Infer temperature drops across a two-layer barrier

Establish the model and evidence

At steady state, energy does not accumulate at the interface between the two layers. The same thermal power must therefore cross each layer. Equal power does not imply equal temperature drops or equal temperature gradients. The material parameters and geometry determine how the total drop is divided.

For the stated one-dimensional model, power is kAΔT/L. Define thermal resistance Rth=L/(kA), with units K/W. Series resistances add, so total power is the imposed boundary-temperature difference divided by R 1+R 2. Recover the interface temperature from either layer’s drop; the two calculations must agree.

Changing layer B thickness and conductivity can produce the same resistance. Consequently one measurement of total thermal power cannot always identify which property changed. The position of the interface temperature and independent geometry measurements help constrain the explanation. The displayed temperature profile is continuous because contact resistance is explicitly neglected.

The model is a steady boundary-value calculation, not a prediction of how quickly the wall cools. Adding material heat capacity would be necessary to describe transient storage. Convection, radiation, edge leakage or imperfect contact would create additional pathways or resistances. These are model-selection questions, not reasons to label the arithmetic incorrect.

Worked evidence

Layer A has R=0.010/(0.20×0.010)=5 K/W. With layer B thickness 0.020 m and conductivity 0.20 W/(m K), RB=10 K/W. A 20 K total drop gives 1.333 W; the interface is 310−1.333×5≈303.333 K.

Predict and investigate

At fixed thickness, predict how increasing layer B conductivity changes total power and the interface temperature.

Capture B thickness 0.020 m with conductivity 0.20 W/(m K), then double conductivity. Separately double both B thickness and conductivity. Explain which pairs share resistance and which profile features differ.

Transfer to changed conditions

A wall has resistances 4 and 6 K/W between reservoirs 320 and 300 K. Find power and the interface temperature after the first layer from the hot side.

Compare the worked transfer

Power=20/10=2 W. The first drop is 2×4=8 K, so the interface is 312 K.

Explain the boundary

Use an interface energy balance to justify equal steady power. Identify a measurement that would reveal the failure of the perfect-contact assumption.

Separate conductivity, specific heat and geometry

Thermal conductivity and specific heat describe different material responses. Conductivity controls a heat-flow rate under a gradient; specific heat controls an energy change per mass and temperature change. Each depends on microscopic arrangement and interactions and may vary with material state or temperature; geometry is not itself an intrinsic material property.

Predict and check: for a stipulated material with c=400 J/(kg K), warming 0.20 kg by 5 K requires 400 J; doubling mass requires 800 J under unchanged c. That does not determine the power through a wall made of the material. In the steady conduction law, doubling area at fixed k, thickness and temperature difference doubles power, while doubling thickness halves it. Record units J/(kg K) versus W/(m K) and explain why equal conductivity does not imply equal heat capacity. (Current clauses 9.5.A.2 and 9.5.B.2.)

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. The numerical cases, text and computed representations are original. All browser evidence is synthetic or calculated, not a record of completed practical work. No hazardous apparatus is required.

Clause-level source/action reconciliation checked 31 October 2026 against the linked current framework and Fall 2026 clarification. Added packets are original virtual comparisons; no new practical or learner-completion claim.