SCIENCE · INVESTIGATE AND EXPLAIN

Infer an interface temperature from a steady heat ledger

Conduction transfers energy through microscopic interactions without requiring bulk material motion. For a uniform layer under steady one-dimensional conditions, heat rate P=kAΔT/L. Define thermal resistance Rth=L/(kA), measured in K/W, so P=ΔT/Rth. This is a rate law, not the heat capacity that sets energy needed to change temperature.

Learning outcome: Compare equal and unequal thermal resistances, then alter only one layer. Retain both temperature drops and equal steady rates; identify a separate radiation or convection term that would invalidate a conduction-only inference.

Starting recap

Declare temperature and time references, system boundaries and model regimes. Use absolute temperatures for ratios; distinguish state, path, oscillator phase and wave propagation.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

At first thickness 0.01 m and second conductivity 0.3 W/(m K), both layer resistances are 0.2 K/W. Heat rate is 100 W and interface temperature 310 K.

Open the experiment

Predict

In a lossless steady series-layer model, the interface requires:

Your prediction and reason

Experiment

Compare equal and unequal thermal resistances, then alter only one layer. Retain both temperature drops and equal steady rates; identify a separate radiation or convection term that would invalidate a conduction-only inference.

Infer an interface temperature from equal steady heat rates

Worked teaching model · synthetic values · Steady one-dimensional conduction through area .5 m²: first layer conductivity .1 W/(m K), selected thickness; second layer thickness .03 m and selected conductivity. Boundary temperatures 330 and 290 K. No contact resistance, storage, convection or radiation in this conduction packet.

First-layer thickness: 0.02 m Second-layer conductivity: 0.2 W/(m K)

Reset keeps your prediction, captures and practice history.

Piecewise linear steady temperature profile

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: temperature (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Piecewise linear steady temperature profile

temperature

distance through layers (m)	temperature (K)
0	330
0.02	307.143
0.05	290

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
First layer thermal resistance	0.4	K/W
Second layer thermal resistance	0.3	K/W
Common steady heat rate	57.1429	W
Interface temperature	307.143	K
First layer temperature drop	22.8571	K
Second layer temperature drop	17.1429	K

Series layers carry equal steady heat rates, not necessarily equal temperature drops.

Each thermal resistance is L/(kA); the common heat rate is Delta T divided by their sum. Equal rates prevent energy accumulation at the interface. This conduction-only packet excludes radiation and convection; adding either requires another boundary account, not silently adjusting a conductivity. Interfacial contact resistance is explicitly zero.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Series layers carry equal steady heat rates, not necessarily equal temperature drops.

Observation to attach to a capture

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

Understand

Declare the state, system and observation

Build the algebraic model

Two layers in series carry the same steady heat rate when their interface stores no energy. The first has conductivity 0.1 W/(m K), area 0.5 m² and selected thickness; the second has thickness 0.03 m and selected conductivity. With first thickness 0.02 m and second conductivity 0.2, resistances are 0.4 and 0.3 K/W. Boundaries 330 and 290 K give P=40/0.7≈57.1429 W.

Check the original evidence packet

The interface temperature is 330−P×0.4≈307.143 K. The first drop is about 22.8571 K and the second 17.1429 K. Equal rates do not mean equal drops. Doubling one resistance increases its share of the temperature difference while reducing the common rate. An interfacial contact resistance, nonsteady energy storage or side loss would require an additional equation rather than silently reusing this two-layer result.

Challenge the domain and transfer

Convection transports energy with moving fluid and can be natural or externally driven. Radiation needs no material medium: for a supplied gray-surface model net radiative power is εσA(T⁴−Ts⁴), with kelvin temperatures. For original ε=0.5,A=1 m²,T=300 K,Ts=290 K and σ=5.67×10⁻⁸ W/(m² K⁴), predict and check about 29.1208 W outward. Reversing the two temperatures reverses the sign; making them equal gives zero net rate despite emission and absorption continuing. That separate radiation packet is not already included in the conduction ledger.

Worked original case

At first thickness 0.01 m and second conductivity 0.3 W/(m K), both layer resistances are 0.2 K/W. Heat rate is 100 W and interface temperature 310 K.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

Two series thermal resistances are 0.3 and 0.5 K/W between boundaries differing by 24 K. Find steady heat rate in W.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. Packets are supplied or generated teaching evidence, not measured apparatus. No heating, pressure vessels, physical oscillators, sound exposure or construction is requested. Public checks and saved reports do not establish supervised practical skill, equipment safety or independent mastery.

Apply

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

Why are thermal conductivity and heat capacity not interchangeable parameters?

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: Test a phase boundary before solving a final temperature Next: Compare route-dependent heat with endpoint-dependent internal energy

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.