SCIENCE · INVESTIGATE AND EXPLAIN

Keep geothermal gradient distinct from heat flux

Temperature, temperature gradient and heat flux are different quantities. Fourier conduction in one dimension gives flux opposite the temperature gradient. With depth positive downward and temperature increasing downward, heat flows upward. Its magnitude is conductivity multiplied by gradient in kelvin per meter.

Learning outcome: Hold gradient fixed while changing conductivity, then hold conductivity fixed while changing gradient. Keep flux direction and magnitude separate. Explain why a straight shallow geotherm should not be extrapolated to the core.

Starting recap

Keep reference frames, material properties, path geometry and actual catalog evidence distinct. Infer only what the retained observations and stated model support.

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Try

With gradient 20 K/km and conductivity 3 W/(m K), upward flux is 0.060 W/m²=60 mW/m². Temperature at 10 km is 210°C. Multiplying by 20 without converting kilometers would overstate flux by a factor of 1000.

Open the experiment

Predict

Can two rocks share a temperature gradient but carry different conductive heat fluxes?

Your prediction and reason

Experiment

Hold gradient fixed while changing conductivity, then hold conductivity fixed while changing gradient. Keep flux direction and magnitude separate. Explain why a straight shallow geotherm should not be extrapolated to the core.

Distinguish temperature gradient, heat flux and transport mechanism

Worked teaching model · synthetic values · Steady 1-D shallow conductive layer with constant gradient and conductivity, depth positive downward. Upward heat-flux magnitude k×gradient/1000. Surface temperature 10°C, temperature at 10 km=10+10×gradient. No internal heat production in this layer and no extrapolation to whole mantle.

Temperature increase per depth: 20 K/km Thermal conductivity: 3 W/(m K)

Reset keeps your prediction, captures and practice history.

A shallow temperature gradient is not the same quantity as heat flux or mantle temperature.

Heat moves by conduction through this stationary layer. Mantle convection, advected magma, radiogenic heating and secular cooling require additional energy terms. Extending this shallow line to core depths is physically unsupported.

Complete conditional ledger; units belong to each row
Quantity	Value	Unit
Upward heat-flux magnitude	60	mW/m²
Temperature at 10 km	210	°C
Gradient in SI length units	0.02	K/m
Conductivity	3	W/(m K)

Numerical reconstruction points: temperature (°C), depth (km)
x	y
10	0
210	10

A shallow temperature gradient is not the same quantity as heat flux or mantle temperature.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Define the evidence and frame

2. Connect mechanism and model

The supplied shallow layer has constant conductivity and a constant 10–30 K/km gradient, with surface temperature 10°C. Converting kilometers to meters gives upward flux kG/1000 W/m², or kG mW/m². At 10 km, the temperature is 10+10G °C. A temperature difference of one kelvin equals a difference of one Celsius degree, but the units on gradient still include length.

This steady layer contains no internal heat production. Actual crust can generate heat through radioactive decay; cooling, magmatic advection and hydrothermal transport also alter heat budgets. The deeper mantle transfers substantial heat by convection while conducting through boundary layers. The near-surface conductive gradient cannot be extended linearly through the whole planet.

Heat loss and buoyancy interact with plate tectonics, but a measured surface heat flux alone does not uniquely partition radioactive heating, secular cooling and local fluid transport. A defensible interpretation states the layer, direction convention and omitted terms before linking a gradient to a geological mechanism.

3. Reconstruct the worked packet

4. Predict and investigate

Predict first: Can two rocks share a temperature gradient but carry different conductive heat fluxes?

5. Test a changed case

Apply the evidence to a changed case

With gradient 30 K/km and conductivity 2 W/(m K), find upward flux and temperature at 10 km.

Retain the evidence frame, calculations and assumptions before opening public feedback.

Compare a worked response

Flux is 60 mW/m², matching the worked flux despite a different gradient. Temperature at 10 km is 310°C. Equal flux does not imply equal temperature profiles.

6. Retain a defensible evidence notebook

Preserve first predictions, full units, geometry, source provenance, assumptions and rejected alternatives. Synthetic models are not field measurements. The explicitly labeled USGS packet is actual catalog evidence, but working through a public example is not independent practical competence or an operational hazard assessment.

Mechanism and data reference, checked 30 October 2026. Original teaching prose, diagrams and exercises; actual USGS rows retain their agency and provenance.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

With gradient 30 K/km and conductivity 2 W/(m K), find upward flux and temperature at 10 km.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Connect elastic moduli to P and S wave support Next: Audit actual USGS phase records before interpreting the interior

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