IB Physics HL Thermal Energy Transfers — Complete Cheatsheet
Every formula, definition, trick, and trap you need for IB Physics HL Topic B.1. Hand-built by an IBO-certified Singapore tutor with 15+ years of IB experience.
Thermal Energy Transfers is the gateway topic of the Theme B: Thermal & Particle Physics block in IB Physics. It anchors every later topic in the block — gas laws, thermodynamics, even the stellar luminosity calculations in the astrophysics paper — so a clean grasp of it pays off all the way to E5. The HL syllabus expects you to handle three transfer mechanisms (conduction, convection, radiation), switch confidently between $Q = mc\Delta T$ and $Q = mL$, and apply Stefan-Boltzmann and Wien's laws to stars and to the Earth.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.1 into one revisable page. Each section pairs the IB-style formula box with the highest-leverage trick and the trap that wastes most marks in mock exams. Scroll to the worked example to see the IB mark-scheme rhythm of M1/A1/R1, and use the FAQ block at the bottom to bullet-proof the concepts examiners love to probe.
TrickDensity of a gas is roughly 1000× smaller than that of a solid or liquid because the molecules are around 10× further apart in each direction ($10^3$ in volume).
TrapDensity is a scalar — no direction. Don't mix unit systems either: $\text{g cm}^{-3}$ and $\text{kg m}^{-3}$ differ by a factor of 1000.
NoteAt a phase change density jumps discontinuously — most dramatically when liquid becomes gas (volume rises by ~1000×).
The three states of matter. A solid holds an ordered lattice; a liquid keeps molecules close but mobile; a gas is sparse with rapid random motion.
From the Photon question bank
What examiners actually test on this concept:
ρ = m/VStates of matterGas vs solid densityg cm⁻³ vs kg m⁻³
Density is a scalar — mind the units.frequent g cm⁻³ and kg m⁻³ differ by a factor of 1000.
Gas density is ~1000× smaller than a solid or liquid because the molecules sit ~10× further apart in every direction.
At a phase change density jumps discontinuously — most dramatically liquid → gas (volume up ~1000×).
§2 — Temperature Scales & Average KE Topic B.1
Key formulae
K from °C:$T(\text{K}) = t({}^\circ\text{C}) + 273$
Difference:$\Delta T (\text{K}) = \Delta t ({}^\circ\text{C})$
Average KE:$\bar{E}_k = \tfrac{3}{2} k_B T$, where $k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1}$
Trick$\Delta T$ is identical in K and °C. So in $Q = mc\Delta T$ either scale works — just be consistent.
TrapNever add 273 to a temperature difference. Add 273 only when converting an absolute temperature from °C to K.
NoteAt $T = 0\,\text{K}$ all molecular motion ceases ($\bar{E}_k = 0$). It is physically impossible to reach a temperature below 0 K.
From the Photon question bank
What examiners actually test on this concept:
T(K) = t + 273ΔT same in K and °CĒ_k = 3/2 k_B TAbsolute zero
Add 273 only to an ABSOLUTE temperature, never to a difference.frequent $\Delta T$ is identical in K and °C.
Average translational KE per molecule is $\tfrac32 k_B T$ — proportional to the absolute temperature.
At $0$ K molecular KE $\to 0$; you cannot go below absolute zero.
§3 — Internal Energy Topic B.1
Definition: internal energy $U$ is the total random kinetic energy plus the total intermolecular potential energy of every molecule in the system.
Heating a substance raises $U$.
Doing mechanical work on a gas (compression) raises $U$.
TrapTemperature measures the average KE per molecule. Internal energy is the total energy of all molecules. A large cold lake can hold more $U$ than a hot cup of tea — this is a perennial multiple-choice trick.
NoteDuring a phase change temperature is constant (KE constant), so all the energy goes into PE. Internal energy still increases.
TrickIn a mixing problem set "energy lost by hot = energy gained by cold": $m_\text{hot} c_\text{hot}(T_\text{hot} - T_f) = m_\text{cold} c_\text{cold}(T_f - T_\text{cold})$. Solve for the unknown final temperature $T_f$.
TrapIf the calorimeter or container has significant mass you must include its heat capacity too — this is the easiest A1 to lose in a real-world experiment question.
NoteOn a heating curve a steeper slope means a smaller specific heat capacity — slope $= P/(mc)$ when constant power $P$ is supplied.
From the Photon question bank
What examiners actually test on this concept:
Q = mcΔTMethod of mixturesInclude the containerHeating-curve slope
Method of mixtures: energy lost by the hot body = energy gained by the cold; solve for $T_f$. frequent
Include the calorimeter's heat capacity in real-world problems — the easiest A1 to drop.
On a heating curve, a steeper slope means a smaller $c$ (slope $= P/mc$ for constant power).
For water: $L_f = 334\,\text{kJ kg}^{-1}$, $L_v = 2260\,\text{kJ kg}^{-1}$.
Phase-change vocabulary
Melting
solid → liquid
Freezing
liquid → solid
Boiling
liquid → gas (at boiling point)
Condensing
gas → liquid
Evaporation
liquid → gas (below boiling point)
Trick$L_v > L_f$ always — vaporisation has to break essentially all intermolecular bonds, while melting only partially disrupts the lattice.
TrapDuring a phase change the temperature is constant; KE does not change. Molecules do not speed up while ice melts or water boils.
NoteOn a cooling curve for water in a freezer: steep drop (water $c$), plateau at 0 °C (latent heat of fusion), then a gentler drop (ice has roughly half the $c$ of water).
Heating curve. Sloped sections warm one phase ($Q=mc\Delta T$); the flat plateaus are phase changes at constant temperature ($Q=mL$) — the boiling plateau is longer because $L_v > L_f$.
From the Photon question bank
What examiners actually test on this concept:
Q = mLL_v > L_fConstant T at phase changeHeating / cooling curves
During a phase change the temperature is constant — energy goes into PE, molecules don't speed up. very frequent
$L_v > L_f$ always: vaporisation breaks essentially all bonds; melting only disrupts the lattice.
A plateau length is $\propto mL$; use $Q = mL$ (no $\Delta T$) for that segment.
§6 — Conduction (Fourier's Law) Topic B.1
Key formula
Fourier:$\dfrac{\Delta Q}{\Delta t} = k A \dfrac{\Delta T}{\Delta x}$
$k$ — thermal conductivity (W m${}^{-1}$ K${}^{-1}$)
$A$ — cross-sectional area (m$^2$)
$\Delta T / \Delta x$ — temperature gradient (K m${}^{-1}$)
TrickHigh $k$ = good conductor. Metals have very high $k$ thanks to free electrons. Air has very low $k$ — that is why insulation works by trapping pockets of still air.
TrapConduction can occur in any state of matter — it is just most efficient in solids (especially metals). Don't write that conduction is impossible in liquids or gases.
NoteAt steady state the rate $\Delta Q/\Delta t$ is the same at every cross-section of a uniform bar — energy in equals energy out.
Conduction through a bar. Heat flows down the temperature gradient at a rate $\Delta Q/\Delta t = kA\,\Delta T/\Delta x$; a bigger gradient or higher conductivity $k$ means faster flow.
From the Photon question bank
What examiners actually test on this concept:
Fourier ΔQ/Δt = kAΔT/ΔxHigh k = conductorSteady stateOccurs in any state
High $k$ = good conductor (metals, via free electrons); still air has very low $k$, so insulation works by trapping it. frequent
At steady state the heat-flow rate is equal at every cross-section of a uniform bar.
Conduction occurs in all states — just most efficiently in solids; don't claim it's impossible in fluids.
§7 — Convection & Radiation Topic B.1
Convection
Bulk movement of fluid driven by density differences. Heated fluid expands, density drops, and it rises; cooler denser fluid sinks. Convection cannot occur in solids.
Radiation
EM waves (mainly infrared). No medium required. Every body above 0 K radiates. Dark, rough surfaces are good emitters and absorbers; shiny, smooth surfaces are poor radiators.
TrickThree mechanisms only: conduction (collisional KE transfer), convection (bulk fluid motion + density gradient), radiation (EM waves, vacuum-friendly).
TrapA thermos flask suppresses all three: double-walled vacuum (kills conduction and convection) plus silvered walls (low emissivity radiation). Quote all three in any "explain" question.
NoteEvaporation acts like a 4th cooling mechanism — the fastest molecules escape from the surface, lowering the average KE of those left behind. This happens at any temperature, unlike boiling.
A convection current. Heated fluid expands, becomes less dense and rises; it cools at the top, grows denser and sinks — a self-sustaining loop that needs a fluid.
From the Photon question bank
What examiners actually test on this concept:
Convection needs a fluidRadiation needs no mediumGood emitters/absorbersThermos flask
Convection = bulk fluid motion from density differences (impossible in solids); radiation = EM/IR and works in a vacuum. frequent
Dark, rough surfaces are good emitters and absorbers; shiny, smooth surfaces are poor radiators.
A thermos suppresses all three: vacuum gap (conduction + convection) + silvered walls (radiation) — quote all three.
Peak shift:$\lambda_\text{max} T = 2.9 \times 10^{-3}\,\text{m K}$
Apparent brightness (inverse square law)
Brightness:$b = \dfrac{L}{4\pi d^2}$
TrickDouble the temperature and luminosity rises by $2^4 = 16\times$ ($L \propto T^4$). Combined area + temperature changes scale as $L \propto r^2 T^4$.
TrapWien's law uses the peak wavelength — not the average. And the units are strict: $\lambda$ in metres, $T$ in kelvin.
NoteTwo stars with equal apparent brightness: the more distant one is intrinsically brighter (greater $L$). Two stars with the same colour ($\lambda_\text{max}$) have the same surface temperature.
Black-body spectra. As temperature rises the whole curve lifts ($L\propto T^4$) and the peak shifts to shorter wavelengths (Wien's law) — hotter bodies glow bluer.
Only radiation operates — discount conduction & convection
"Why doesn't temperature rise?"
Phase change — energy goes into PE, not KE
Worked Example — IB-Style Mixing with a Phase Change
Question (HL Paper 2 style — 6 marks)
An ice cube of mass $50\,\text{g}$ at $-10\,{}^\circ\text{C}$ is dropped into a well-insulated cup containing $250\,\text{g}$ of water at $30\,{}^\circ\text{C}$. Take $c_\text{ice} = 2100\,\text{J kg}^{-1}\,\text{K}^{-1}$, $c_\text{water} = 4200\,\text{J kg}^{-1}\,\text{K}^{-1}$, $L_f = 3.34 \times 10^{5}\,\text{J kg}^{-1}$. Find the final temperature of the mixture.
Solution
Energy released by water cooling from $30\,{}^\circ\text{C}$ to a final $T_f$: $Q_1 = 0.250 \times 4200 \times (30 - T_f)$ J. (M1)
Energy absorbed by ice in three stages — warm to 0 °C, melt, then warm liquid water to $T_f$:
$Q_2 = 0.050 \times 2100 \times 10 + 0.050 \times 3.34 \times 10^{5} + 0.050 \times 4200 \times (T_f - 0)$. (M1)(M1)
Insulated cup ⇒ apply energy conservation $Q_1 = Q_2$:
$1050(30 - T_f) = 1050 + 16700 + 210 T_f$. (R1)
Examiner's note: The two most common errors are (i) forgetting to warm the ice from $-10\,{}^\circ\text{C}$ to $0\,{}^\circ\text{C}$ before melting, and (ii) leaving out $Q = mL$ for the actual melting. Always split the absorbing side into three stages whenever the substance crosses a phase change.
Common Student Questions
Why is temperature constant during a phase change?
During melting or boiling all the energy supplied goes into breaking intermolecular bonds — this is potential energy, not kinetic energy. Since temperature is a measure of average KE per molecule, $T$ stays flat on the heating curve while the latent heat is being absorbed. Internal energy still rises, just through PE instead of KE.
Can I use Celsius in $Q = mc\Delta T$ and Stefan-Boltzmann?
For $Q = mc\Delta T$ either kelvin or Celsius works, because a temperature difference is identical in both scales. For Stefan-Boltzmann ($L = \sigma A T^4$) and Wien's law you must use kelvin — these involve absolute temperature raised to a power. Substitute Celsius into Stefan-Boltzmann and your answer can be wrong by orders of magnitude.
What's the difference between temperature and internal energy?
Temperature is the average random kinetic energy per molecule — intensive. Internal energy $U$ is the total random KE plus total intermolecular PE summed across every molecule — extensive. A swimming pool at $20\,{}^\circ\text{C}$ has vastly more internal energy than a cup of tea at $80\,{}^\circ\text{C}$, even though the tea is much hotter. IB multiple-choice questions love this distinction.
Why is the latent heat of vaporisation always larger than fusion?
Going liquid → gas requires breaking essentially every intermolecular bond and doing work pushing back the surrounding atmosphere as the gas expands. Melting only partially disrupts the lattice — molecules stay close together. For water $L_v$ is roughly $7\times$ larger than $L_f$. Quote both numbers ($L_f = 334\,\text{kJ kg}^{-1}$, $L_v = 2260\,\text{kJ kg}^{-1}$) when an "explain" question asks why.
How does a thermos flask work?
A thermos suppresses all three transfer mechanisms simultaneously. The double-walled vacuum blocks conduction and convection (no medium between the walls). The silvered walls have very low emissivity, so they emit and absorb almost no infrared. Only a tiny conduction path remains via the narrow neck and stopper, so contents stay hot or cold for hours. Mention all three mechanisms in a "describe" answer.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Thermal Energy Transfers library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — every concept explained with diagrams, derivations, and worked intuition.
Tutorial booklet — IB-style questions sequenced from foundation through full HL difficulty.
Tutorial Solutions — full mark-scheme-style solutions with R1/M1/A1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
