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First teaching 2023

First exams 2025

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Thermal Conduction (SL IB Physics)

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Katie M

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Katie M

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Thermal Conduction

  • Thermal energy can be transferred from a hotter area to a cooler area through one of the following mechanisms:
    • Conduction
    • Convection
    • Radiation

Thermal conduction, convection and radiation in a mug of tea

Thermal Energy Transfers in a Mug of Tea, for IB Physics Revision Notes

  • Objects will always lose heat until they are in thermal equilibrium with their surroundings
    • For example, a mug of hot tea will cool down until it reaches room temperature

Conduction

  • Conduction is the main method of thermal energy transfer in solids
  • Conduction occurs when:

Two solids of different temperatures come in contact with one another, thermal energy is transferred from the hotter object to the cooler object

  • Metals are the best thermal conductors
    • This is because they have a high number of free electrons
  • Non-metals, such as plastic or glass, are poor at conducting heat
    • Poor conductors of heat tend to also be poor conductors of electricity
    • This suggests a link between the mechanisms behind both types of conduction
  • Liquids and gases are even poorer thermal conductors
    • This is because the atoms are further apart

Conduction of Heat in a Metal

Conduction of Heat in a Metal, for IB Physics Revision Notes

During conduction, the atoms in a solid vibrate and collide with each other

  • Conduction can occur through two mechanisms:
    • Atomic vibrations
    • Free electron collisions
  • When a substance is heated, the atoms, or ions, start to move around, or vibrate, more
    • The atoms at the hotter end of the solid will vibrate more than the atoms at the cooler end
    • As they do so, they bump into each other, transferring energy from atom to atom
    • These collisions transfer internal energy until thermal equilibrium is achieved throughout the substance
    • This occurs in all solids, metals and non-metals alike
  • Metals are especially good at conducting heat due to their high number of delocalised electrons
    • These can collide with the atoms, increasing the rate of transfer of vibrations through the material
    • This allows metals to achieve thermal equilibrium faster than non-metals

Worked example

Determine which of the following metals is likely to be the best thermal conductor, and which is likely to be the worst. 

Metal Density / g cm−3 Relative atomic mass
Copper 8.96 63.55
Steel 7.85 55.85
Aluminium 2.71 26.98


Assume that each metal contributes one free electron per atom.

Answer:

Step 1: Use dimensional analysis to determine the equation for the number of free electrons

  • Units for number of free electrons per cubic centimetre, [n] = cm−3
  • Units for density, [ρ] = g cm−3
  • Units for Avogadro's number, [NA] = mol−1
  • Units for relative atomic mass, [A] = g mol−1

[n]a = [ρ]b [NA]c [A]d

(cm−3)a = (g cm−3)b (mol−1)c (g mol−1)d

  • The only unit present on both sides is cm−3, therefore:

a = b = 1

  • No other units are present on both sides, so:

c + d = 0

b + d = 0

∴ d = −1, c = 1

Step 2: Write out the equation for the number of free electrons per cubic centimetre

[n]1 = [ρ]1 [NA]1 [A]−1

n space equals space fraction numerator rho N subscript A over denominator A end fraction

Step 3: Calculate the number of free electrons in each metal

  • Avogadro constant, NA = 6.02 × 1023 mol−1 (this is given in the data booklet)

Copper:  n space equals space fraction numerator 8.96 space cross times space open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 63.55 end fraction space equals space 8.49 space cross times space 10 to the power of 22 space cm to the power of negative 3 end exponent

Steel:  n space equals space fraction numerator 7.85 space cross times space open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 55.85 end fraction space equals space 8.46 space cross times space 10 to the power of 22 space cm to the power of negative 3 end exponent

Aluminium: n space equals space fraction numerator 2.71 space cross times space open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 26.98 end fraction space equals space 6.05 space cross times space 10 to the power of 22 space cm to the power of negative 3 end exponent

Step 4: Rank the metals from best thermal conductor to worst

  • Best thermal conductor = copper (highest number of free electrons)
  • Worst thermal conductor = aluminium (lowest number of free electrons)

Examiner Tip

Regarding the worked example above, dimensional analysis is a vital skill in IB physics. This question, however, could also have been tackled by finding the number of atoms (and therefore free electrons) per gram and multiplying this value by the density to find the number of free electrons per cubic centimetre.

Remember, if a question mentions thermal energy transfers and metals, the answer will likely be about conduction!

Temperature Gradient Equation

Thermal Conductivity

  • The conductivity of a material can be quantified by its thermal conductivity
  • Thermal conductivity is defined as

The ability of a substance to transfer heat via conduction

  • It is denoted by the symbol k and has units of W m−1 K−1
  • The thermal conductivities of some common materials are shown in the table below
Substance Thermal conductivity / W m−1 K−1
air 0.024
rubber 0.13
water 0.6
ice 1.6
iron 80
copper 400
silver 429
diamond 1600

  • Excellent thermal conductors...
    • Have high values of thermal conductivity
    • Transfer thermal energy at a fast rate
    • (Usually) contain a large number of delocalised electrons (diamond being the obvious exception)
  • Poor thermal conductors (insulators)...
    • Have low values of thermal conductivity
    • Transfer thermal energy at a slow rate
    • Contain few delocalised electrons

Temperature Gradient Formula

  • When there is a temperature difference between two points, thermal energy will flow from the region of higher temperature to the region of lower temperature
    • This is known as a temperature gradient
  • The rate of the heat transfer via conduction is given by

fraction numerator increment Q over denominator increment t end fraction space equals space k A fraction numerator increment T over denominator increment x end fraction

  • Where
    • fraction numerator increment Q over denominator increment t end fraction = flow of thermal energy per second (W)
    • k = thermal conductivity of the material (W m−1 K−1)
    • A = cross-sectional area (m2)
    • ΔT = temperature difference (K or °C) 
    • Δx = thickness of the material (m)
  • The flow of thermal energy per second can be considered to be uniform across a temperature gradient, provided A is constant, regardless of material
    • This is analogous to electrical current being constant throughout a series circuit, even though the components may have different resistances

Conduction of thermal energy through a solid

2-1-10-temperature-gradient-diagram-png-ib-2025-physics

Thermal energy flows down a temperature gradient. The rate of energy transfer depends on the properties of the material

Worked example

A composite rod is made of three rods; steel, aluminium and copper. Each rod has the same length and cross-section, as shown in the diagram.

2-1-10-composite-rod-worked-example-ib-2025-physics

The steel end is held at 100°C and the copper end is held at 0°C.

Determine the temperatures at the steel-aluminium junction and the aluminium-copper junction.

Assume that the rods are perfectly insulated from the surroundings.

  • Thermal conductivity of steel = 60 W m−1 K−1
  • Thermal conductivity of aluminium = 240 W m−1 K−1
  • Thermal conductivity of copper = 400 W m−1 K−1

Answer:

Step 1: Analyse the scenario and set up an equation

  • As the rods have identical dimensions, the amount of heat flowing through each rod per second is uniform and must be the same
  • Therefore, rate of energy transfer in steel = rate of energy transfer in aluminium

fraction numerator increment Q subscript s over denominator increment t end fraction space equals space fraction numerator increment Q subscript a over denominator increment t end fraction space equals space fraction numerator increment Q subscript c over denominator increment t end fraction

k subscript s A fraction numerator increment T subscript s over denominator increment x end fraction space equals space k subscript a A fraction numerator increment T subscript a over denominator increment x end fraction space equals space k subscript c A fraction numerator increment T subscript c over denominator increment x end fraction

k subscript s increment T subscript s space equals space k subscript a increment T subscript a space equals space k subscript c increment T subscript c

Step 2: Form two simultaneous equations and substitute in the values of ΔT and k

  • Temperature difference in steel:  increment T subscript s space equals space open parentheses 100 space minus space T subscript 1 close parentheses
  • Temperature difference in aluminium:  increment T subscript a space equals space open parentheses T subscript 1 space minus space T subscript 2 close parentheses
  • Temperature difference in copper:  increment T subscript c space equals space open parentheses T subscript 2 space minus space 0 close parentheses

k subscript s open parentheses 100 space minus space T subscript 1 close parentheses space equals space k subscript a open parentheses T subscript 1 space minus space T subscript 2 close parentheses space space space space space rightwards double arrow space space space space space 60 open parentheses 100 space minus space T subscript 1 close parentheses space equals space 240 open parentheses T subscript 1 space minus space T subscript 2 close parentheses  eq. (1)

k subscript a open parentheses T subscript 1 space minus space T subscript 2 close parentheses space space equals space k subscript c open parentheses T subscript 2 space minus space 0 close parentheses space space space space space rightwards double arrow space space space space space 240 open parentheses T subscript 1 space minus space T subscript 2 close parentheses space space equals space 400 open parentheses T subscript 2 space minus space 0 close parentheses  eq. (2)

Step 3: Expand and simplify eq. (1)

6000 space minus space 60 T subscript 1 space equals space 240 T subscript 1 space minus space 240 T subscript 2

300 T subscript 1 space minus space space 240 T subscript 2 space equals space 6000

5 T subscript 1 space minus space space 4 T subscript 2 space equals space 100

Step 4: Expand and simplify eq. (2)

240 T subscript 1 space minus space 240 T subscript 2 space space equals space 400 T subscript 2

240 T subscript 1 space equals space space 640 T subscript 2

T subscript 2 space equals space space 3 over 8 T subscript 1

Step 5: Determine the temperature at the steel-aluminium junction T1

5 T subscript 1 space minus space space 4 open parentheses 3 over 8 T subscript 1 close parentheses space equals space 100

7 over 2 T subscript 1 space equals space 100

T subscript 1 space equals space 200 over 7 space equals space 28.6 space equals space 29 space degree straight C

Step 6: Determine the temperature at the aluminium-copper junction T2

T subscript 2 space equals space space 3 over 8 cross times 200 over 7 space equals space 10.7 space equals space 11 space degree straight C

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.