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Thermal Conduction (HL IB Physics)

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Katie M

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13 June 2024

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, [N_A] = mol⁻¹
Units for relative atomic mass, [A] = g mol⁻¹

[n]^a = [ρ]^b [N_A]^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 [N_A]¹ [A]⁻¹

$n = \frac{ρ N_{A}}{A}$

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

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

Copper: $n = \frac{8.96 \times (6.02 \times 10^{23})}{63.55} = 8.49 \times 10^{22} {cm}^{- 3}$

Steel: $n = \frac{7.85 \times (6.02 \times 10^{23})}{55.85} = 8.46 \times 10^{22} {cm}^{- 3}$

Aluminium: $n = \frac{2.71 \times (6.02 \times 10^{23})}{26.98} = 6.05 \times 10^{22} {cm}^{- 3}$

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

$\frac{∆ Q}{∆ t} = k A \frac{∆ T}{∆ x}$

Where
- $\frac{∆ Q}{∆ t}$ = 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

$\frac{∆ Q_{s}}{∆ t} = \frac{∆ Q_{a}}{∆ t} = \frac{∆ Q_{c}}{∆ t}$

$k_{s} A \frac{∆ T_{s}}{∆ x} = k_{a} A \frac{∆ T_{a}}{∆ x} = k_{c} A \frac{∆ T_{c}}{∆ x}$

$k_{s} ∆ T_{s} = k_{a} ∆ T_{a} = k_{c} ∆ T_{c}$

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

Temperature difference in steel: $∆ T_{s} = (100 - T_{1})$
Temperature difference in aluminium: $∆ T_{a} = (T_{1} - T_{2})$
Temperature difference in copper: $∆ T_{c} = (T_{2} - 0)$

$k_{s} (100 - T_{1}) = k_{a} (T_{1} - T_{2}) \Rightarrow 60 (100 - T_{1}) = 240 (T_{1} - T_{2})$ eq. (1)

$k_{a} (T_{1} - T_{2}) = k_{c} (T_{2} - 0) \Rightarrow 240 (T_{1} - T_{2}) = 400 (T_{2} - 0)$ eq. (2)

Step 3: Expand and simplify eq. (1)

$6000 - 60 T_{1} = 240 T_{1} - 240 T_{2}$

$300 T_{1} - 240 T_{2} = 6000$

$5 T_{1} - 4 T_{2} = 100$

Step 4: Expand and simplify eq. (2)

$240 T_{1} - 240 T_{2} = 400 T_{2}$

$240 T_{1} = 640 T_{2}$

$T_{2} = \frac{3}{8} T_{1}$

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

$5 T_{1} - 4 (\frac{3}{8} T_{1}) = 100$

$\frac{7}{2} T_{1} = 100$

$T_{1} = \frac{200}{7} = 28.6 = 29 ° C$

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

$T_{2} = \frac{3}{8} \times \frac{200}{7} = 10.7 = 11 ° C$

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