CourseCourseQ BankQuestion BankDocsDocuments

B.1.11—Conduction

Syllabus
First assessment 2025
Objective
—
Level
HL

Explain Conduction Microscopically

Conduction

In conduction, particles in a hotter region have greater average kinetic energy. Through collisions and intermolecular forces, they transfer energy to neighbouring particles in the cooler region.

What moves and what does not

Energy propagates through the material, but the material does not need to undergo bulk flow. In a solid, particles usually vibrate about fixed positions while transferring energy to neighbours.

Compare with other mechanisms

Conduction needs matter and microscopic contact. Convection transfers energy through bulk motion of a fluid. Radiation transfers energy by electromagnetic waves and can cross a vacuum.

Common trap

Conduction is not the same as particles travelling from the hot end to the cold end. The net transfer is through local interactions.

B.1.11 Exam Analysis

Assessment in practice

2 marks
How it is assessed

The evidence repeats a two-mark structured prompt asking for the microscopic mechanism of conduction through a wall.

Command terms

Describe

What earns marks

For conduction in a solid, mention particle or atomic vibrations and energy transfer through collisions/interactions between adjacent particles. If the material is metallic and the question invites more detail, include mobile electrons colliding with atoms/ions. Do not describe bulk fluid motion.

Watch for

Saying that the particles themselves flow from hot to cold, or giving a convection explanation.

Representative question

Question 1

[Maximum number: 2]

Describe the mechanism of heat transfer by conduction.

The diagram shows a wall separating the inside of a room from the outside. The temperature of the room is kept constant by a heater.

The following data are available:

 Thickness of wall =0.25 m Area of wall =18 m2 Thermal conductivity of wall =1.3Wm−1 K−1 Constant room temperature =22∘C Constant outside temperature =13∘C\begin{aligned} \text { Thickness of wall } & =0.25 \mathrm{~m} \\ \text { Area of wall } & =18 \mathrm{~m}^{2} \\ \text { Thermal conductivity of wall } & =1.3 \mathrm{Wm}^{-1} \mathrm{~K}^{-1} \\ \text { Constant room temperature } & =22^{\circ} \mathrm{C} \\ \text { Constant outside temperature } & =13^{\circ} \mathrm{C} \end{aligned}