Thermal energy transfers

B.1 — Thermal Energy Transfers

IB Physics HL • Topic B: The particulate nature of matter • SL and HL core content

Thermal physics explains how energy is stored inside matter and how it transfers between objects. This lesson uses the particle model to describe solids, liquids, and gases, introduces internal energy, and develops key quantitative ideas such as density, specific heat capacity, latent heat, and black-body radiation.

Learning objectives

Describe solids, liquids, and gases using the particle model.
Define internal energy and explain kinetic vs potential contributions.
Define density and solve density problems.
Use Celsius and Kelvin temperature scales and convert between them.
Relate Kelvin temperature to the average kinetic energy of ideal gas molecules.
Explain heating with and without phase change and interpret heating curves.
Use $Q = mc\Delta T$ and apply thermal equilibrium (energy conservation).
Use $Q = mL$ for phase change energy calculations.
Explain conduction, convection, and radiation, including key applications.
Apply black-body concepts, intensity, Wien’s law, and Stefan–Boltzmann law.

B.1.1 Molecular theory in solids, liquids, and gases

Matter is made of a very large number of tiny particles (atoms or molecules). These particles are in constant random motion and interact through intermolecular forces. The arrangement, motion, and energy of the particles determine the physical properties of a substance.

Solids

Fixed volume; do not flow.
Particles closely packed; crystalline solids have a regular arrangement, while amorphous solids do not.
Particles vibrate about fixed equilibrium positions. Their average kinetic energy increases with temperature.
Strong interactions hold particles near their equilibrium positions.

When a solid is heated:

Particles gain kinetic energy.
Vibration amplitude increases.
For most solids, average separation increases slightly as vibration amplitude increases.
The solid expands (thermal expansion).

Typically, solids expand less than liquids or gases for the same temperature increase. The amount depends on the material and the constraints on it.

Liquids

Flow and take the shape of their container.
Particles close together but arranged randomly.
Particles move randomly and are not confined to fixed positions. Kinetic energy depends on temperature; changing phase at constant temperature does not increase average kinetic energy.
Particles can move around each other.
Intermolecular interactions remain important, but particles can rearrange and move past one another.

When a liquid is heated:

Particles gain kinetic energy.
Average separation increases.
The arrangement and intermolecular potential energy may also change.
The liquid expands (thermal expansion).

Typically, liquids expand more than solids for the same temperature increase; water has an important exception between 0 °C and 4 °C.

Gases

Flow and fill the container completely.
Volume changes strongly with pressure and temperature.
Particles far apart and randomly arranged.
Particles move randomly in all directions; their average translational kinetic energy depends on absolute temperature.
Intermolecular forces are negligible in the ideal-gas model, except during collisions.

When a gas is heated:

Average particle speeds increase as temperature rises.
At constant pressure, the gas expands and the average particle separation increases. In a rigid container, volume stays fixed and pressure rises.
At constant pressure, gases typically expand more than solids or liquids for the same temperature increase.

Internal energy

Particles in a substance have two forms of microscopic energy:

Random kinetic energy (due to particle motion)
Intermolecular potential energy (due to forces between particles)

Definition — internal energy

The internal energy of a system is the sum of the total random kinetic energy of its particles and the total intermolecular potential energy of its particles.

Internal energy diagram — Fig B.1.2 — Internal energy: kinetic + potential contributions

The internal energy of a system can change due to:

a change in kinetic energy (usually a temperature change)
a change in potential energy (often a phase change)

For the same substance:

At the melting point, solid and liquid have the same temperature; melting increases intermolecular potential energy.
At the boiling point, liquid and gas have the same temperature; vaporisation increases intermolecular potential energy.
For a fixed mass heated through melting and boiling, internal energy increases along the heating process. It is not determined by phase alone.

Intermolecular potential energy

With zero defined at infinite separation, the potential energy of a bound pair near equilibrium is negative.
Bringing particles closer from a large separation lowers potential energy until the equilibrium separation is reached.
Separating bound particles beyond equilibrium increases potential energy toward zero.
Work must be done to separate particles against attraction (increases potential energy).
At infinite separation, potential energy approaches $0$ . At extremely small separations, strong repulsion can make it positive; zero is not the maximum possible value.

Density

Definition — density

Density is mass per unit volume.

$\displaystyle \rho = \frac{m}{V}$

Unit: $\text{kg m}^{-3}$

For a given particle mass, smaller average spacing usually means greater density. For the same substance:

Solid usually has higher density than liquid. Water is an exception: ice is less dense than liquid water.
Gas has the lowest density.

Density comparison — Fig B.1.3 — Density and particle spacing

Worked example 1 — Density (IB style)

During an experiment, the mass of a copper cube with side length $20.0\,\text{mm}$ is measured to be $72\,\text{g}$ . Calculate the density of the copper.

Show answer and full working

Step 1: Convert to SI units $m = 72\,\text{g} = 0.072\,\text{kg}$ Side length $a = 20.0\,\text{mm} = 0.0200\,\text{m}$

Step 2: Volume of the cube
$\displaystyle V = a^3 = (0.0200)^3 = 8.00\times 10^{-6}\,\text{m}^3$

Step 3: Use the density equation
$\displaystyle \rho = \frac{m}{V} = \frac{0.072}{8.00\times 10^{-6}} = 9.00\times 10^3\,\text{kg m}^{-3}$

Answer: $\rho = 9.00\times 10^3\,\text{kg m}^{-3}$

B.1.2 Temperature scales

Temperature

Temperature describes how hot or cold a system is. Microscopically, temperature is related to the random motion of particles. Increasing temperature means increasing the average kinetic energy of particles.

Temperature does not depend on mass or the number of particles; it depends on the average microscopic kinetic energy.

Celsius and Kelvin scales

Celsius scale: based on water freezing at $0^\circ\text{C}$ and boiling at $100^\circ\text{C}$ (at standard atmospheric pressure).
Kelvin scale: absolute temperature scale, starts at absolute zero ( $0\,\text{K}$ ), no negative values.

Conversion between Celsius and Kelvin

$\displaystyle T(\text{K}) = \theta(^\circ\text{C}) + 273.15$

Celsius and Kelvin scales — Fig B.1.4 — Celsius vs Kelvin scales

Temperature changes

A change of $1^\circ\text{C}$ is the same size as a change of $1\,\text{K}$ : $\displaystyle \Delta T(\text{K}) = \Delta \theta(^\circ\text{C})$

Worked example 2 — Temperature change

Water is heated from $18^\circ\text{C}$ to $100^\circ\text{C}$ . Calculate the change in temperature in kelvin.

Show answer and full working

$\displaystyle \Delta \theta = 100 - 18 = 82^\circ\text{C}$
$\displaystyle \Delta T = 82\,\text{K}$
Answer: $82\,\text{K}$

Measuring temperature

Temperature is measured using thermometers (liquid-in-glass, digital, etc.). Digital thermometers often use a thermistor, a component whose resistance changes with temperature.

Kelvin temperature and average kinetic energy (ideal gas)

For an ideal gas, the average kinetic energy per atom is:

$\displaystyle \overline{E_k} = \frac{3}{2}k_B T$

where $k_B = 1.38\times 10^{-23}\,\text{J K}^{-1}$ (Boltzmann constant). This shows $\overline{E_k} \propto T$ , and the classical ideal-gas model predicts zero translational kinetic energy at $0\,\text{K}$ . This extrapolation does not describe quantum zero-point motion; real gases cease to behave ideally before this limit.

Worked example 3 — Average kinetic energy

Calculate the average kinetic energy of helium atoms at $22^\circ\text{C}$ .

Show answer and full working

Step 1: Convert to kelvin
$\displaystyle T = 22 + 273.15 \approx 295\,\text{K}$

Step 2: Use the equation
$\displaystyle \overline{E_k} = \frac{3}{2}k_B T$

Step 3: Substitute
$\displaystyle \overline{E_k} = \frac{3}{2}(1.38\times 10^{-23})(295) \approx 6.11\times 10^{-21}\,\text{J}$

Answer: $\overline{E_k} \approx 6.11\times 10^{-21}\,\text{J}$ per atom

Temperature and gas pressure (constant volume)

For a fixed mass of gas at constant volume:

Lower temperature → lower average kinetic energy → less frequent collisions with the walls and less momentum transferred per collision → lower pressure.
Extrapolating a pressure–temperature graph suggests pressure becomes zero at about $-273^\circ\text{C}$ ( $0\,\text{K}$ ).

B.1.3 Changing temperature and changing phase

Heating a substance: two cases

1) Heating without a change of phase

Energy supplied increases average kinetic energy.
Particles move faster.
Temperature increases.

2) Heating with a change of phase

Energy supplied does not increase average kinetic energy.
In the IB particle model, energy increases intermolecular potential energy; it is not used to increase average kinetic energy.
Energy changes the particle arrangement and overcomes attractive interactions. Separation increases greatly during vaporisation; melting need not increase volume.
For a pure substance changing phase in equilibrium at constant pressure, temperature stays constant during the phase change.

Heating curves

A heating curve is a graph of temperature against time (or energy supplied). Sloping sections correspond to temperature change (kinetic energy increases). Flat sections correspond to phase change (potential energy increases, temperature constant).

Specific heat capacity

Definition — specific heat capacity

The specific heat capacity $c$ is the energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\,\text{K}$ .

$\displaystyle Q = mc\Delta T$

Units: $c$ in $\text{J kg}^{-1}\text{K}^{-1}$

Worked example 4 — Specific heat capacity

How much energy is required to increase the temperature of $2.0\,\text{kg}$ of copper by $1\,\text{K}$ ? (Specific heat capacity of copper: $c = 385\,\text{J kg}^{-1}\text{K}^{-1}$ .)

Show answer and full working

$\displaystyle Q = mc\Delta T = (2.0)(385)(1) = 770\,\text{J}$
Answer: $770\,\text{J}$

Worked example 5 — Specific heat capacity

How much energy is required to increase the temperature of $1.0\,\text{kg}$ of copper by $2\,\text{K}$ ?

Show answer and full working

$\displaystyle Q = mc\Delta T = (1.0)(385)(2) = 770\,\text{J}$
Answer: $770\,\text{J}$

Thermal equilibrium and energy conservation

When two objects at different temperatures are in thermal contact, thermal energy transfers from hot to cold until they reach the same final temperature. This state is thermal equilibrium.

If energy losses to the surroundings are negligible:

$\displaystyle \text{Energy lost by hot object} = \text{Energy gained by cold object}$

Worked example 6 — Thermal equilibrium

A copper cube of mass $0.072\,\text{kg}$ is heated to $970^\circ\text{C}$ and plunged into $920\,\text{g}$ of water at $18^\circ\text{C}$ . Show that the equilibrium temperature is approximately $25^\circ\text{C}$ . $c_{\text{Cu}} = 385\,\text{J kg}^{-1}\text{K}^{-1}$ , $c_{\text{water}} = 4200\,\text{J kg}^{-1}\text{K}^{-1}$

Model assumptions: negligible energy loss to the surroundings, negligible container heat capacity, and constant specific heat capacities. Real copper has a temperature-dependent heat capacity.

Show answer and full working

Step 1: Write known values Copper: $m_{\text{Cu}} = 0.072\,\text{kg}$ , $T_{\text{Cu,i}} = 970^\circ\text{C}$ Water: $m_w = 920\,\text{g} = 0.920\,\text{kg}$ , $T_{w,i}=18^\circ\text{C}$ Let the equilibrium temperature be $T_f$ (in $^\circ\text{C}$ ). (Note: only temperature differences are used, so conversion to kelvin is not required.)

Step 2: Energy balance
$\displaystyle m_{\text{Cu}}c_{\text{Cu}}(970 - T_f) = m_wc_w(T_f - 18)$

Step 3: Substitute values
$\displaystyle (0.072)(385)(970 - T_f) = (0.920)(4200)(T_f - 18)$

Step 4: Solve Left side: $(0.072)(385)=27.72$ Right side: $(0.920)(4200)=3864$ $\displaystyle 27.72(970 - T_f) = 3864(T_f - 18)$
$\displaystyle 26888.4 - 27.72T_f = 3864T_f - 69552$
$\displaystyle 96440.4 = 3891.72T_f$
$\displaystyle T_f \approx 24.8^\circ\text{C}$

Answer: $T_f \approx 25^\circ\text{C}$ , which agrees with the statement in the question.

Phase change

A phase change occurs when a substance changes state (solid $\rightarrow$ liquid, liquid $\rightarrow$ gas, etc.). During a phase change:

For a pure substance at constant pressure, temperature remains constant during an equilibrium phase change.
Energy changes intermolecular potential energy.
The equation $Q=mc\Delta T$ does not apply.

Phase change diagram — Fig B.1.5 — Temperature constant during phase change

Specific latent heat

Definition — specific latent heat

The specific latent heat $L$ is the energy required to change the phase of $1\,\text{kg}$ of a substance at constant temperature.

$\displaystyle Q = mL$

Types:

Specific latent heat of fusion $L_f$ : solid $\rightarrow$ liquid
Specific latent heat of vaporisation $L_v$ : liquid $\rightarrow$ gas

Worked example 7 — Latent heat of fusion

To make a gold ring, $1.6\,\text{g}$ of gold is melted. Calculate the energy required. Specific latent heat of fusion of gold: $L_f = 63,000\,\text{J kg}^{-1}$ .

Assume the gold is already at its melting point. If it starts below this temperature, first add the energy needed to heat it to the melting point.

Show answer and full working

Step 1: Convert mass to kilograms
$\displaystyle m = 1.6\,\text{g} = 1.6\times 10^{-3}\,\text{kg}$

Step 2: Use $Q=mL$
$\displaystyle Q = (1.6\times 10^{-3})(63,000) = 100.8\,\text{J}$

Answer: $Q \approx 1.01\times 10^2\,\text{J}$

Worked example 8 — Latent heat and power

An electric heater supplying thermal power of $1200\,\text{W}$ is used to vaporise $220\,\text{g}$ of water at $100^\circ\text{C}$ . Calculate the time taken to vaporise all the water, assuming negligible energy losses. Use $L_v(\text{water}) \approx 2.26\times 10^6\,\text{J kg}^{-1}$ .

Show answer and full working

Step 1: Convert mass to kilograms
$\displaystyle m = 220\,\text{g} = 0.220\,\text{kg}$

Step 2: Energy required for vaporisation
$\displaystyle Q = mL_v = (0.220)(2.26\times 10^6) = 4.97\times 10^5\,\text{J}$

Step 3: Use power equation $P=\frac{Q}{t}$
$\displaystyle t = \frac{Q}{P} = \frac{4.97\times 10^5}{1200} \approx 4.14\times 10^2\,\text{s}$

Answer: $t \approx 4.14\times 10^2\,\text{s} \approx 6.9\,\text{min}$ (assuming all power goes into boiling).

Key points (B.1.3)

Temperature changes correspond to changes in kinetic energy.
Phase changes correspond to changes in potential energy.
$Q=mc\Delta T$ applies only when temperature changes.
$Q=mL$ applies only during phase changes.
Energy is conserved in thermal interactions (if losses are negligible).

B.1.4 Thermal energy transfer

Thermal energy transfer occurs when there is a temperature difference. Energy flows spontaneously from higher temperature to lower temperature. There are three main mechanisms: conduction, convection, and radiation.

Conduction

Conduction transfers energy through a material due to particle collisions and (in metals) the movement of free electrons. It is most effective in solids because particles are close together.

Molecular explanation:

Heating one end increases particle kinetic energy.
Particles vibrate more and collide with neighbours.
Kinetic energy is passed along without bulk movement of the material.

Conduction in metals:

Metals contain free (delocalised) electrons.
Electrons transfer energy rapidly by collisions.
This makes metals excellent thermal conductors.

Rate of energy transfer by conduction

The rate of energy transfer through a slab of material is:

$\displaystyle \frac{\Delta Q}{\Delta t} = \frac{kA\Delta T}{\Delta x}$

where $k$ is thermal conductivity ( $\text{W m}^{-1}\text{K}^{-1}$ ), $A$ is cross-sectional area, $\Delta x$ is thickness, and $\Delta T$ is the temperature difference across the material.

Applications: metal cookware, heat sinks, and insulation (low $k$ materials reduce conduction).

Convection

Convection transfers thermal energy by the bulk movement of fluid (liquids and gases only).

Molecular explanation:

Heating increases particle kinetic energy.
Particles spread out → density decreases.
Warm, less dense fluid rises.
Cool, denser fluid sinks to replace it.
This creates a convection current.

Natural convection occurs due to density differences. Forced convection uses an external device (fan/pump).

Applications: boiling water, atmospheric circulation, room heating by radiators.

Convection currents diagram — Fig B.1.6 — Convection currents in a fluid

Thermal radiation

Thermal radiation is energy transfer by electromagnetic waves (mainly infrared). Unlike conduction and convection:

No medium is required.
Radiation can occur through a vacuum.

All objects above $0\,\text{K}$ emit thermal radiation. Surface properties matter:

Matt black surfaces are generally good absorbers and emitters of thermal radiation.
Polished metallic surfaces are generally poor emitters and absorbers and good reflectors. Visible colour alone does not determine infrared behaviour.

Applications: infrared heaters, thermos flasks (shiny surfaces), thermal imaging.

Comparison of methods

Conduction: needs medium; particle collisions / free electrons; most effective in solids.
Convection: needs medium; bulk fluid movement; liquids and gases.
Radiation: no medium; electromagnetic waves; works in vacuum.

B.1.5 Black-body emission

Thermal radiation from stars

Stars emit thermal radiation due to the high temperatures of particles within them, radiating in all directions. As radiation travels away, the same power spreads over a larger area, so intensity decreases with distance.

Intensity

Definition — intensity

Intensity is power per unit area.

$\displaystyle I = \frac{P}{A}$

Units: $\text{W m}^{-2}$

Radiation from a star (inverse-square spread)

At distance $r$ from a star, the radiation is spread over a sphere:

$\displaystyle A = 4\pi r^2$

If the star’s total power output (luminosity) is $L$ , then:

$\displaystyle I = \frac{L}{4\pi r^2}$

Worked example 9 — Intensity of the Sun at Earth

Calculate the intensity of radiation from the Sun at Earth’s orbit. Use typical values: distance $r = 1.50\times 10^{11}\,\text{m}$ and solar luminosity $L = 3.83\times 10^{26}\,\text{W}$ .

Show answer and full working

$\displaystyle I = \frac{L}{4\pi r^2} = \frac{3.83\times 10^{26}}{4\pi(1.50\times 10^{11})^2}$
$\displaystyle I \approx 1.35\times 10^3\,\text{W m}^{-2}$
Answer: $I \approx 1.35\times 10^3\,\text{W m}^{-2}$ (about the solar constant)

Black bodies

A black body is an ideal object that:

absorbs all incident electromagnetic radiation,
emits radiation at all wavelengths,
has an emission distribution that depends only on temperature.

No real object is perfectly black, but stars are good approximations.

Black-body emission spectrum

A black-body emission spectrum is a graph of spectral intensity (intensity per unit wavelength) against wavelength. The area under the curve represents total intensity. Key features:

Radiation at all wavelengths
A peak intensity at a particular wavelength
The curve is not symmetrical; spectral intensity approaches zero as wavelength approaches zero and also at very long wavelengths
Higher temperature → greater total emitted power and peak shifts to shorter wavelengths

Black-body spectrum — Fig B.1.7 — Black-body emission spectrum

Wien’s displacement law

The peak wavelength is inversely proportional to absolute temperature:

$\displaystyle \lambda_{\mathrm{max}}T = b_W$

where $b_W \approx 2.90\times 10^{-3}\,\text{m K}$ .

Worked example 10 — Temperature of a star

A star has peak wavelength $500\,\text{nm}$ . Calculate its surface temperature.

Show answer and full working

Step 1: Convert to metres
$\displaystyle \lambda_{\mathrm{max}} = 500\,\text{nm} = 5.00\times 10^{-7}\,\text{m}$

Step 2: Use Wien’s law
$\displaystyle T = \frac{b_W}{\lambda_{\mathrm{max}}} = \frac{2.90\times 10^{-3}}{5.00\times 10^{-7}} = 5.80\times 10^3\,\text{K}$

Answer: $T \approx 5.8\times 10^3\,\text{K}$

Luminosity and the Stefan–Boltzmann law

Luminosity $L$ is the total energy emitted per second (power output) by a star.

$\displaystyle L = \sigma A T^4$

where $\sigma = 5.67\times 10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$ and $A$ is surface area. For a spherical star with radius $R$ :

$\displaystyle A = 4\pi R^2 \qquad \Rightarrow \qquad L = 4\pi R^2\sigma T^4$

Apparent brightness

Apparent brightness $b$ is the power per unit area received by an observer:

$\displaystyle b = \frac{L}{4\pi d^2}$

Key distinction: luminosity is intrinsic; apparent brightness depends on luminosity and distance.

Worked example 11 — Apparent brightness (IB style)

Sirius is the brightest star in the night sky. Suppose Sirius has luminosity $L = 9.7\times 10^{27}\,\text{W}$ and its distance is $d = 8.14\times 10^{16}\,\text{m}$ . Calculate its apparent brightness at Earth.

Show answer and full working

$\displaystyle b = \frac{L}{4\pi d^2} = \frac{9.7\times 10^{27}}{4\pi(8.14\times 10^{16})^2}$
$\displaystyle b \approx 1.16\times 10^{-7}\,\text{W m}^{-2}$
Answer: $b \approx 1.16\times 10^{-7}\,\text{W m}^{-2}$

Worked example 12 — Multi-step IB problem

A star has diameter $1.4\times 10^9\,\text{m}$ and emits radiation with peak wavelength $240\,\text{nm}$ . Its apparent brightness at Earth is $b = 1.0\times 10^{-9}\,\text{W m}^{-2}$ . Calculate the distance of the star from Earth. Use: $b = \frac{L}{4\pi d^2}$ , $L = 4\pi R^2\sigma T^4$ , $\lambda_{\mathrm{max}}T=b_W$ with $b_W=2.90\times 10^{-3}\,\text{m K}$ .

Show answer and full working

Step 1: Find the temperature (Wien)
$\displaystyle \lambda_{\mathrm{max}} = 240\,\text{nm} = 2.40\times 10^{-7}\,\text{m}$
$\displaystyle T = \frac{2.90\times 10^{-3}}{2.40\times 10^{-7}} = 1.21\times 10^4\,\text{K}$

Step 2: Find the radius
$\displaystyle R = \frac{\text{diameter}}{2} = \frac{1.4\times 10^9}{2} = 7.0\times 10^8\,\text{m}$

Step 3: Find luminosity (Stefan–Boltzmann)
$\displaystyle L = 4\pi R^2\sigma T^4$
$\displaystyle L = 4\pi(7.0\times 10^8)^2(5.67\times 10^{-8})(12083.3)^4 \approx 7.44\times 10^{27}\,\text{W}$

Step 4: Use apparent brightness to find distance
$\displaystyle b = \frac{L}{4\pi d^2}\Rightarrow d = \sqrt{\frac{L}{4\pi b}}$
$\displaystyle d = \sqrt{\frac{7.44\times 10^{27}}{4\pi(1.0\times 10^{-9})}} \approx 7.70\times 10^{17}\,\text{m}$

Answer: $d \approx 7.7\times 10^{17}\,\text{m}$

IB exam tips

Always state whether energy goes into kinetic (temperature change) or potential (phase change).
Use $Q=mc\Delta T$ only when temperature changes; use $Q=mL$ during phase changes.
Convert grams and millimetres to SI units before using formulas.
For star problems: often combine Wien $\rightarrow$ Stefan–Boltzmann $\rightarrow$ inverse-square brightness.

Essential equations — B.1

Use SI units. Absolute temperatures in kinetic-theory and radiation equations must be in kelvin.

$\displaystyle \rho=\frac{m}{V}$

$\displaystyle \overline{E_k}=\frac{3}{2}k_B T$

$\displaystyle Q=mc\Delta T$

$\displaystyle Q=mL$

$\displaystyle \frac{\Delta Q}{\Delta t}=kA\frac{\Delta T}{\Delta x}$

$\displaystyle L=\sigma AT^4$

$\displaystyle b=\frac{L}{4\pi d^2}$

$\displaystyle \lambda_{\mathrm{max}}T=2.90\times10^{-3}\,\text{m K}$

Symbol reminder: $L$ means specific latent heat in $Q=mL$ , but luminosity in radiation equations. The context and units distinguish them. Use $b_W$ for Wien’s constant when $b$ denotes apparent brightness.

Radiation from real surfaces

A black body has emissivity $\varepsilon=1$ . A real surface emits less power than a black body of the same area and temperature:

$\displaystyle P=\varepsilon \sigma AT^4$

For a small object exchanging radiation with large surroundings at temperature $T_s$ , with the same effective emissivity for absorption and emission:

$\displaystyle P_{\text{net,out}}=\varepsilon \sigma A(T^4-T_s^4)$

The net transfer is zero when the object and surroundings have the same temperature, even though each continues to emit and absorb radiation.

Practice — check your understanding

1. Heating and phase change

Explain why the temperature of a pure substance remains constant while it melts at constant pressure, even though energy continues to enter it.

Show answer

Energy increases intermolecular potential energy by changing the particle arrangement and overcoming attractive interactions. Average kinetic energy does not increase, so temperature remains constant. Internal energy nevertheless increases.

2. Heating water

Calculate the energy required to heat $0.500\,\text{kg}$ of water from $20.0^\circ\text{C}$ to $80.0^\circ\text{C}$ . Use $c=4200\,\text{J kg}^{-1}\text{K}^{-1}$ and neglect heat losses.

Show answer and working

$\displaystyle \Delta T=80.0-20.0=60.0\,\text{K}$

$\displaystyle Q=mc\Delta T=(0.500)(4200)(60.0)=1.26\times10^5\,\text{J}$

3. Melting ice

Calculate the energy needed to melt $0.200\,\text{kg}$ of ice already at $0^\circ\text{C}$ . Use $L_f=3.34\times10^5\,\text{J kg}^{-1}$ .

Show answer and working

$\displaystyle Q=mL_f=(0.200)(3.34\times10^5)=6.68\times10^4\,\text{J}$

No temperature rise is involved in this stage.

4. Conduction through a wall

A wall has area $12.0\,\text{m}^2$ , thickness $0.200\,\text{m}$ and thermal conductivity $0.800\,\text{W m}^{-1}\text{K}^{-1}$ . Its two surfaces differ in temperature by $15.0\,\text{K}$ . Find the steady rate of thermal energy transfer.

Show answer and working

$\displaystyle \frac{\Delta Q}{\Delta t}=\frac{kA\Delta T}{\Delta x}$

$\displaystyle \frac{\Delta Q}{\Delta t}=\frac{(0.800)(12.0)(15.0)}{0.200}=720\,\text{W}$

Doubling thickness halves this rate, provided the other quantities stay fixed.

5. Black-body comparisons

Two stars have equal radii. Star A has twice the absolute surface temperature of star B. Compare their luminosities and peak wavelengths. How does apparent brightness change when distance from either star doubles?

Show answer and reasoning

$\displaystyle \frac{L_A}{L_B}=\left(\frac{T_A}{T_B}\right)^4=2^4=16$

$\displaystyle \frac{\lambda_{max,A}}{\lambda_{max,B}}=\frac{T_B}{T_A}=\frac{1}{2}$

$\displaystyle b\propto \frac{1}{d^2}$

Star A has 16 times the luminosity and half the peak wavelength. Doubling distance reduces apparent brightness to one quarter, assuming unobstructed radiation spreading uniformly in all directions.

Faith connection — God’s care in creation and in our lives

“He healeth the broken in heart, and bindeth up their wounds.”Psalm 147:3 (KJV)

Creation — a world sustained by God

As Christians, we recognise the order of creation as a reason to worship God. The Sun supplies energy that warms Earth, and the thermal properties of water help moderate changes in temperature. Studying these processes invites us to appreciate God’s wisdom and to care responsibly for the world He has entrusted to us. Physical laws describe how these processes work; this reflection expresses our Christian response to them.

The Fall — a world marked by suffering

Scripture teaches that sin has damaged our relationship with God and with one another. We experience suffering, selfishness and brokenness, and people around us may lack warmth, shelter or practical care. Scientific knowledge gives us opportunities to serve them through safer homes, wiser use of energy and thoughtful stewardship. Thermal processes themselves are not moral failings; our choices about how we use resources matter.

Redemption — receiving the love of Jesus

God has not abandoned us. Jesus Christ came to save us through His death and resurrection. Just as Psalm 147 speaks of God’s care for the brokenhearted, the gospel invites us to bring our burdens and sins to Him. Turn to Jesus in repentance and faith, receive Him as your personal God, Saviour and Lord, and trust His grace rather than your own achievements (John 3:16; Ephesians 2:8–10).

Restoration — hope that leads to action

God promises a renewed creation in which suffering and death will end (Revelation 21:4). While we wait, we can show Christ’s love through generosity, compassion and care for creation. Connect with other believers, read and study the Bible, and spend time in prayer so that your faith and knowledge of God continue to grow.

Reflect and pray

Reflect: Who needs practical help or encouragement this week? How could your understanding of thermal energy help you make a thoughtful choice about energy use?

Prayer: Lord Jesus, thank You for Your care in creation and for the salvation You offer. Forgive my sins, heal what is broken in my life, and help me trust You as my God, Saviour and Lord. Teach me through Your Word and guide me to serve others with love. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Thermal energy transfers

B.1 — Thermal Energy Transfers

Learning objectives

B.1.1 Molecular theory in solids, liquids, and gases

Solids

Liquids

Gases

Internal energy

Definition — internal energy

Intermolecular potential energy

Density

Definition — density

Worked example 1 — Density (IB style)

B.1.2 Temperature scales

Temperature

Celsius and Kelvin scales

Conversion between Celsius and Kelvin

Temperature changes

Worked example 2 — Temperature change

Measuring temperature

Kelvin temperature and average kinetic energy (ideal gas)

Worked example 3 — Average kinetic energy

Temperature and gas pressure (constant volume)

B.1.3 Changing temperature and changing phase

Heating a substance: two cases

Heating curves

Specific heat capacity

Definition — specific heat capacity

Worked example 4 — Specific heat capacity

Worked example 5 — Specific heat capacity

Thermal equilibrium and energy conservation

Worked example 6 — Thermal equilibrium

Phase change

Specific latent heat

Definition — specific latent heat

Worked example 7 — Latent heat of fusion

Worked example 8 — Latent heat and power

Key points (B.1.3)

B.1.4 Thermal energy transfer

Conduction

Rate of energy transfer by conduction

Convection

Thermal radiation

Comparison of methods

B.1.5 Black-body emission

Thermal radiation from stars

Intensity

Definition — intensity

Radiation from a star (inverse-square spread)

Worked example 9 — Intensity of the Sun at Earth

Black bodies

Black-body emission spectrum

Wien’s displacement law

Worked example 10 — Temperature of a star

Luminosity and the Stefan–Boltzmann law

Apparent brightness

Worked example 11 — Apparent brightness (IB style)

Worked example 12 — Multi-step IB problem

IB exam tips

Essential equations — B.1

Radiation from real surfaces

Practice — check your understanding

1. Heating and phase change

2. Heating water

3. Melting ice

4. Conduction through a wall

5. Black-body comparisons

Faith connection — God’s care in creation and in our lives

Creation — a world sustained by God

The Fall — a world marked by suffering

Redemption — receiving the love of Jesus

Restoration — hope that leads to action

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