Distinguish heat-transfer rate from stored energy

Energy can pass through a system without accumulating

The hot outer surface of this two-layer assembly stays at 80 °C; the cold outer surface stays at 20 °C. The temperature profile is already established. Predict whether energy can keep crossing the layers even though no point on the profile is warming.

At steady state, energy enters each layer at the same rate it leaves. Net storage is zero, but the transfer rate can be nonzero. A maintained hot source supplies energy and a maintained cold sink removes it. These are surface temperatures, not air temperatures outside an ordinary wall.

Separate the transfer mechanisms

Conduction transfers energy through microscopic interactions without bulk transport of material through the assembly. Convection involves moving fluid carrying energy and exchanging it with surfaces. Radiation transfers energy through electromagnetic waves and can cross a vacuum. A real object can exchange energy by all three mechanisms at once.

This investigation includes conduction only, with no lateral losses, internal heat generation or contact resistance. The layers have fixed conductivity and common area 0.5 m². The OpenStax heat-transfer reference gives the physical context for these mechanisms. The synthetic assembly here is not an insulation product rating or a prediction of a real room’s heat loss.

Relate a temperature gradient to a rate

For a uniform slab under steady one-dimensional conditions, P = kA(Th − Tc)/L. P is heat-transfer rate in watts, k is thermal conductivity in W/(m K), A is cross-sectional area in m² and L is thickness in metres. Th and Tc are the temperatures of that slab’s two faces. Their difference can be expressed in kelvin or Celsius degrees without adding an offset.

In words, a larger conductivity, area or face-temperature difference increases the rate; a greater thickness decreases it when those other quantities stay fixed. The temperature difference must be across the particular slab being calculated, not automatically the entire assembly.

Thermal conductivity is different from specific heat. Conductivity connects a temperature gradient to a transfer rate. Specific heat connects a thermal energy change to temperature change within a phase. The model specifies conductivity but does not specify layer heat capacities, so it cannot calculate how long these layers take to warm up.

Check the default numbers independently

Layer 1 is 0.020 m thick with conductivity 0.2 W/(m K). Layer 2 is initially 0.030 m thick with conductivity 0.1 W/(m K). The calculated interface temperature is 65 °C.

For layer 1, the drop is 80 − 65 = 15 K. Therefore P₁ = 0.2 × 0.5 × 15 / 0.020 = 75 W. For layer 2, the drop is 65 − 20 = 45 K, giving P₂ = 0.1 × 0.5 × 45 / 0.030 = 75 W. Different drops and different slopes are consistent with the same rate.

The next lesson derives the interface temperature instead of taking it as given. Here, calculate each layer rate from the ledger as an independent check. If the rates differed in this steady model, energy would accumulate at the interface, contradicting its assumptions.

Distinguish watts, joules and flux

A watt is a joule per second. At a constant 75 W for 600 s, the energy transferred is Q = Pt = 45000 J = 45 kJ. This is energy that crosses the whole assembly. Adding 45 kJ for each layer to report 90 kJ would count the same transfer twice.

Heat flux is rate per area: j = P/A, in W/m². Here j = 75/0.5 = 150 W/m². Flux is useful when comparing different areas, but it is not a total power. Over ten minutes each square metre would transmit 90 kJ under these conditions; this assembly has only half a square metre.

The ten-minute energy calculation assumes the source and sink continue maintaining their temperatures and the properties remain fixed. If the source cools, the instantaneous rate changes; multiplying its initial rate by a long duration would overestimate the transfer.

Read a spatial profile

The graph’s horizontal coordinate is distance through the layers, in millimetres. It is not elapsed time. Each segment is linear because each uniform layer carries a constant rate with no internal source. At the interface, temperature is continuous because perfect thermal contact is assumed. A change in conductivity changes the slope; it need not make a jump in temperature.

Change the second layer thickness. The plotted width changes on the same distance scale, and the model calculates a new steady profile. It does not animate the transient journey from the previous profile. A real comparison could use temperature sensors at known positions and a measured heater power after stabilization, with lateral losses and contact effects accounted for.

Transfer to a single slab

A slab has k = 0.3 W/(m K), area 0.20 m², thickness 0.040 m and a maintained face-temperature difference of 24 K. Calculate rate, flux and transferred energy over five minutes. State one condition needed for that time multiplication.

Check the conduction-rate transfer

P = 0.3 × 0.20 × 24 / 0.040 = 36 W. Flux is 36/0.20 = 180 W/m². Five minutes is 300 s, giving Q = 36 × 300 = 10800 J = 10.8 kJ. The rate must remain constant over those five minutes, with maintained boundaries and the stated steady, uniform-property assumptions.