What sets the rate through a wall?
You will be able to: Relate conduction rate to material, area, thickness and temperature difference.
Why does thicker insulation slow energy loss?
Two walls have the same material and area, but one is twice as thick. With the same temperatures on their faces, energy must pass along a longer temperature gradient through the thicker wall. The steady conduction rate is smaller.
A useful starting point: Find the shared final temperature →
Words and symbols before equations
- Rate H=Q/Δt
- Energy transferred per time, in watts (W=J/s); not total energy.
- Conductivity k
- Material coefficient in W/(m·K); distinct from Boltzmann constant k_B.
- Area A
- Wall area perpendicular to energy flow, in m².
- Thickness L
- Distance through the wall along the transfer direction, in m.
What this picture assumes
Uniform slab, A=2 m², fixed face temperatures 40 °C and 20 °C. Constant k, steady one-dimensional conduction, no internal energy generation or side losses. Graph shows temperature versus actual distance through the selected slab.
Connect the picture to the physics
For a uniform slab at steady state with fixed face temperatures, H=kA(T_hot−T_cold)/L. A larger temperature difference, greater area or higher conductivity increases the rate. Greater thickness reduces it.
The temperature profile is linear in the ideal uniform slab with constant k and no internal heat generation. The slope magnitude is ΔT/L. “Steady” means the profile and rate are not changing in time; energy still flows.
The equation isolates conduction. Real walls can have multiple layers, surface convection, radiation and thermal bridges. The model assumes fixed face temperatures rather than automatically equating them to room-air temperatures. Specific heat describes energy needed to change temperature; conductivity describes transfer rate through a gradient.
A worked example, step by step
A slab has k=0.5 W/(m·K), A=2 m², L=0.10 m and face temperatures 40 °C and 20 °C. Find H and energy transferred in 60 s.
- ΔT=20 K.
- H=(0.5)(2)(20)/0.10=200 W.
- Q=HΔt=(200)(60)=12000 J.
- Doubling L with the same face temperatures halves H to 100 W.
Watts measure a rate, not an amount of energy. Conductivity k and specific heat c describe different properties.
At steady state, is energy flow zero?
Compare with an explanation
No. Temperatures are unchanging because energy passes through at a constant rate without accumulating.
Predict. Change one thing. Explain.
Change thickness and conductivity while holding area and face temperatures fixed. Compare rate and temperature gradient. The horizontal distance scale expands for thicker slabs; this is a steady-state calculation, not a warm-up animation.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
H=200 W from 40 °C face toward 20 °C face. In 60 s, Q=12000 J. Temperature gradient=-200 K/m. The horizontal axis ends at the chosen thickness 0.1 m.
Uniform slab, A=2 m², fixed face temperatures 40 °C and 20 °C. Constant k, steady one-dimensional conduction, no internal energy generation or side losses. Graph shows temperature versus actual distance through the selected slab.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant particle, temperature or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA slab conducts 80 W. With the same material and face temperatures, its area is doubled and thickness tripled. (a) Give the area factor. (b) Give the thickness factor. (c) Find the new rate. (d) Find energy transferred in 30 s at the new rate.
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Compare with the answer and four-point rubric
- 1 point: Area contributes a factor 2.
- 1 point: Thickness contributes a factor 1/3.
- 1 point: H_new=80(2/3)=53.3 W.
- 1 point: Q≈53.3(30)=1600 J.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does conductivity describe?
How readily a material transfers energy through a temperature gradient.
RECALL 2How does thickness enter?
Rate is inversely proportional to thickness for fixed face temperatures.
RECALL 3Why multiply rate by time?
To calculate the total energy transferred during a constant-rate interval.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What sets the rate through a wall?
- H=Q/Δt=kA(T_hot−T_cold)/L.
- Q=HΔt for constant H.
- Temperature-profile slope magnitude = ΔT/L.
Remember: Watts measure a rate, not an amount of energy. Conductivity k and specific heat c describe different properties.
Conditions: Uniform slab, A=2 m², fixed face temperatures 40 °C and 20 °C. Constant k, steady one-dimensional conduction, no internal energy generation or side losses. Graph shows temperature versus actual distance through the selected slab.
Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.5.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.5, objectives 9.5.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the first AP Physics 2 unit; College Board numbers it Unit 9, continuing after AP Physics 1. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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