SL & HL The Particulate Nature of Matter BETA B.1

Thermal Energy Transfers

This topic explores the microscopic properties of solids, liquids, and gases to explain macroscopic thermal behaviors. It covers the quantitative analysis of temperature changes, phase transitions, and the mechanisms of conduction, convection, and thermal radiation.

Part of IB Physics (2025-2030 syllabus) — Standard and Higher Level.

Key points

  • Matter is modeled as a collection of particles whose arrangement, spacing, and motion define the solid, liquid, and gas phases.
  • Temperature is a macroscopic measure of the average random translational kinetic energy of the particles within a substance, where absolute zero (0 K=273.15C0\ \text{K} = -273.15^\circ\text{C}) corresponds to minimum molecular kinetic energy.
  • Internal energy (UU) of a system is the total random kinetic energy plus the total potential energy of all its constituent particles.
  • Specific heat capacity (cc) dictates the energy transfer required to alter a system's temperature without a change of state.
  • During a phase transition, temperature remains constant because thermal energy is used exclusively to alter intermolecular potential energy rather than kinetic energy.
  • Thermal energy transfers from regions of higher temperature to regions of lower temperature via three main mechanisms: conduction (particle collisions and free electron diffusion), convection (buoyant density currents in fluids), and thermal radiation (electromagnetic wave emission).
  • A black body is a perfect absorber and emitter of radiation (e=1e = 1); its emission spectrum peaks at a wavelength that decreases as its absolute temperature increases.

Subtopic by subtopic

Molecular theory of solids, liquids and gases

The kinetic model treats all matter as an enormous number of tiny particles in constant random motion, with the phase of a substance set by how the particles are spaced, ordered and able to move.

In a solid, strong intermolecular forces lock particles into a closely packed, often regular arrangement; each particle can only vibrate about a fixed position, so solids keep both their shape and volume.

In a liquid, the particles remain almost as close together, but they carry enough energy to slide past their neighbours, so liquids flow and take the shape of their container while keeping a nearly fixed volume.

In a gas, particles are typically separated by around ten particle diameters, travel rapidly in straight lines between collisions, and feel negligible intermolecular forces except at the instant of collision, so a gas expands to fill any container and has a much lower density.

  • Solid: strong forces, vibration only, fixed shape and volume.
  • Liquid: forces hold particles close but not in place; fixed volume, variable shape.
  • Gas: negligible forces, fast random motion, no fixed shape or volume.

A concrete comparison: 1 kg1\ \text{kg} of steam occupies roughly a thousand times the volume of 1 kg1\ \text{kg} of liquid water, which the model explains purely through particle spacing.

You should be able to use particle arrangement, spacing and motion to compare densities and explain compressibility and flow in each phase.

Temperature, internal energy and the Kelvin scale

Temperature is the macroscopic property that decides the direction of thermal energy transfer: energy always flows spontaneously from the hotter body to the colder one until both reach the same temperature (thermal equilibrium).

Microscopically, the absolute temperature of a substance is directly proportional to the average random translational kinetic energy of its particles:

Eˉk=32kBT\bar{E}_k = \frac{3}{2} k_B T

Doubling the Kelvin temperature therefore doubles the average kinetic energy per particle.

Internal energy is a different quantity: it is the total of the random kinetic energies *and* the intermolecular potential energies of every particle in the system.

A swimming pool at 25C25^\circ\text{C} has a far greater internal energy than a cup of tea at 90C90^\circ\text{C}, simply because it contains vastly more particles, yet thermal energy would still flow from the tea to the pool.

The Kelvin scale is the absolute scale required for these proportionalities. Convert using:

T(K)=θ(C)+273T(\text{K}) = \theta(^\circ\text{C}) + 273

Absolute zero, 0 K=273.15C0\ \text{K} = -273.15^\circ\text{C}, is the temperature at which particle kinetic energy reaches its minimum possible value.

Note that a temperature *difference* has the same numerical value in kelvin and in degrees Celsius, but any formula containing TT itself (such as Eˉk=32kBT\bar{E}_k = \frac{3}{2} k_B T) demands Kelvin.

Specific heat capacity

Specific heat capacity cc measures how much thermal energy must be supplied per kilogram of a substance to raise its temperature by one kelvin, with no change of phase. The defining relationship is:

Q=mcΔTQ = mc\Delta T

where QQ is the energy transferred, mm the mass and ΔT\Delta T the temperature change.

Substances differ enormously: water's value, about 4180 J kg1K14180\ \text{J kg}^{-1}\text{K}^{-1}, is roughly ten times that of copper, which is why coastal climates are moderated by the sea and why water is used as a coolant in car engines and central heating systems.

In the laboratory, cc is usually found electrically: an immersion heater of known power PP warms a measured mass for a time tt, so that Pt=mcΔTPt = mc\Delta T once energy losses are accounted for.

Real experiments always lose some energy to the surroundings and to the container, so the measured value of cc tends to come out too large unless insulation or a cooling correction is used.

You must be able to:

  • rearrange Q=mcΔTQ = mc\Delta T for any variable
  • combine it with electrical power (Q=PtQ = Pt)
  • handle mixing problems where energy lost by the hot object equals energy gained by the cold one.

Phase changes and latent heat

During melting, freezing, boiling or condensing, a substance changes phase at a constant temperature. The energy involved is given by:

Q=mLQ = mL

where LL is the specific latent heat: of fusion LfL_f for solid–liquid changes, of vaporization LvL_v for liquid–gas changes.

For water, Lv2.26×106 J kg1L_v \approx 2.26 \times 10^6\ \text{J kg}^{-1} is nearly seven times Lf3.33×105 J kg1L_f \approx 3.33 \times 10^5\ \text{J kg}^{-1}, because vaporization must separate the molecules almost completely and do work pushing back the atmosphere.

The constant temperature is explained microscopically by energy bookkeeping. Particles in a solid oscillate in deep potential energy wells created by strong electrostatic attraction; supplying thermal energy normally increases the amplitude of these vibrations, raising the random kinetic energy and hence the temperature.

During melting, however, the supplied energy is used to do work against the attractive forces, pulling the particles slightly further apart and raising their intermolecular potential energy while their average kinetic energy stays constant.

This is why a heating curve shows a horizontal plateau at each phase change: the energy is spent climbing out of the potential energy well, not speeding the particles up.

On a heating or cooling curve, apply Q=mcΔTQ = mc\Delta T on the sloped sections and Q=mLQ = mL on the plateaus, and be ready to chain several stages together in one calculation.

GraphGraph with axes energy added and T / °C. meltingboilingenergy addedT / °C
Heating curve: as energy is added at a steady rate the temperature rises, but stays constant during melting and boiling while the substance changes state.

Conduction, convection and thermal radiation

Thermal energy moves from hot to cold by three distinct mechanisms.

Conduction occurs mainly in solids: energetic particles pass on kinetic energy through collisions with neighbours, and in metals free electrons diffuse rapidly through the lattice, making metals excellent conductors. The rate of conduction through a slab is given by:

ΔQΔt=kAΔTΔx\frac{\Delta Q}{\Delta t} = kA\frac{\Delta T}{\Delta x}

Each of the following increases the rate:

  • a larger area AA
  • a bigger temperature difference ΔT\Delta T
  • a thinner layer Δx\Delta x
  • a higher thermal conductivity kk

Double glazing works by trapping a low-kk air gap.

Convection occurs only in fluids: fluid warmed at the bottom expands, becomes less dense, and rises, while cooler denser fluid sinks to replace it, setting up a circulating current. Sea breezes and the cycling of water in a heated pan are everyday examples.

Thermal radiation is the emission of electromagnetic waves (mostly infrared at everyday temperatures) and needs no medium, which is how the Sun heats the Earth. Every object above absolute zero radiates.

The emitted power follows the Stefan-Boltzmann law: a black body radiates a total power (luminosity)

L=σAT4L = \sigma A T^4

while a real surface radiates P=eσAT4P = e\sigma A T^4, where the emissivity ee compares the surface to a perfect black body (e=1e = 1).

The black-body spectrum peaks at a wavelength given by Wien's law, λmaxT=2.9×103 m K\lambda_{\text{max}} T = 2.9 \times 10^{-3}\ \text{m K}, and the apparent brightness of a distant source falls with distance as b=L4πd2b = \frac{L}{4\pi d^2}, where LL is its total radiated power (luminosity).

Formulae

ρ=mV\rho = \frac{m}{V}

To find the density of a substance from its mass mm and volume VV. Comparing the density of the same substance in different phases reflects how closely its particles are spaced.

Q=mcΔTQ = mc\Delta T

To calculate the thermal energy transferred when a substance of mass mm undergoes a temperature change ΔT\Delta T without changing its state.

Q=mLQ = mL

To calculate the thermal energy transferred during a phase change (fusion or vaporization) at constant temperature.

L=σAT4L = \sigma A T^4

The Stefan-Boltzmann law: LL is the total radiated power (luminosity) emitted as electromagnetic radiation by a black body of surface area AA at absolute temperature TT. A real (grey) body radiates P=eσAT4P = e\sigma A T^4, where the emissivity e<1e < 1.

b=L4πd2b = \frac{L}{4\pi d^2}

To find the apparent brightness bb (power received per unit area) at a distance dd from a source of luminosity LL. Brightness falls off as the inverse square of the distance.

Eˉk=32kBT\bar{E}_k = \frac{3}{2} k_B T

To relate the average random translational kinetic energy of a particle to the absolute (Kelvin) temperature TT of the substance, where kB=1.38×1023 J K1k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}.

ΔQΔt=kAΔTΔx\frac{\Delta Q}{\Delta t} = kA\frac{\Delta T}{\Delta x}

To calculate the rate of thermal energy conduction through a material of thermal conductivity kk, cross-sectional area AA and thickness Δx\Delta x when its faces differ in temperature by ΔT\Delta T.

λmaxT=2.9×103 m K\lambda_{\text{max}} T = 2.9 \times 10^{-3}\ \text{m K}

Wien's displacement law: to find the wavelength at which a black body's emission spectrum peaks, or to estimate a body's surface temperature from its peak wavelength.

Definitions

Internal Energy
The sum of the random kinetic energy and potential energy of all the molecules in a substance.
Temperature
A macroscopic quantity proportional to the average random translational kinetic energy of the constituent particles of a system, measured on an absolute scale.
Specific Heat Capacity
The quantity of thermal energy required per unit mass of a substance to increase its temperature by one Kelvin (or one degree Celsius) without any phase change.
Specific Latent Heat
The quantity of thermal energy required per unit mass to change the state of a substance at a constant temperature.
Emissivity
The ratio of the power radiated per unit area by a surface to the power radiated per unit area by a black body at the same temperature; it ranges from 00 for a perfect reflector to 11 for a perfect black body.
Black Body
An idealized object that absorbs all electromagnetic radiation incident on it and, for a given temperature, emits the maximum possible radiated power across all wavelengths.

Worked examples

1

An electrical heater of power 150 W150\ \text{W} is used to heat 0.50 kg0.50\ \text{kg} of a liquid. The liquid's temperature increases from 20C20^\circ\text{C} to 80C80^\circ\text{C} in 400 s400\ \text{s}. If 20%20\% of the energy from the heater is lost to the surroundings, calculate the specific heat capacity of the liquid.

  1. 1
    Calculate the total electrical energy delivered by the heater: Etotal=P×t=150 W×400 s=60000 JE_{\text{total}} = P \times t = 150\ \text{W} \times 400\ \text{s} = 60000\ \text{J}.
  2. 2
    Determine the useful energy absorbed by the liquid, accounting for the 20%20\% loss (meaning 80%80\% is useful): Q=0.80×60000 J=48000 JQ = 0.80 \times 60000\ \text{J} = 48000\ \text{J}.
  3. 3
    Calculate the temperature change of the liquid: ΔT=80C20C=60 K\Delta T = 80^\circ\text{C} - 20^\circ\text{C} = 60\ \text{K}.
  4. 4
    Rearrange the specific heat capacity formula: c=QmΔTc = \frac{Q}{m\Delta T}.
  5. 5
    Substitute the values to find cc: c=48000 J0.50 kg×60 K=1600 J kg1K1c = \frac{48000\ \text{J}}{0.50\ \text{kg} \times 60\ \text{K}} = 1600\ \text{J\ kg}^{-1}\text{K}^{-1}.

Answer: 1600 J kg1K11600\ \text{J\ kg}^{-1}\text{K}^{-1}

2

A solid block of ice of mass 0.20 kg0.20\ \text{kg} at 0C0^\circ\text{C} is placed in 0.80 kg0.80\ \text{kg} of water at 25C25^\circ\text{C}. Determine the final temperature of the mixture, assuming no thermal energy is lost to the environment. (Specific heat capacity of water cw=4180 J kg1K1c_w = 4180\ \text{J\ kg}^{-1}\text{K}^{-1}, specific latent heat of fusion of ice Lf=3.33×105 J kg1L_f = 3.33 \times 10^5\ \text{J\ kg}^{-1})

  1. 1
    Determine if the warm water has enough energy to melt all the ice. The energy required to melt all the ice is Qmelt=miceLf=0.20 kg×3.33×105 J kg1=66600 JQ_{\text{melt}} = m_{\text{ice}} L_f = 0.20\ \text{kg} \times 3.33 \times 10^5\ \text{J\ kg}^{-1} = 66600\ \text{J}.
  2. 2
    The energy released if the warm water cooled all the way to 0C0^\circ\text{C} is Qlimit=mwatercwΔT=0.80 kg×4180 J kg1K1×25 K=83600 JQ_{\text{limit}} = m_{\text{water}} c_w \Delta T = 0.80\ \text{kg} \times 4180\ \text{J\ kg}^{-1}\text{K}^{-1} \times 25\ \text{K} = 83600\ \text{J}. Since 83600 J>66600 J83600\ \text{J} > 66600\ \text{J}, all the ice melts and the final temperature TfT_f will be above 0C0^\circ\text{C}.
  3. 3
    Set up the conservation of energy equation: Heat gained by ice=Heat lost by water\text{Heat gained by ice} = \text{Heat lost by water}.
  4. 4
    Heat gained by ice includes melting it and then heating the melted ice water from 0C0^\circ\text{C} to TfT_f: Qgained=miceLf+micecw(Tf0)Q_{\text{gained}} = m_{\text{ice}} L_f + m_{\text{ice}} c_w (T_f - 0).
  5. 5
    Heat lost by warm water: Qlost=mwatercw(25Tf)Q_{\text{lost}} = m_{\text{water}} c_w (25 - T_f).
  6. 6
    Equate the expressions: 66600+0.20×4180×Tf=0.80×4180×(25Tf)66600 + 0.20 \times 4180 \times T_f = 0.80 \times 4180 \times (25 - T_f).
  7. 7
    Simplify the equation: 66600+836Tf=3344(25Tf)    66600+836Tf=836003344Tf66600 + 836 T_f = 3344 (25 - T_f) \implies 66600 + 836 T_f = 83600 - 3344 T_f.
  8. 8
    Group the terms of TfT_f: 4180Tf=170004180 T_f = 17000.
  9. 9
    Solve for TfT_f: Tf=1700041804.1CT_f = \frac{17000}{4180} \approx 4.1^\circ\text{C}.

Answer: 4.1C4.1^\circ\text{C}

3

A polished metal sphere of radius 0.040 m0.040\ \text{m} has a surface temperature of 500 K500\ \text{K} and an emissivity of 0.600.60. Calculate the power it radiates to its surroundings. (Stefan-Boltzmann constant σ=5.67×108 W m2K4\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4})

  1. 1
    Calculate the surface area of the sphere: A=4πr2=4π×(0.040 m)2=2.01×102 m2A = 4\pi r^2 = 4\pi \times (0.040\ \text{m})^2 = 2.01 \times 10^{-2}\ \text{m}^2.
  2. 2
    Calculate the fourth power of the absolute temperature: T4=(500 K)4=6.25×1010 K4T^4 = (500\ \text{K})^4 = 6.25 \times 10^{10}\ \text{K}^4.
  3. 3
    Write down the Stefan-Boltzmann law for a non-black body: P=eσAT4P = e\sigma A T^4.
  4. 4
    Substitute the values: P=0.60×5.67×108 W m2K4×2.01×102 m2×6.25×1010 K4P = 0.60 \times 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4} \times 2.01 \times 10^{-2}\ \text{m}^2 \times 6.25 \times 10^{10}\ \text{K}^4.
  5. 5
    Evaluate to obtain the radiated power: P42.8 W43 WP \approx 42.8\ \text{W} \approx 43\ \text{W}.

Answer: 43 W\approx 43\ \text{W}

Common mistakes

  • ×Confusing temperature and internal energy. Remember that two bodies can have the exact same temperature but vastly different internal energies due to differences in mass or phase.
  • ×Adding thermal energy to a system during a phase change and expecting a temperature rise. During a state change, the temperature remains strictly constant as the average kinetic energy of the particles does not change.
  • ×Forgetting to convert mass values to standard SI units (kilograms) when working with specific heat capacities specified in J kg1K1\text{J\ kg}^{-1}\text{K}^{-1}.
  • ×Failing to convert Celsius to Kelvin when applying the Stefan-Boltzmann radiation law (PT4P \propto T^4). While temperature differences ΔT\Delta T are identical in Celsius and Kelvin, absolute values must be in Kelvin.

Exam tips

  • When asked to *distinguish* between internal energy and temperature, clarify that temperature is a measure of the average random kinetic energy of particles, while internal energy is the sum total of both kinetic and potential energies across all particles.
  • If a question asks you to *explain* why the specific latent heat of vaporization is always greater than the specific latent heat of fusion for the same substance, discuss how vaporization requires completely breaking intermolecular bonds and performing work against atmospheric pressure, whereas melting only requires weakening those bonds.
  • When *determining* values from heating or cooling curves, identify horizontal plateaus as phase changes where Q=mLQ = mL, and sloped lines as temperature changes where Q=mcΔTQ = mc\Delta T.
  • To score high marks on thermal radiation descriptions, explicitly state that *all* objects above absolute zero emit electromagnetic radiation, and distinguish the emissivity of a perfect black body (e=1e = 1) from other reflective surfaces (e<1e < 1).

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