Mean heat flow through the ocean floor is higher than through the continents, although oceanic crust is almost free of radioactive heat sources. Which explanation fits?

Ocean floor is young on average and loses heat as new plate cools, while old, thick continents rest their heat flow on radiogenic crust

Oceanic crust holds more uranium, thorium and potassium per unit volume than granitic continental crust, so each column produces more heat

Sea water raises the thermal conductivity of oceanic rock so much that every column conducts more heat at the same temperature gradient

Continental lithosphere is thinner than oceanic lithosphere on average, so heat has a shorter path to escape through the continents

A crustal column has uniform radiogenic heat production A and conductivity k. The surface temperature is T₀ and the surface heat flow is q₀, with z measured downward. In steady state the temperature at depth z is

T₀ + q₀ z/k − A z²/(2k), because the flux falls as q₀ − A z and is integrated once more

T₀ + q₀ z/k + A z²/(2k), because the heat produced adds to the flux from below

T₀ + (q₀ − A) z/k, because the heat production reduces the flux by a constant amount

T₀ + q₀ z/k − A z/k, because the heat production lowers the gradient in proportion to z

T₀ + q₀ z/k − A z²/(2k), because the flux falls as q₀ − A z and is integrated once more

Assume uniform thermal conductivity. Which of the following statements about steady-state continental geotherms are correct?

With uniform heat production A, the upward heat flux decreases linearly with depth as q₀ − A z

For the same surface heat flow, a crust with more radiogenic heat production has a lower temperature at the Moho

The reduced heat flow is the part of the surface heat flow that enters the heat-producing layer from below

With uniform heat production A, the upward heat flux decreases linearly with depth as q₀ − A z

For the same surface heat flow, a crust with more radiogenic heat production has a lower temperature at the Moho

The reduced heat flow is the part of the surface heat flow that enters the heat-producing layer from below

With no heat production the geotherm is a curved line whose temperature gradient decreases steadily with increasing depth

Oceanic lithosphere cools by half-space cooling. The age of a section of sea floor increases from 16 Myr to 64 Myr. What happens to the depth of a given isotherm and to the surface heat flow?

The isotherm depth doubles and the heat flow falls to one half

The isotherm depth quadruples and the heat flow falls to one quarter

The isotherm depth doubles and the heat flow doubles as well

The isotherm depth doubles and the heat flow falls to one half

The isotherm depth stays the same and the heat flow falls to one half

For ocean floor older than about 70 Myr, observed depth stops increasing as √age and the measured heat flow stays above the half-space-cooling prediction. Which change in the thermal model accounts for both observations?

Replace the half-space with a plate of fixed thickness held at the mantle temperature at its base, so heat is supplied from below

Allow the thermal diffusivity to rise steadily with age, so that old lithosphere cools faster than young lithosphere does

Replace the half-space with a plate of fixed thickness held at the mantle temperature at its base, so heat is supplied from below

Add hydrothermal circulation that removes progressively more heat from old ocean floor than from young ocean floor

Add a radiogenic heat source that grows steadily with age within the oceanic crust and so slows the subsidence of old sea floor

The potential temperature of the mantle is defined as

the temperature at the base of the thermal lithosphere, where the geotherm leaves the conductive regime

the mean of the temperatures at the top and the bottom of the mantle

the temperature at which dry peridotite begins to melt at the surface

Suppose the heat that can be conducted down the adiabat of the outer core is larger than the heat that actually crosses the core–mantle boundary. Which conclusion is correct?

Thermal convection cannot run, but compositional buoyancy from inner-core freezing can still drive a dynamo

The outer core must freeze at once, because no heat at all is being removed from the fluid across its boundary

Thermal convection cannot run, but compositional buoyancy from inner-core freezing can still drive a dynamo

The outer core can no longer convect by any mechanism whatever, so the geodynamo must switch off immediately

Convection becomes more vigorous, because a fluid whose temperature lies below its adiabat is unstably stratified

Which of the following statements about the mantle adiabat are correct?

The adiabatic gradient equals α g T / c_p

A well-mixed convecting layer with no strong boundary layers lies close to an adiabat

For otherwise equal properties, the adiabatic gradient is larger where the absolute temperature is higher

The adiabatic gradient equals α g T / c_p

A well-mixed convecting layer with no strong boundary layers lies close to an adiabat

The adiabatic gradient in the upper mantle is about 10 K per km, comparable with the conductive gradient in the continental lithosphere

For otherwise equal properties, the adiabatic gradient is larger where the absolute temperature is higher

Which of the following can bring a parcel of mantle peridotite to melting?

Adiabatic decompression as the parcel rises beneath a spreading ridge

Addition of water released from a subducting slab, which lowers the solidus

Arrival of mantle several hundred kelvin hotter than normal, as in a plume

Adiabatic decompression as the parcel rises beneath a spreading ridge

Addition of water released from a subducting slab, which lowers the solidus

Arrival of mantle several hundred kelvin hotter than normal, as in a plume

Raising the pressure on a dry parcel at constant temperature, as when it is carried downward

Heat Flow: Geotherms, Half-Space Cooling, Convection, the Adiabat and Mantle Melting

This chapter answers the heading “Heat flow” of Part B2 of the Geology and Geophysics (GG) paper: “Elements of conduction and convection within the mantle and the core, Half-space cooling, calculation of simple geotherms, geothermal gradient, equilibrium geotherms, thermal models in plate tectonics, adiabat, and melting in the mantle.” It teaches Fourier’s law and the heat equation, steady-state geotherms with and without radiogenic heat, the half-space cooling solution and the plate model, convection and the Rayleigh number in the mantle and core, the adiabat, and how the mantle melts. The numericals are geotherm temperatures, the error-function temperature profile, lithosphere thickness from age, the adiabatic gradient and the Rayleigh number.

1. Heat in the Earth: sources, Fourier’s law and surface heat flow

The Earth loses about 45 TW of heat through its surface, a mean heat flow of about 0.09 W/m² (87 mW/m²). The mean is about 65 mW/m² on continents and about 100 mW/m² on the ocean floor, which has almost no radioactive crust but is young and still cooling. The heat comes from the decay of uranium, thorium and potassium, from secular cooling (stored heat of formation released as the planet cools) and from the latent heat and gravitational energy released as the inner core grows. How the total divides among these is still debated, so the exam asks about mechanisms, not percentages.

Fourier’s law for conduction is q = −k dT/dz: heat flux (W/m²) runs down the temperature gradient and is proportional to it, the constant being the thermal conductivity k (typically 2 to 3 W m⁻¹ K⁻¹ for crustal rock). With z measured downward and upward flow counted positive, the magnitude is q = k dT/dz, so the geothermal gradient is dT/dz = q/k. A heat flow of 60 mW/m² with k = 2.5 W/m/K gives 24 K/km, close to the observed near-surface gradient of 20 to 30 K/km. Fourier’s law plus energy conservation gives the heat equation ∂T/∂t = κ ∂²T/∂z² + A/(ρ c_p), where κ = k/(ρ c_p) is the thermal diffusivity (about 10⁻⁶ m²/s for rock) and A is the radiogenic heat production per unit volume (W/m³), of order 1 µW/m³ in continental crust and orders of magnitude less in mantle peridotite.

⚠️ Convert units before substituting

1 heat-flow unit (HFU) = 1 µcal cm⁻² s⁻¹ = 41.84 mW/m². The commonest numerical slip is mixing mW/m² with W/m². Convert mW/m² to W/m², km to m and µW/m³ to W/m³ first, then check that the answer has the units of temperature.

2. Simple geotherms: steady state with and without radiogenic heat, and the equilibrium geotherm

A geotherm is the temperature–depth curve. At steady state (∂T/∂t = 0) the heat equation reduces to k d²T/dz² = −A. Let T₀ be the surface temperature and q₀ the surface heat flow.

No heat production (A = 0): the gradient is constant and T(z) = T₀ + q₀ z/k, a straight line of slope q₀/k.
Uniform A: the upward flux shrinks with depth as q(z) = q₀ − A z, since the heat produced above does not come from below. Integrating twice gives T(z) = T₀ + q₀ z/k − A z²/(2k). The gradient, (q₀ − A z)/k, decreases with depth.
Exponentially decaying A = A₀ e^(−z/h): T(z) = T₀ + q_r z/k + (A₀ h²/k)(1 − e^(−z/h)), with q₀ = q_r + A₀ h. The reduced heat flow q_r is the flux entering the radiogenic layer from below, and h is about 10 km in many heat-flow provinces.

Worked example. q₀ = 60 mW/m², k = 3.0 W/m/K, A = 1.0 µW/m³, T₀ = 0 °C, depth 35 km (a Moho depth). Without heat production T = 0.060 × 35 000 / 3.0 = 700 °C. With it, subtract A z²/(2k) = 1.0×10⁻⁶ × (35 000)² / 6.0 ≈ 204 °C, so T ≈ 496 °C, and the flux reaching 35 km is 60 − 35 = 25 mW/m². For the same surface heat flow, crustal heat production lowers the deep temperature, because less heat has to come from below.

An equilibrium geotherm is the steady-state geotherm a lithosphere reaches when heat arriving from below plus heat produced inside equals heat lost at the surface. A disturbance such as crustal thickening, rifting or intrusion decays over a time of order L²/κ: for L = 100 km and κ = 10⁻⁶ m²/s this is 10¹⁶ s, about 300 Myr, so old lithosphere can stay out of equilibrium for a very long time. At equilibrium a lower q₀ means a cooler, thicker lithosphere. Cratons have q₀ near 40 mW/m² and the base of their thermal lithosphere, where the geotherm meets the mantle adiabat (near the 1300 °C isotherm), lies at 200 km or more; hot extended terranes have thin lithosphere.

3. Half-space cooling and the oceanic lithosphere

New ocean floor forms at a ridge at the mantle temperature T_m and then cools from the top. Model it as a half-space z ≥ 0 at uniform T_m at t = 0, with the surface held at T₀ (sea water) afterwards. Conduction without sources, ∂T/∂t = κ ∂²T/∂z², is solved by the similarity variable z/(2√(κt)) and gives T(z, t) = T₀ + (T_m − T₀) erf( z / (2√(κt)) ), with erf(0) = 0 and erf(∞) = 1. Here t is the age of the sea floor, the distance from the ridge divided by the spreading rate.

Values of the error function
η = z / 2√(κt)	erf(η)	Meaning
0.5	0.520	Temperature is 52% of the way from T₀ to T_m
1.0	0.843	84% of the way
1.163	0.900	90%: conventional base of the thermal lithosphere
2.0	0.995	Essentially at mantle temperature

Isotherm depth z ∝ √(κt): quadrupling the age doubles the depth of every isotherm. With the 90% criterion the thickness is z_L = 2.32 √(κt). For κ = 10⁻⁶ m²/s and 100 Myr (3.15×10¹⁵ s) this is 2.32 × 56.1 km ≈ 130 km.
Surface heat flow q(t) = k (T_m − T₀)/√(π κ t) ∝ t^(−1/2): for typical constants about 0.4 to 0.5 W/m² divided by √(age in Myr), so near 100 mW/m² at 20 Myr.
Sea-floor depth grows as √t because the cooling column contracts: d ≈ 2500 + 350 √t metres (t in Myr) for ages up to about 70 Myr, so ocean floor deepens from the ridge crest outward.

Worked example. Take T_m = 1300 °C, T₀ = 0 °C and an age at which 2√(κt) = 80 km. At z = 40 km, η = 0.5 and erf(0.5) = 0.5205, so T = 0.5205 × 1300 ≈ 677 °C. At z = 80 km, η = 1 and T = 0.843 × 1300 ≈ 1096 °C.

4. Thermal models in plate tectonics: the plate model, ridges, slabs and continents

Beyond about 70 Myr, sea-floor depth and heat flow stop following √t and level off. The plate model treats the lithosphere as a plate of fixed thickness (of order 100 km) held at T_m at its base, so heat is supplied from below. It agrees with half-space cooling for young ages and approaches a steady state at old ages, with a linear geotherm, constant heat flow and constant depth. Young ocean floor also shows measured heat flow below the prediction, because sea water circulating through cracks in the crust carries heat away (hydrothermal circulation) and conduction probes do not see it.

Half-space cooling against the plate model
Feature	Half-space cooling	Plate model
Lower boundary	None: semi-infinite	Fixed plate base at T_m
Thickness with age	Grows as √t without limit	Approaches a constant
Old-age heat flow	Keeps falling as t^(−1/2)	Levels off to a constant
Old-age depth	Keeps deepening as √t	Levels off

Subduction zones: a slab sinks faster than it warms by conduction, so it stays cold, which is why it shows as a fast-velocity anomaly in tomography. How deep the cold survives scales with the thermal parameter, the age of the plate times its vertical sinking rate.
Continents: the equilibrium geotherm of Section 2, with radiogenic crust. Ridges and plumes: hot mantle rises, melts, and the new plate then cools as above.

⚠️ Do not apply half-space cooling to old ocean floor

The √t and t^(−1/2) laws are for young sea floor. At large ages the half-space formulas predict ever-thicker lithosphere and ever-smaller heat flow, but real plates level off near 100 km, which is the whole reason the plate model exists.

5. Conduction and convection in the mantle and the core

Heat moves by conduction when viscous drag and thermal diffusion win, and by convection when buoyancy wins. The contest is measured by the Rayleigh number Ra = ρ g α ΔT d³ / (κ η), where α is the thermal expansion coefficient, ΔT the temperature drop across a layer of thickness d and η the dynamic viscosity. Convection begins when Ra exceeds a critical value of order 10³ (the exact value depends on the boundary conditions). Ra grows as d³ and falls as 1/η. For the whole mantle (η of order 10²¹ Pa s) Ra is of order 10⁷, far above critical, so convection is vigorous; its surface expression is plate motion at a few cm/yr.

Mantle: hot material rises in plumes and cold lithosphere sinks at subduction zones. The interior is well mixed and close to an adiabat; conduction does the work in thin thermal boundary layers, the lithosphere at the top and the D″ layer just above the core–mantle boundary (CMB). In vigorous convection the Nusselt number (total heat flux divided by the purely conductive flux) grows roughly as Ra^(1/3).
Outer core: a liquid iron alloy of very low viscosity, so Ra is enormous and the fluid is nearly adiabatic. Iron conducts heat well, so a large part of the heat leaving the core may be carried by conduction down the adiabat. If conduction alone could carry all the heat across the CMB, thermal convection would stop, but freezing of the inner core releases latent heat and light elements, and the resulting compositional convection can still drive the geodynamo.
Inner core: solid, growing by freezing of the outer core, and heat passes through it mainly by conduction.

🎯 A solid mantle that flows

On a human timescale the mantle is rigid and carries S waves. Over millions of years solid-state creep lets it flow like a fluid of viscosity 10²⁰ to 10²² Pa s. Because η sits in the denominator of Ra, a layer that is more than twenty orders of magnitude more viscous than water can still convect once it is thousands of kilometres thick.

6. The adiabat: temperature in the convecting mantle and core

An adiabat is the temperature change of a parcel compressed or decompressed without heat exchange (constant entropy). From (∂T/∂P)_S = αT/(ρ c_p) and dP = ρ g dz, the adiabatic gradient is (dT/dz)_S = α g T / c_p. For the mantle with α = 3×10⁻⁵ K⁻¹, g = 10 m/s², T = 1600 K and c_p = 1200 J/kg/K it is 4×10⁻⁴ K/m = 0.4 K/km, tiny beside the 20 to 30 K/km of the continental lithosphere. A well-mixed convecting layer therefore lies close to an adiabat, and the gradient is proportional to T.

The potential temperature T_p is the temperature a parcel would have if brought adiabatically to the surface without melting; for ambient mantle it is about 1300 to 1400 °C. Following the adiabat, temperature is of the order of 1500 °C at 410 km and 1650 °C at 660 km. Below the mantle’s lower boundary layer D″, where the temperature rises steeply to the CMB value, the outer core is again near-adiabatic, and the inner-core boundary temperature is of order 5000 K. The geotherm of the lithosphere, steep and conductive, meets the mantle adiabat at the base of the thermal lithosphere.

ℹ️ Phase changes bend the adiabat and the slabs

Olivine changes to a denser, spinel-like high-pressure phase (wadsleyite) near 410 km (positive Clapeyron slope), so the boundary rises in cold slabs. The change near 660 km to bridgmanite plus ferropericlase has a negative Clapeyron slope, so the boundary is pushed deeper in cold slabs and resists their passage.

7. Melting in the mantle

Most of the mantle is solid because its temperature lies below the solidus, where melting begins. The dry peridotite solidus rises steeply with pressure, about 100 to 130 K per GPa in the upper mantle, while the adiabat rises only about 10 to 15 K per GPa (α T/(ρ c_p) = 3×10⁻⁵ × 1600 / (3300 × 1200) ≈ 12 K/GPa). A parcel that rises adiabatically therefore closes on the solidus at roughly 110 K per GPa of ascent, and melting starts where the two curves cross. Pressure and depth are linked by P = ρ g z, so 1 GPa is about 30 km in the upper mantle.

Decompression melting is the main mechanism: passive upwelling beneath mid-ocean ridges (melting starts roughly 60 to 80 km down and builds about 7 km of basaltic crust), in rifts and in plume heads.
Flux melting: water and other volatiles released from a subducting slab lower the solidus by hundreds of kelvin, producing arc magmas.
Heating: a plume is hotter than ambient mantle by a hundred to a few hundred kelvin, so at a hotspot it starts to melt deeper and melts more, giving thicker crust. Heat from intrusions can also melt crust.

⚠️ Pressure alone does not melt a dry mantle

Raising the pressure on a dry parcel raises its solidus, so it moves away from melting. Melting needs a parcel to be decompressed, wetted or heated until it crosses the solidus. The latent heat of melting also cools the rising parcel, so melt fraction builds up only gradually with ascent.

Key takeaways

Fourier’s law q = −k dT/dz gives the geothermal gradient q/k; the heat equation adds κ = k/(ρ c_p) and radiogenic heat A. Convert mW/m², km and µW/m³ to SI before substituting.
Steady geotherms: T = T₀ + q₀ z/k with no heat production, and T = T₀ + q₀ z/k − A z²/(2k) with uniform A, where the flux falls as q₀ − A z. Crustal heat production lowers deep temperatures for a given q₀, and low q₀ means thick, cool lithosphere.
Half-space cooling: T = T₀ + (T_m − T₀) erf(z/2√(κt)); isotherm depth and sea-floor depth grow as √t, heat flow falls as t^(−1/2), and the 90% thickness is 2.32√(κt) (about 130 km at 100 Myr).
The plate model adds a fixed-thickness plate with a hot base, so old sea floor levels off in depth and heat flow instead of following √t; hydrothermal circulation makes measured heat flow of young floor too low, and cold slabs sink faster than they warm.
Ra = ρ g α ΔT d³/(κ η), critical value of order 10³, mantle about 10⁷. The mantle interior is near-adiabatic with conducting boundary layers (lithosphere, D″); the outer core convects vigorously and can be stirred compositionally even if conduction carries all its heat.
The adiabatic gradient is α g T/c_p, about 0.4 K/km in the mantle. Melting happens where the shallow-sloping adiabat crosses the steep solidus (decompression), when volatiles lower the solidus (flux), or when the parcel is hotter (plume), never from added pressure on a dry parcel.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A continental site has a surface heat flow of 50 mW/m² through rock of thermal conductivity 2.5 W/m/K, with negligible heat production. What is the near-surface geothermal gradient?
1. 2 K/km
2. 20 K/km
3. 125 K/km
4. 200 K/km
Show answer
Answer: B — 20 K/km
The gradient is q/k = 0.050 / 2.5 = 0.020 K/m = 20 K/km. Multiplying instead of dividing gives the tempting 125 figure, but the product q·k does not have the units of a gradient, and the other values come from slipping a power of ten in the unit conversion.
Mean heat flow through the ocean floor is higher than through the continents, although oceanic crust is almost free of radioactive heat sources. Which explanation fits?
1. Ocean floor is young on average and loses heat as new plate cools, while old, thick continents rest their heat flow on radiogenic crust
2. Oceanic crust holds more uranium, thorium and potassium per unit volume than granitic continental crust, so each column produces more heat
3. Sea water raises the thermal conductivity of oceanic rock so much that every column conducts more heat at the same temperature gradient
4. Continental lithosphere is thinner than oceanic lithosphere on average, so heat has a shorter path to escape through the continents
Show answer
Answer: A — Ocean floor is young on average and loses heat as new plate cools, while old, thick continents rest their heat flow on radiogenic crust
Heat flow through young ocean floor is dominated by cooling of the new plate, which is why the oceanic mean is higher. The radiogenic option is backwards, since oceanic crust is poor in heat-producing elements, and continental lithosphere is on average thicker, not thinner. The conductivity option confuses hydrothermal circulation (which actually lowers measured conductive heat flow on young floor) with conduction.
A crustal column has uniform radiogenic heat production A and conductivity k. The surface temperature is T₀ and the surface heat flow is q₀, with z measured downward. In steady state the temperature at depth z is
1. T₀ + q₀ z/k + A z²/(2k), because the heat produced adds to the flux from below
2. T₀ + (q₀ − A) z/k, because the heat production reduces the flux by a constant amount
3. T₀ + q₀ z/k − A z/k, because the heat production lowers the gradient in proportion to z
4. T₀ + q₀ z/k − A z²/(2k), because the flux falls as q₀ − A z and is integrated once more
Show answer
Answer: D — T₀ + q₀ z/k − A z²/(2k), because the flux falls as q₀ − A z and is integrated once more
From k d²T/dz² = −A the flux is q(z) = q₀ − A z and the gradient is (q₀ − A z)/k; integrating gives T₀ + q₀ z/k − A z²/(2k). A dimension check removes the rest: A z/k has units of K/m, a gradient and not a temperature, and q₀ − A subtracts W/m³ from W/m², while A z²/(2k) has units of kelvin. Adding the quadratic term would also raise the temperature, the opposite of the effect of heat that is produced above and so does not have to be conducted from below.
Assume uniform thermal conductivity. Which of the following statements about steady-state continental geotherms are correct?
1. With uniform heat production A, the upward heat flux decreases linearly with depth as q₀ − A z
2. For the same surface heat flow, a crust with more radiogenic heat production has a lower temperature at the Moho
3. The reduced heat flow is the part of the surface heat flow that enters the heat-producing layer from below
4. With no heat production the geotherm is a curved line whose temperature gradient decreases steadily with increasing depth
Show answer
Answer: A — With uniform heat production A, the upward heat flux decreases linearly with depth as q₀ − A z; B — For the same surface heat flow, a crust with more radiogenic heat production has a lower temperature at the Moho; C — The reduced heat flow is the part of the surface heat flow that enters the heat-producing layer from below
The flux q₀ − A z, the cooler Moho for a given q₀ (less heat has to come from below) and the definition of reduced heat flow are all correct. With A = 0 the heat equation gives d²T/dz² = 0, so the geotherm is a straight line of constant gradient q₀/k; a gradient that decreases with depth needs heat production.
Oceanic lithosphere cools by half-space cooling. The age of a section of sea floor increases from 16 Myr to 64 Myr. What happens to the depth of a given isotherm and to the surface heat flow?
1. The isotherm depth quadruples and the heat flow falls to one quarter
2. The isotherm depth doubles and the heat flow doubles as well
3. The isotherm depth doubles and the heat flow falls to one half
4. The isotherm depth stays the same and the heat flow falls to one half
Show answer
Answer: C — The isotherm depth doubles and the heat flow falls to one half
Isotherm depth goes as √t and heat flow as t^(−1/2). With t multiplied by 4, √4 = 2 deepens the isotherm by a factor of 2 and the heat flow is divided by 2. Quadrupling both is the tempting mistake of using t instead of √t.
For ocean floor older than about 70 Myr, observed depth stops increasing as √age and the measured heat flow stays above the half-space-cooling prediction. Which change in the thermal model accounts for both observations?
1. Allow the thermal diffusivity to rise steadily with age, so that old lithosphere cools faster than young lithosphere does
2. Replace the half-space with a plate of fixed thickness held at the mantle temperature at its base, so heat is supplied from below
3. Add hydrothermal circulation that removes progressively more heat from old ocean floor than from young ocean floor
4. Add a radiogenic heat source that grows steadily with age within the oceanic crust and so slows the subsidence of old sea floor
Show answer
Answer: B — Replace the half-space with a plate of fixed thickness held at the mantle temperature at its base, so heat is supplied from below
A plate with a hot base stops thickening, so depth and heat flow level off to constant values above the half-space prediction. Extra cooling (faster diffusion or more hydrothermal loss) would push heat flow below the prediction, the wrong direction, and oceanic crust has too little radiogenic heat to matter.
A convecting layer has its viscosity reduced tenfold and its thickness halved, with the same temperature drop and material properties otherwise. By what factor does the Rayleigh number change?
1. It is multiplied by 1.25
2. It is multiplied by exactly 5
3. It is multiplied by about 0.8
4. It is multiplied by exactly 80
Show answer
Answer: A — It is multiplied by 1.25
Ra is proportional to d³/η. Halving d multiplies d³ by 1/8, and dividing η by 10 multiplies Ra by 10, so the net factor is 10/8 = 1.25. Using d instead of d³ gives the tempting 5, the inverse gives 0.8, and 80 multiplies by 8 instead of dividing.
A mantle parcel rises adiabatically. At 5.0 GPa its temperature is 216 K below the dry solidus. As pressure falls, the adiabat cools by 12 K per GPa and the solidus by 120 K per GPa. By how much must the pressure fall before melting begins?
1. 1.6 GPa
2. 1.8 GPa
3. 2.0 GPa
4. 18 GPa
Show answer
Answer: C — 2.0 GPa
The gap closes at (120 − 12) = 108 K per GPa, so 216 / 108 = 2.0 GPa, and melting begins near 3.0 GPa. Ignoring the cooling of the adiabat gives 216/120 = 1.8, adding the two rates gives 216/132 ≈ 1.6, and dividing by the adiabat rate alone gives the impossible 18.
The potential temperature of the mantle is defined as
1. the temperature at the base of the thermal lithosphere, where the geotherm leaves the conductive regime
2. the mean of the temperatures at the top and the bottom of the mantle
3. the temperature at which dry peridotite begins to melt at the surface
4. the temperature a parcel would have if brought adiabatically to the surface without melting
Show answer
Answer: D — the temperature a parcel would have if brought adiabatically to the surface without melting
Potential temperature removes the effect of depth along an adiabat by referring every parcel to the surface. The temperature at the base of the thermal lithosphere is a different quantity (about 1300 °C) and the surface solidus is lower than the potential temperature of ordinary mantle, which is why mantle that rises reaches the solidus at depth.
Suppose the heat that can be conducted down the adiabat of the outer core is larger than the heat that actually crosses the core–mantle boundary. Which conclusion is correct?
1. The outer core must freeze at once, because no heat at all is being removed from the fluid across its boundary
2. Thermal convection cannot run, but compositional buoyancy from inner-core freezing can still drive a dynamo
3. The outer core can no longer convect by any mechanism whatever, so the geodynamo must switch off immediately
4. Convection becomes more vigorous, because a fluid whose temperature lies below its adiabat is unstably stratified
Show answer
Answer: B — Thermal convection cannot run, but compositional buoyancy from inner-core freezing can still drive a dynamo
If conduction along the adiabat can carry all the heat, the fluid is stably stratified thermally and thermal convection stops. A fluid colder than its adiabat is stable, not unstable, which rules out the claim of more vigorous convection. Freezing at the inner-core boundary releases latent heat and light elements, and the compositional buoyancy can sustain convection and the dynamo, so the dynamo need not stop.
Which of the following statements about the mantle adiabat are correct?
1. The adiabatic gradient equals α g T / c_p
2. A well-mixed convecting layer with no strong boundary layers lies close to an adiabat
3. The adiabatic gradient in the upper mantle is about 10 K per km, comparable with the conductive gradient in the continental lithosphere
4. For otherwise equal properties, the adiabatic gradient is larger where the absolute temperature is higher
Show answer
Answer: A — The adiabatic gradient equals α g T / c_p; B — A well-mixed convecting layer with no strong boundary layers lies close to an adiabat; D — For otherwise equal properties, the adiabatic gradient is larger where the absolute temperature is higher
The gradient α g T/c_p is proportional to T, and a well-mixed convecting interior follows it. Its size in the mantle is about 0.4 K/km, far below the 20 to 30 K/km conductive gradient of the lithosphere, so a gradient of 10 K/km is wrong by a factor of about 25.
Which of the following can bring a parcel of mantle peridotite to melting?
1. Adiabatic decompression as the parcel rises beneath a spreading ridge
2. Addition of water released from a subducting slab, which lowers the solidus
3. Arrival of mantle several hundred kelvin hotter than normal, as in a plume
4. Raising the pressure on a dry parcel at constant temperature, as when it is carried downward
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Answer: A — Adiabatic decompression as the parcel rises beneath a spreading ridge; B — Addition of water released from a subducting slab, which lowers the solidus; C — Arrival of mantle several hundred kelvin hotter than normal, as in a plume
Decompression, volatile fluxing and excess temperature each move a parcel across the solidus. Raising the pressure on a dry parcel raises its solidus and moves the parcel away from melting, so it cannot cause melt.
A continental column has surface temperature 10 °C, surface heat flow 60 mW/m², thermal conductivity 3.0 W/m/K and uniform radiogenic heat production 1.0 µW/m³. Using the steady-state solution with z measured downward, find the temperature at 30 km depth, in °C.
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Answer: 460
T = T₀ + q₀ z/k − A z²/(2k). q₀ z/k = 0.060 × 30 000 / 3.0 = 600 K. A z²/(2k) = 1.0×10⁻⁶ × (30 000)² / 6.0 = 150 K. T = 10 + 600 − 150 = 460 °C. Forgetting the heat-production term gives 610, and forgetting the factor 2 gives 310.
Oceanic lithosphere follows half-space cooling with mantle temperature 1300 °C and surface temperature 0 °C. At the age considered, 2√(κt) = 80 km. Using erf(0.5) = 0.5205, find the temperature at 40 km depth, to the nearest °C.
Numerical answer — type the value.
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Answer: 677
η = z / (2√(κt)) = 40 / 80 = 0.5. T = T₀ + (T_m − T₀) erf(η) = 0 + 1300 × 0.5205 = 676.65, which rounds to 677 °C. Using z/√(κt) = 0.5 × 2 = 1 by mistake would give erf(1) and about 1096 °C.
The base of the thermal lithosphere is taken where the temperature is 90% of the way from the surface value to the mantle value, which for half-space cooling is z = 2.32 √(κt). For κ = 1.0×10⁻⁶ m²/s and an age of 100 Myr (1 Myr = 3.15×10¹³ s), find the thickness in km, to the nearest km.
Numerical answer — type the value.
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Answer: 130
κt = 1.0×10⁻⁶ × 100 × 3.15×10¹³ = 3.15×10⁹ m², so √(κt) = 56.1 km. z = 2.32 × 56.1 = 130.2 km, which rounds to 130 km. Forgetting the square root would give an absurd 10⁹ scale, and using the factor 2 alone gives 112 km.
In a mantle layer, α = 2.5×10⁻⁵ K⁻¹, g = 10 m/s², c_p = 1000 J/kg/K and the temperature is taken as 1600 K throughout. Using the adiabatic gradient α g T / c_p, find the temperature rise in kelvin over a depth interval of 500 km.
Numerical answer — type the value.
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Answer: 200
Gradient = 2.5×10⁻⁵ × 10 × 1600 / 1000 = 4.0×10⁻⁴ K/m = 0.40 K/km. Over 500 km the rise is 0.40 × 500 = 200 K. Leaving the depth in km while the gradient is in K/m would be the usual unit slip.
A mantle layer has ρ = 3000 kg/m³, g = 10 m/s², α = 3×10⁻⁵ K⁻¹, ΔT = 1000 K, thickness d = 1000 km, κ = 1×10⁻⁶ m²/s and η = 1×10²¹ Pa s. Its Rayleigh number is N × 10⁵. Find N.
Numerical answer — type the value.
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Answer: 9
Ra = ρ g α ΔT d³ / (κ η). Numerator: 3000 × 10 × 3×10⁻⁵ × 1000 = 900, times d³ = (10⁶ m)³ = 10¹⁸ gives 9×10²⁰. Denominator: 10⁻⁶ × 10²¹ = 10¹⁵. Ra = 9×10⁵, so N = 9. Using d = 1000 (km) without converting to metres is the usual error.