.0541\,\mathrm{m}^{3}$, fraction $0.992$. The value at $100\,\mathrm{MPa}$ is the least reliable: water vapour at such pressure is dense and far from ideal (and much of the water would in fact still be dissolved).

Exercise 5.8 ★★

In a basin, the density of the rocks increases with depth $z$ as $\rho(z) = 2300 + 60\,z$ ($\mathrm{kg}/\mathrm{m}^{3}$, $z$ in $\mathrm{km}$) down to $10\,\mathrm{km}$. (a) Where is the level of neutral buoyancy of a basaltic melt of $2650\,\mathrm{kg}/\mathrm{m}^{3}$? (b) The density model of this chapter shows that one per cent of dissolved water lowers the density of a basaltic melt by about $70\,\mathrm{kg}/\mathrm{m}^{3}$. Where does a melt with one more per cent of water stop? (c) What is the buoyancy force per cubic metre of the dry melt $2\,\mathrm{km}$ below its level of neutral buoyancy?

Solution

Solution of Exercise 5.8.

(a) $2300 + 60z = 2650$ gives $z = 5.8\,\mathrm{km}$. (b) The melt has $2580\,\mathrm{kg}/\mathrm{m}^{3}$: $z = 280/60 = 4.7\,\mathrm{km}$, about $1.2\,\mathrm{km}$ higher. (c) At $z = 7.8\,\mathrm{km}$ the rock has $2300 + 60\times7.8 = 2770\,\mathrm{kg}/\mathrm{m}^{3}$; the force is $(2770 - 2650)\times9.81 = 1.2\,\mathrm{kN}$ per cubic metre, upward.

Exercise 5.9 ★★

Suppose a granitic diapir of radius $1\,\mathrm{km}$ and density deficit $300\,\mathrm{kg}/\mathrm{m}^{3}$ rises through lower crust of viscosity $10^{19}\,\mathrm{Pa}\,\mathrm{s}$. Compute its Stokes velocity (rigid sphere) in millimetres per year and the time to rise $10\,\mathrm{km}$. Compare with the time scale $a^2/\kappa$ of its cooling, with $\kappa = 1 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$. What do you conclude?

Solution

Solution of Exercise 5.9.

$v = 2\times300\times9.81\times(1000)^2/(9\times10^{19}) = 6.5 \times 10^{-11}\,\mathrm{m}/\mathrm{s}$, or $6.5 \times 10^{-11}\times3.156 \times 10^{7} = 2.1 \times 10^{-3}\,\mathrm{m}$ a year, $2.1\,\mathrm{mm}/\mathrm{yr}$. Rising $10\,\mathrm{km}$ takes $10^{4}/2.1 \times 10^{-3} = 4.8\,\mathrm{Myr}$. The cooling time scale is $(1000)^2/10^{-6} = 10^{12}\,\mathrm{s} = 32\,\mathrm{kyr}$: the body solidifies after rising about $32000\times0.0021 \approx 70\,\mathrm{m}$. A magmatic diapir cannot cross such a crust; it needs hotter, softer surroundings.

Exercise 5.10 ★★★

Water dissolves in a melt both as molecules, $x_m = K_1 f$, and as hydroxyl groups, $x_{\mathrm{OH}}^2 = K_2 x_m$ (the oxygen of the melt in large excess), so that the total dissolved water is $x = x_m + x_{\mathrm{OH}}/2$. (a) Express $x$ as a function of $f$ and show that it behaves as $\sqrt f$ at low $f$ and as $f$ at high $f$. (b) At what fugacity are the two forms of water equally abundant? (c) With $K_1 = 1 \times 10^{-4}\,\mathrm{MPa}^{-1}$ and $K_2 = 0.2$, compute that fugacity, and the share of water dissolved as hydroxyl at $100\,\mathrm{MPa}$.

Solution

Solution of Exercise 5.10.

(a) $x_{\mathrm{OH}} = \sqrt{K_2x_m} = \sqrt{K_1K_2f}$, so $x = K_1f + \frac12\sqrt{K_1K_2f}$. The ratio of the first term to the second is $2\sqrt{K_1f/K_2}$: it tends to zero when $f \to 0$ (the square root dominates) and grows without bound with $f$ (the linear term dominates). (b) $K_1 f = \frac12\sqrt{K_1K_2 f}$ gives $K_1^2f^2 = K_1K_2f/4$, so $f = K_2/(4K_1)$. (c) $f = 0.2/(4 \times 10^{-4}) = 500\,\mathrm{MPa}$. At $100\,\mathrm{MPa}$: $x_m = 0.0100$, $\frac12\sqrt{10^{-4}\times0.2\times100} = \frac12\times0.0447 = 0.0224$, so $x = 0.0324$ and the hydroxyl share is $0.0224/0.0324 = 69\,\%$.

Exercise 5.11 ★★★

A granite pluton of $1000\,\mathrm{km}^{3}$ is assembled over $4\,\mathrm{Myr}$. (a) Compute its mean magma supply in $\mathrm{km}^{3}$ per year and in $\mathrm{m}^{3}$ per second. (b) A single dyke $1\,\mathrm{km}$ long and $2\,\mathrm{m}$ wide, with a mean velocity of $0.8\,\mathrm{m}/\mathrm{s}$, feeds it: how long would the dyke take to deliver the whole pluton? What fraction of the $4\,\mathrm{Myr}$ is that? (c) If the pluton is built of sheets $20\,\mathrm{m}$ thick and $10\,\mathrm{km}$ by $10\,\mathrm{km}$ wide, how many sheets are needed, and how much time separates two of them on average? (d) With $\lambda = 0.65$ and $\kappa = 0.7 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$, how long does a sheet take to solidify? Is the pluton ever a single large body of magma?

Solution

Solution of Exercise 5.11.

(a) $1000/4 \times 10^{6} = 2.5 \times 10^{-4}\,\mathrm{km}^{3}$ a year, that is $2.5 \times 10^{5}\,\mathrm{m}^{3}$ a year, or $2.5 \times 10^{5}/3.156 \times 10^{7} = 0.0079\,\mathrm{m}^{3}/\mathrm{s}$, about eight litres a second. (b) The dyke carries $1000\times2\times0.8 = 1600\,\mathrm{m}^{3}/\mathrm{s}$, and delivers $10^{12}\,\mathrm{m}^{3}$ in $6.25 \times 10^{8}\,\mathrm{s}$, about 20 years: $5 \times 10^{-6}$ of the $4\,\mathrm{Myr}$. (c) Each sheet holds $20\times10^{4}\times10^{4} = 2 \times 10^{9}\,\mathrm{m}^{3} = 2\,\mathrm{km}^{3}$, so 500 sheets, one every $4 \times 10^{6}/500 = 8000$ years on average. (d) $t = (10)^2/(4\times0.65^2\times0.7 \times 10^{-6}) = 8.5 \times 10^{7}\,\mathrm{s}$, 2.7 years. Each sheet solidifies thousands of years before the next arrives: unless the supply comes in rapid bursts, the pluton is never one large body of magma, as the dating of large plutons suggests.

Exercise 5.12 ★★★

Far from a thin dyke of width $b$ (at distances $x \gg b$), its heat acts as an instantaneous plane source: $T - T_0 = (T_m - T_0)\,b\,e^{-x^2/(4\kappa t)}/\sqrt{4\pi\kappa t}$. (a) Show that at distance $x$ the temperature is highest at $t^* = x^2/(2\kappa)$, and that this highest rise is $\Delta T_{\max} = (T_m - T_0)\,b/(x\sqrt{2\pi e})$. (b) For a dyke $2\,\mathrm{m}$ wide with $T_m - T_0 = 1100\,\mathrm{K}$, at what distance does the rock warm by at most $100\,\mathrm{K}$, and when ($\kappa = 0.7 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$)? (c) Why are the aureoles of thin dykes so narrow?

Solution

Solution of Exercise 5.12.

(a) At fixed $x$, $\ln(T - T_0) = \text{const} - \frac12\ln t - x^2/(4\kappa t)$, whose derivative $-1/(2t) + x^2/(4\kappa t^2)$ vanishes at $t^* = x^2/(2\kappa)$. Then $e^{-x^2/(4\kappa t^*)} = e^{-1/2}$ and $\sqrt{4\pi\kappa t^*} = x\sqrt{2\pi}$, so $\Delta T_{\max} = (T_m - T_0)\,b\,e^{-1/2}/(x\sqrt{2\pi}) = (T_m - T_0)\,b/(x\sqrt{2\pi e})$. (b) $\sqrt{2\pi e} = 4.13$: $x = 1100\times2/(100\times4.13) = 5.3\,\mathrm{m}$, reached after $t^* = 5.3^2/(2\times0.7 \times 10^{-6}) = 2.0 \times 10^{7}\,\mathrm{s}$, about 230 days (at only 2.7 dyke widths the plane-source approximation is rough). (c) The heating falls off as $b/x$: a thin dyke carries little heat, and the rock is warmed appreciably only within a few dyke widths.

5.8 Problem: The sill under the plateau

Problem 5.1

Weekend problem — following a basaltic magma from its feeder dyke to a sill, its water, and the time the sill takes to solidify

Beneath an imaginary plateau, sediments (shale and sandstone) of mean density $2450\,\mathrm{kg}/\mathrm{m}^{3}$ rest at $3.0\,\mathrm{km}$ depth on a crystalline basement of density $2800\,\mathrm{kg}/\mathrm{m}^{3}$. A basaltic magma rose from $15\,\mathrm{km}$ through a dyke $2.0\,\mathrm{m}$ wide and spread along the base of the sediments as a dolerite sill $100\,\mathrm{m}$ thick. Data: intrusion temperature $1150\,{}^{\circ}\mathrm{C}$; melt viscosity law $\log_{10}\eta = -4.55 + 5510/(T - 611)$ ($T$ in kelvins); melt density with its water $2710\,\mathrm{kg}/\mathrm{m}^{3}$; $20\,\%$ of crystals by volume (Einstein–Roscoe, $R = 1.35$); dissolved water $0.40\,\%$, solubility $s = 0.34$ per cent per $\sqrt{\mathrm{MPa}}$; temperature of the host at the sill $100\,{}^{\circ}\mathrm{C}$; $\kappa = 0.7 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$, $\rho c = 2.5\,\mathrm{MJ}/(\mathrm{m}^{3}\,\mathrm{K})$, latent heat $900\,\mathrm{MJ}/\mathrm{m}^{3}$, the magma being treated as solidifying at its intrusion temperature; $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$; one year is $3.156 \times 10^{7}\,\mathrm{s}$.

Part I — The magma.

1. Compute $\log_{10}\eta$ and $\eta$ of the melt at $1150\,{}^{\circ}\mathrm{C}$ ($1423\,\mathrm{K}$).
2. Same at $1100\,{}^{\circ}\mathrm{C}$. By what factor does the viscosity change over these $50\,\mathrm{K}$?
3. Compute the viscosity of the magma with its crystals at $1150\,{}^{\circ}\mathrm{C}$.
4. At what crystal fraction would the Einstein–Roscoe relation make the magma jam? Compare with the field estimate of the end of bulk flow.
5. Compute the density difference between the magma and the basement, and between the magma and the sediments.
6. Compute the lithostatic pressure at the level of the sill.

Part II — Up the dyke.

7. Compute the buoyant pressure gradient $\Delta\rho\,g$ that drives the magma through the basement.
8. Compute the mean velocity of the magma in the dyke.
9. Compute the flux per metre of dyke, and the total flux if the dyke is $1.0\,\mathrm{km}$ long.
10. Compute the Reynolds number of the flow. Is it laminar?
11. How long does the magma take to rise from $15\,\mathrm{km}$ to $3.0\,\mathrm{km}$?
12. Compare with the time scale $(w/2)^2/\kappa$ over which the dyke loses its heat. Can it stay open?
13. Explain why the magma stopped at $3.0\,\mathrm{km}$ and spread sideways, using the chapter’s vocabulary.

Part III — The water.

14. Compute the saturation pressure of the melt’s water, and the depth in the sediments where it is reached.
15. How much water could the melt hold at the level of the sill? Does the sill magma exsolve water?
16. Vesicles in the chilled margin of the sill contain mostly CO$_2$. Explain.
17. If all $0.40\,\%$ of water came out at the surface, at $1150\,{}^{\circ}\mathrm{C}$ and $0.1\,\mathrm{MPa}$, what volume fraction of the magma would the vapour occupy (ideal gas)?

Part IV — Cooling and solidification.

18. Ignoring latent heat, compute the contact temperature of the sill.
19. Ignoring latent heat, compute the temperature $10\,\mathrm{m}$ above the top contact $10$ years after the intrusion.
20. Same model: compute the temperature at the middle of the sill after $10$ years and $50$ years.
21. Same model: $80\,\mathrm{m}$ above the top contact, compute the temperature after $300$, $363$ and $420$ years. Does the rock there ever reach $300\,{}^{\circ}\mathrm{C}$?
22. Compute $L\sqrt\pi/(\rho c\,(T_m - T_0))$ for the sill.
23. Evaluate $e^{-\lambda^2}/(\lambda(1 + \operatorname{erf}\lambda))$ for $\lambda = 0.64$, $0.65$ and $0.66$ ($\operatorname{erf}0.64 = 0.6346$, $\operatorname{erf}0.65 = 0.6420$, $\operatorname{erf}0.66 = 0.6494$), and give $\lambda$ to two decimals.
24. With latent heat, compute the contact temperature while the sill solidifies. Why is it higher than in question 18?
25. Compute the time scale $(b/2)^2/\kappa$ in years.
26. Compute the time the sill takes to solidify, in years, the key number of this problem, and explain why it is shorter than the time scale of the last question even though latent heat must be removed.

Solution

Solution of Problem 5.1.

1. $\log_{10}\eta = -4.55 + 5510/(1423 - 611) = -4.55 + 6.786 = 2.24$; $\eta = 172\,\mathrm{Pa}\,\mathrm{s}$.

2. $-4.55 + 5510/762 = -4.55 + 7.231 = 2.68$, $\eta = 480\,\mathrm{Pa}\,\mathrm{s}$: a factor of $480/172 = 2.8$ for $50\,\mathrm{K}$.

3. $172\times(1 - 1.35\times0.20)^{-2.5} = 172\times0.73^{-2.5} = 172\times2.20 = 378\,\mathrm{Pa}\,\mathrm{s}$.

4. $\phi = 1/1.35 = 0.74$. The field evidence places the end of bulk flow at a melt fraction of $30\text{ to }50\,\%$, that is $50\text{ to }70\,\%$ of crystals: earlier, because real crystals are not spheres of one size.

5. Basement: $2800 - 2710 = 90\,\mathrm{kg}/\mathrm{m}^{3}$, the magma is lighter. Sediments: $2450 - 2710 = -260\,\mathrm{kg}/\mathrm{m}^{3}$, the magma is denser.

6. $p = 2450\times9.81\times3000 = 72.1\,\mathrm{MPa}$.

7. $90\times9.81 = 883\,\mathrm{Pa}/\mathrm{m}$.

8. $\bar u = 883\times2.0^2/(12\times378) = 3532/4536 = 0.78\,\mathrm{m}/\mathrm{s}$.

9. $q = 0.78\times2.0 = 1.56\,\mathrm{m}^{2}/\mathrm{s}$; over $1.0\,\mathrm{km}$, about $1560\,\mathrm{m}^{3}/\mathrm{s}$.

10. $Re = 2710\times0.78\times2.0/378 = 11$: laminar.

11. $12000/0.78 = 15\,400\,\mathrm{s}$, about $4.3\,\mathrm{h}$.

12. $(1.0)^2/0.7 \times 10^{-6} = 1.43 \times 10^{6}\,\mathrm{s}$, 16.5 days, far longer than $4.3\,\mathrm{h}$: the dyke stays open while the magma flows.

13. At $3.0\,\mathrm{km}$ the magma meets sediments lighter than itself: the base of the sediments is its level of neutral buoyancy. Above it the magma would be denser than its surroundings, so it spreads sideways along the contact as a sill.

14. $p_s = (0.40/0.34)^2 = 1.38\,\mathrm{MPa}$, reached at $1.38 \times 10^{6}/(2450\times9.81) = 58\,\mathrm{m}$ below the surface.

15. $0.34\times\sqrt{72.1} = 0.34\times8.49 = 2.9\,\%$, far more than the $0.40\,\%$ present: the sill magma exsolves no water.

16. CO$_2$ is much less soluble than water: it saturates and forms bubbles deep in the ascent, so the first gas of a basalt is mostly CO$_2$.

17. Per kilogram, $n = 4.0/18.015 = 0.222\,\mathrm{mol}$, $V = 0.222\times8.314\times1423/ 10^{5} = 0.0263\,\mathrm{m}^{3}$; the melt occupies $0.996/2710 = 3.68 \times 10^{-4}\,\mathrm{m}^{3}$; the fraction is $0.0263/(0.0263 + 0.000368) = 0.986$, $98.6\,\%$.

18. $(1150 + 100)/2 = 625\,{}^{\circ}\mathrm{C}$.

19. After 10 years, $2\sqrt{\kappa t} = 2\sqrt{0.7 \times 10^{-6}\times3.156 \times 10^{8}} = 29.7\,\mathrm{m}$. The point is at $x = 60\,\mathrm{m}$ from the middle: $T = 100 + 525\,[\operatorname{erf}(-10/29.7) + \operatorname{erf}(110/29.7)] = 100 + 525\,(-0.365 + 1.000) = 433\,{}^{\circ}\mathrm{C}$.

20. At the middle, $T = 100 + 1050\operatorname{erf}(50/2\sqrt{\kappa t})$: after 10 years $1050\operatorname{erf}(1.68) = 1050\times0.983$, $T = 1132\,{}^{\circ}\mathrm{C}$; after 50 years, $2\sqrt{\kappa t} = 66.5\,\mathrm{m}$, $1050\operatorname{erf}(0.752) = 1050\times0.713$, $T = 848\,{}^{\circ}\mathrm{C}$.

21. At $x = 130\,\mathrm{m}$: $294\,{}^{\circ}\mathrm{C}$ after 300 years, $295\,{}^{\circ}\mathrm{C}$ after 363 years (the maximum), $295\,{}^{\circ}\mathrm{C}$ after 420 years. Without latent heat, the rock $80\,\mathrm{m}$ above the sill never quite reaches $300\,{}^{\circ}\mathrm{C}$.

22. $9 \times 10^{8}\times1.772/(2.5 \times 10^{6}\times1050) = 0.608$.

23. $\lambda = 0.64$: $e^{-0.4096}/(0.64\times1.6346) = 0.6639/1.0461 = 0.635$; $0.65$: $0.6554/1.0673 = 0.614$; $0.66$: $0.6469/1.0886 = 0.594$. The value $0.608$ lies between $0.65$ and $0.66$, close to $0.65$: $\lambda = 0.65$.

24. $100 + 1050/(1 + 0.642) = 100 + 640 = 740\,{}^{\circ}\mathrm{C}$. The latent heat released at the solidification front keeps the magma side at $1150\,{}^{\circ}\mathrm{C}$ longer, so the contact receives more heat.

25. $(50)^2/0.7 \times 10^{-6} = 3.57 \times 10^{9}\,\mathrm{s}$, 113 years.

26. $t = (100)^2/(16\times0.65^2\times0.7 \times 10^{-6}) = 10^{4}/4.73 \times 10^{-6} = 2.11 \times 10^{9}\,\mathrm{s}$, so $2.11 \times 10^{9}/3.156 \times 10^{7}$: the sill is solid after $67\,$ years. It is shorter than $(b/2)^2/\kappa$ by the factor $1/(4\lambda^2) = 0.59$: that time scale only sets the order of magnitude, and both walls of the sill draw heat into a cold host at once, fast enough to remove the latent heat as well.