Compare solid layers in a controlled conduction test
Use matched boundaries and supplied transfer rates to compare conducting and insulating layers.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Identify energy transfer between different temperatures.
- Calculate a supplied energy-transfer rate.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which comparison most directly controls the conduction path?
What you will learn
- Describe a second conductor/insulator experiment with matched layer conditions.
- Interpret supplied energy-per-time data without confusing conductivity and thermal storage.
Keep the thermal path controlled
A described teacher-supervised test places each suitable solid layer between controlled warmer and cooler sides. Match the layer thickness, contact area and maintained temperature difference, and use suitable thermal contact. Compare a temperature response or, with appropriate apparatus, a separately measured transfer rate. Shield or account for other routes so the test meaning concerns the layer.
An insulating layer transfers less energy per second than a better-conducting layer under the same boundary conditions. A receiver's temperature response can also depend on its mass and heat capacity, so match the receiver or use an appropriate energy account. Simply feeling samples is not a quantitative or safe measurement method.
Interpret what the supplied data measure
If a supplied controlled test transfers 120 J through A in 60 s and 30 J through B in 60 s, rates are 2.0 W and 0.50 W. With the stated equal geometry, temperature difference, contacts and other transfers accounted for, A is the better conductor in that comparison. The numbers are illustrative data rather than a performed measurement.
The comparison does not determine a material's absolute internal energy or assert that high heat capacity means high conductivity. Thermal storage and transfer are different properties. Nor does a lower rate mean transfer has stopped entirely. Actual quantitative conductivity requires a suitable measured model and uncertainty analysis beyond this qualitative syllabus demonstration.
Sources for this lesson
Pause and explain
A layer transfers 30 J in 60 s. What is its average transfer rate?
Worked example
Supplied matched layers transfer 120 J and 30 J respectively in 60 s. Find rates and compare.
Show the worked solution
- A: 120/60 = 2.0 W.
- B: 30/60 = 0.50 W.
- Under the stated matched boundary conditions, A gives the greater conduction rate.
Answer A 2.0 W; B 0.50 W; A is better conducting in the supplied controlled comparison.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe the matched-layer experiment and explain why receiver mass should be controlled for a temperature-response comparison.
Give me a hint
Both the path and the receiver affect the observation.
Compare my reasoning
- Match layer geometry, contacts and warmer/cooler boundary temperatures.
- Use the same receiver or match its thermal properties.
- Compare responses while accounting for other transfer paths and practical limits.
Control the path and receiver before comparing the layer materials.
Look for these in your work
- Name path controls.
- Explain the receiver's effect.
A and B transfer 120 J and 30 J in 60 s through matched paths. Calculate the rate difference.
Give me a hint
Divide each amount by the same duration before subtracting.
Compare my reasoning
- A transfers 2.0 W.
- B transfers 0.50 W.
- The difference is 2.0 − 0.50 = 1.50 W.
A exceeds B by 1.50 W in the supplied comparison.
Look for these in your work
- Calculate both rates.
- Retain the controlled-comparison scope.
A thin layer transfers more quickly than a thick layer of another material. Can the observation rank their intrinsic conduction properties alone?
Give me a hint
Thickness changes the path as well as material.
Compare my reasoning
- The thicknesses are unmatched.
- Path geometry can affect transfer rate.
- Match geometry and boundary conditions before assigning the difference solely to material.
No material-only ranking follows from unmatched thicknesses.
Look for these in your work
- Identify the confounding path change.
- Propose a matched comparison.
Common mistakes
- Match thickness, area, contact and maintained temperature difference.
- Receiver thermal storage can affect temperature response.
- Poor conduction is not zero conduction.
What can you explain now?
Does a large heat capacity alone prove a material is a good thermal conductor?
Compare with the explanation
No.
Heat capacity describes energy storage for a temperature rise; conductivity concerns transfer through a path.
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Make it stick
Compare a controlled layer test and distinguish thermal transfer from thermal storage.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.