IB Physics · Theme B The particulate nature of matter
B.1 Thermal energy transfers
Summary to follow. 17 syllabus statements · 37 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 17 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Molecular (particle) model of matter
Molecular (particle) model of matter
The model in which all matter is made of very small particles (atoms or molecules) that are in continuous random motion and exert forces on one another. The forces are attractive at typical separations and become strongly repulsive if the particles are pushed very close together. The differences between solids, liquids and gases are explained by how strongly the particles are held, how far apart they are on average, and how they move.
Solid (molecular model)
Particles are closely packed, held by strong intermolecular forces in fixed mean positions, often in a regular arrangement, and vibrate about those positions. A solid therefore has a fixed shape and a fixed volume, and is very difficult to compress.
Liquid (molecular model)
Particles are still close together, with a mean separation similar to that in the solid, so a liquid is almost incompressible and its density is close to that of the solid. The intermolecular forces are still significant, but the particles are not held in fixed positions: they move randomly and can slide past one another. A liquid therefore has a fixed volume but takes the shape of its container.
Gas (molecular model)
Particles are far apart compared with their own size (typically about ten molecular diameters at room temperature and atmospheric pressure) and move randomly at high speed in straight lines between collisions. Intermolecular forces are negligible except during collisions. A gas has no fixed shape or volume, fills its container, is easily compressed, and has a density roughly a thousand times smaller than that of the liquid.
Students often think The particles of a solid are completely still; they start to move only when the solid melts. In fact Yes. They vibrate continuously about fixed mean positions; the vibration becomes more vigorous as the temperature rises.
Students often think The particles themselves expand, grow or get lighter when a substance is heated or turns into a gas. In fact No. The particles themselves stay the same size and mass; what changes is their average separation, their arrangement and their motion.
Density ρ
Density ρ
The mass per unit volume of a substance, ρ = m/V. SI unit: kilogram per cubic metre (kg m⁻³). Density is a property of the material (at a given temperature and pressure), not of the size or shape of a sample: cutting a block in half halves both m and V and leaves ρ unchanged. Conversion: 1 g cm⁻³ = 1000 kg m⁻³, because 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³.
Students often think Areas and volumes change in the same ratio as lengths (and lengths in the same ratio as volumes), so 1 m³ = 100 cm³ and doubling a radius doubles the surface area. In fact No. Areas scale as the square and volumes as the cube of the length factor: 1 m³ = (100 cm)³ = 10⁶ cm³, a sphere of twice the radius has four times the surface area, and a volume per molecule 1600 times larger means a separation only about ∛1600 ≈ 12 times larger.
Students often think Density can be found by dividing the volume by the mass. In fact Density is mass divided by volume, ρ = m/V, in kg m⁻³. Volume divided by mass is a different quantity (volume per kilogram, in m³ kg⁻¹).
Celsius scale
Celsius scale
A temperature scale, symbol θ and unit the degree Celsius (°C), on which the melting point of pure ice is 0 °C and the boiling point of pure water at standard atmospheric pressure is very close to 100 °C. Because its zero is not absolute zero, ratios of Celsius temperatures have no physical meaning and Celsius temperatures cannot be used in laws that need absolute temperature.
Kelvin (absolute) temperature scale
The SI temperature scale, symbol T and unit the kelvin (K), whose zero is absolute zero, the lowest possible temperature (−273.15 °C). No temperature below 0 K exists, so kelvin temperatures are never negative. One kelvin is the same size as one degree Celsius. The kelvin temperature must be used in E̅_k = (3/2)k_B T, in L = σAT⁴ and in λ_max T = 2.9 × 10⁻³ m K.
Conversion between Kelvin and Celsius
T/K = θ/°C + 273.15, usually taken as T/K = θ/°C + 273 in IB calculations. So 27 °C = 300 K and 77 K = −196 °C. To go from kelvin to Celsius, subtract 273; to go from Celsius to kelvin, add 273.
Students often think Converting between the scales always means adding 273, whichever way the conversion goes. In fact No. Celsius to kelvin: add 273. Kelvin to Celsius: subtract 273.
Students often think The kelvin is just another name for the degree Celsius, so a temperature has the same numerical value on both scales. In fact No. The two scales have the same size of degree but different zeros: 0 °C = 273 K. Only a temperature CHANGE has the same number on both scales.
Temperature change ΔT
Temperature change ΔT
The difference between two temperatures. Because the kelvin and the degree Celsius are the same size, a temperature change has the same numerical value on both scales: a rise from 20 °C to 35 °C is a rise of 15 °C and also of 15 K (from 293 K to 308 K). The 273 cancels when one temperature is subtracted from the other, so 273 is never added to a temperature change.
Students often think Every Celsius value, including a temperature change ΔT, must have 273 added to it before it is used in a physics equation. In fact No. A temperature change has the same value in K as in °C, because the 273 cancels when one temperature is subtracted from another: (θ₂ + 273) − (θ₁ + 273) = θ₂ − θ₁.
Students often think The kelvin and the degree Celsius are different-sized units, so a temperature change in °C is not equal to the same number of kelvin. In fact No. One kelvin is exactly the same size as one degree Celsius, so any temperature change has the same numerical value in K and in °C.
Boltzmann constant k_B
Boltzmann constant k_B
The constant that links the kelvin temperature of a system to the average energy of its particles: k_B = 1.38 × 10⁻²³ J K⁻¹. It equals the molar gas constant R divided by the Avogadro constant N_A.
Average kinetic energy of particles and temperature
The kelvin temperature is a measure of the average random kinetic energy of the particles of a substance: E̅_k = (3/2)k_B T. E̅_k is the translational kinetic energy per particle averaged over all the particles; it is directly proportional to the ABSOLUTE temperature, so doubling T (in K) doubles E̅_k. It does not depend on the amount of substance or on the mass of the particles, and at a given temperature every ideal-gas particle has the same average kinetic energy.
Students often think Any temperature scale can be used in a physics equation or ratio, so a Celsius temperature that doubles means the particle energy doubles. In fact No. These relationships need the absolute (kelvin) temperature, because their zero is absolute zero. Celsius values give wrong answers and meaningless ratios.
Students often think The average kinetic energy of a particle is simply k_B T, 'the thermal energy'. In fact No. E̅_k = (3/2)k_B T. The factor 3/2 arises because the translational motion has three independent directions, each contributing ½k_B T on average.
Internal energy
Internal energy
The total intermolecular potential energy arising from the forces between the molecules plus the total random kinetic energy of the molecules arising from their random motion. SI unit: joule (J). It does not include the kinetic energy of the body moving as a whole or its gravitational potential energy. It depends on the amount of substance, its temperature and its phase: at a phase change the internal energy changes while the temperature does not.
Intermolecular potential energy
The energy stored because of the forces between molecules; it depends on their separation and arrangement. Energy must be supplied to pull molecules away from one another against the attractive forces, so the intermolecular potential energy increases when a solid melts or a liquid boils, and decreases (energy is released) when a gas condenses or a liquid freezes.
Random kinetic energy of molecules
The kinetic energy of the molecules due to their random, disordered motion (vibration in a solid; translation, and in molecules also rotation and vibration, in a liquid or gas). Its average per molecule rises with temperature. It is distinct from the ordered kinetic energy of a body moving as a whole.
Students often think Internal energy is the total kinetic energy of the molecules, which is what the temperature measures. In fact No. Internal energy is the total random kinetic energy of the molecules PLUS the total intermolecular potential energy.
Students often think Internal energy is all the energy a body has, including the kinetic energy of its motion as a whole and its gravitational potential energy. In fact No. Internal energy concerns only the random motion of molecules and the forces between them. The kinetic energy of the body as a whole and its gravitational potential energy are separate, macroscopic stores.
Thermal energy transfer (heating), Q
Thermal energy transfer (heating), Q
Energy transferred from one body to another because of a temperature difference between them. SI unit: joule (J). The resultant transfer is always from the body at the higher temperature to the body at the lower temperature, whatever their internal energies, masses or sizes. "Heat" names this transfer, not something a body contains; bodies contain internal energy.
Thermal equilibrium
The state reached when two bodies in thermal contact are at the same temperature, so there is no resultant thermal energy transfer between them. Equal temperature, not equal internal energy, is the condition: a bath of water and a small coin in it are in equilibrium at the same temperature although the bath has far more internal energy.
Students often think Temperature and internal energy (or 'heat') are the same thing: a body with more energy is hotter, energy flows from the body with more energy, and equal energy inputs give equal temperatures. In fact No. Temperature measures the average random kinetic energy per particle; internal energy is a total that also depends on the amount of substance and its phase. A large, cool body can have far more internal energy than a small, hot one, and equal energy inputs to different materials give different temperatures.
Students often think Cold is a substance or form of energy that flows from cold objects into warm ones. In fact No. Only energy is transferred, and the resultant transfer is always from the higher temperature to the lower. A body cools because it loses energy, not because it gains 'cold'.
Phase change
Phase change
A change of state (solid, liquid, gas) in which energy is transferred to or from the substance at constant temperature. The energy changes the intermolecular potential energy as the arrangement and separation of the particles change; the average random kinetic energy, and hence the temperature, stays the same until the change is complete.
Melting and freezing
Melting is the change from solid to liquid at the melting point; energy is absorbed. Freezing (solidification) is the reverse change from liquid to solid at the same temperature; the same energy per kilogram is released. Both take place at constant temperature.
Boiling and condensing
Boiling is the change from liquid to gas that takes place throughout the liquid, with bubbles of vapour forming inside it, at the boiling point (which depends on the external pressure); energy is absorbed and the temperature stays constant while the liquid boils. Condensing is the change from gas (vapour) to liquid; energy is released.
Evaporation
The change from liquid to gas that takes place only at the surface of a liquid, at any temperature, not only at the boiling point. The molecules that escape are those near the surface with enough kinetic energy to overcome the attraction of their neighbours, so they are the faster ones; the average kinetic energy of the molecules left behind falls and the liquid cools unless energy is supplied to it.
Students often think Whenever energy is supplied to a substance its molecules speed up, including while it is melting or boiling. In fact No. At a phase change the temperature, and therefore the average random kinetic energy, is constant. The energy supplied increases the intermolecular potential energy.
Students often think If the temperature is not changing, no energy is going into the substance, so a phase change needs no energy of its own. In fact No. A substance absorbs energy throughout melting or boiling (Q = mL) even though its temperature is constant; this latent heat increases its internal energy.
Specific heat capacity c
Specific heat capacity c
The energy required to raise the temperature of 1 kg of a substance by 1 K without a change of phase. Q = mcΔT. SI unit: joule per kilogram per kelvin (J kg⁻¹ K⁻¹). A substance with a large c has a small temperature rise for a given energy input per kilogram; for water c ≈ 4.18 × 10³ J kg⁻¹ K⁻¹.
Specific latent heat L
The energy required to change the phase of 1 kg of a substance at constant temperature. Q = mL. SI unit: joule per kilogram (J kg⁻¹). The same energy per kilogram is released in the reverse change.
Specific latent heat of fusion L_f
The energy required to change 1 kg of a substance from solid to liquid (melting) at constant temperature; for ice at 0 °C, L_f = 3.34 × 10⁵ J kg⁻¹. "Fusion" here means melting.
Specific latent heat of vaporization L_v
The energy required to change 1 kg of a substance from liquid to gas at constant temperature; for water at 100 °C, L_v = 2.26 × 10⁶ J kg⁻¹, almost seven times L_f for ice, because the molecules must be separated almost completely.
Students often think A temperature and a temperature change are the same thing, so the final temperature can be used as ΔT and a calculated ΔT is the new temperature. In fact No. ΔT is the change in temperature, final minus initial. The final temperature is the initial temperature plus ΔT.
Students often think A substance with a large specific heat capacity 'takes in heat easily', so it heats up faster and gets hotter. In fact No. A larger c means MORE energy is needed per kilogram per kelvin, so for the same energy input and mass the temperature rise is SMALLER.
Mechanisms of thermal energy transfer
Mechanisms of thermal energy transfer
Thermal energy is transferred by three primary mechanisms: conduction (energy passed between neighbouring particles, and by free electrons in metals, without bulk movement of the material), convection (bulk movement of a fluid caused by density differences) and thermal radiation (emission of electromagnetic waves from a surface). Conduction and convection need a material medium; thermal radiation does not and can cross a vacuum.
Students often think Heat naturally rises, so thermal energy moves upwards and carries warm air or water up with it. In fact No. Hot fluid rises, because it is less dense than the cooler fluid around it. Energy itself can be transferred in any direction: conduction and radiation go downwards and sideways as easily as upwards.
Students often think Convection can take place in solids, for example when a metal rod heated at one end gets hot at the other end. In fact No. Convection needs bulk movement of the material, which can happen only in fluids (liquids and gases). Energy passes through solids by conduction.
Conduction
Conduction
Transfer of thermal energy through a material from a region of higher temperature to one of lower temperature without bulk movement of the material. Particles in the hotter region have a greater average kinetic energy; through collisions and the forces between neighbouring particles, they pass kinetic energy to neighbours with less, so energy spreads along the temperature gradient. In metals, free electrons also carry energy, which is why metals are good conductors.
Students often think In conduction the hot particles move along the rod from the hot end to the cold end, carrying their energy with them. In fact No. In a solid the particles vibrate about fixed positions. Energy is passed from particle to particle through collisions and the forces between neighbours; the particles themselves stay where they are.
Thermal conductivity k
Thermal conductivity k
A property of a material that measures how well it conducts thermal energy: the rate of energy transfer per unit cross-sectional area per unit temperature gradient. SI unit: watt per metre per kelvin (W m⁻¹ K⁻¹). Metals have large k (copper about 390 W m⁻¹ K⁻¹); insulators such as wood, glass wool and still air have small k.
Temperature gradient ΔT/Δx
The change in temperature per unit distance through a material, measured along the direction of energy flow. SI unit: kelvin per metre (K m⁻¹). For a uniform, lagged bar in steady state it equals the temperature difference between its ends divided by its length.
Rate of thermal energy transfer by conduction
ΔQ/Δt = −kA(ΔT/Δx), where k is the thermal conductivity of the material, A the cross-sectional area through which energy flows and ΔT/Δx the temperature gradient. SI unit of ΔQ/Δt: watt (W). The minus sign shows that energy flows in the direction of decreasing temperature: where T falls along +x, ΔT/Δx is negative and ΔQ/Δt is positive, meaning energy flows in the +x direction. The rate is proportional to k, to A and to the temperature gradient, so it doubles if the thickness of a slab is halved.
Students often think The diameter can be put straight into A = πr² as if it were the radius. In fact No. r = d/2. Using the diameter as the radius makes the area four times too large.
Students often think The minus sign shows that something is decreasing, such as the rate of transfer falling with time or along the conductor. In fact No. The minus sign fixes the direction of the energy flow: towards lower temperature. Where the temperature falls along +x, ΔT/Δx is negative and ΔQ/Δt is positive, so energy flows in the +x direction.
Convection
Convection
Transfer of thermal energy in a fluid (liquid or gas) by bulk movement of the fluid itself. A heated region of fluid expands, so its density falls; the denser, cooler fluid around it sinks below it and pushes it upwards (buoyancy). Convection cannot occur in solids, and it cannot occur in a vacuum.
Convection current
The continuous circulation set up in a fluid heated from below (or cooled from above): warm, less dense fluid rises, cools, becomes denser and sinks, and is replaced by cooler fluid moving in to be heated. A fluid heated only from the top forms no convection current, because the warm, less dense fluid is already above the cool fluid.
Thermal radiation
Thermal radiation
Energy transferred by electromagnetic waves emitted from the surface of a body because of its temperature. Every body above absolute zero emits it, whatever its temperature; at everyday temperatures it is mainly infrared. It needs no medium and travels through a vacuum at the speed of light.
Black body
An idealized body that absorbs all the electromagnetic radiation falling on it and is also the best possible emitter at any temperature. Its emission depends only on its temperature and surface area, which is why stars are modelled as black bodies.
Luminosity L
The total power emitted as electromagnetic radiation by a body over its whole surface. SI unit: watt (W). For a star it can also be expressed as a multiple of the luminosity of the Sun, L⊙ (L⊙ ≈ 3.8 × 10²⁶ W).
Stefan–Boltzmann law
The luminosity of a black body is L = σAT⁴, where A is its whole surface area (4πr² for a sphere), T its absolute (kelvin) temperature and σ the Stefan–Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Because L ∝ T⁴, doubling the kelvin temperature multiplies the power by 16.
Students often think The area of a sphere or star is πr², the area of the circle it appears as. In fact The whole surface area of a sphere, 4πr², because radiation leaves the whole surface and spreads over a whole sphere. πr² is only the area of the flat disc a sphere presents when viewed from one direction.
Students often think Luminosity (and brightness) depends only on temperature, so the hotter star is always the more luminous and the brighter. In fact No. L = σAT⁴ depends on surface area as well as temperature, and apparent brightness also depends on distance. A cool red giant can be far more luminous than a hot white dwarf.
Apparent brightness b
Apparent brightness b
The power of the radiation received from a source per unit area perpendicular to the direction of the radiation, at the observer. SI unit: watt per square metre (W m⁻²). It depends on both the luminosity of the source and its distance, so a star can look bright because it is luminous or because it is near.
Students often think Apparent brightness and luminosity are the same thing, so equally bright stars emit equal power and a more luminous star always looks brighter. In fact No. Luminosity L is the total power emitted by the star (W). Apparent brightness b is the power received per square metre at the observer (W m⁻²), which also depends on distance.
Students often think How bright a star looks depends only on how far away it is: nearer stars always look brighter. In fact No. It depends on both luminosity and distance, b = L/4πd².
Apparent brightness–luminosity relationship
Apparent brightness–luminosity relationship
b = L/4πd². The power L emitted by a point source spreads uniformly over the surface of a sphere of radius d, area 4πd², so the brightness falls as the inverse square of the distance: doubling d reduces b to one quarter. Rearranged, L = 4πd²b gives the luminosity of a star from its measured brightness and distance.
Solar luminosity L⊙
The luminosity of the Sun, about 3.8 × 10²⁶ W, used as a unit for stellar luminosities: a star of luminosity 40 L⊙ emits 40 times the power the Sun does. It must be converted to watts before b = L/4πd² is used to give b in W m⁻².
Students often think Apparent brightness is inversely proportional to distance, so a star twice as far away looks half as bright. In fact No. It falls as the inverse SQUARE of distance: at twice the distance the same power is spread over four times the area, so b is one quarter.
Students often think The inverse-square law for brightness is b = L/d², with the power spread over an area d². In fact No. b = L/4πd². The power spreads over the surface of a sphere of radius d, whose area is 4πd², not d².
Black-body spectrum
Black-body spectrum
The graph of emitted intensity against wavelength for a black body. It is continuous (all wavelengths are emitted), rises from zero at short wavelengths to a single peak and falls away more gradually at long wavelengths. For a hotter body the whole curve is higher at every wavelength, the area under it (the total power) is greater and the peak is at a shorter wavelength.
Wien's displacement law
The wavelength at which a black body emits most intensely, λ_max, is inversely proportional to its absolute temperature: λ_max T = 2.9 × 10⁻³ m K. Measuring λ_max gives the surface temperature: a star with peak at 500 nm has T ≈ 5800 K, while a body at room temperature (300 K) peaks at about 10 μm, in the infrared. Hotter stars look blue-white, cooler stars look red.
Students often think Hotter objects emit at longer wavelengths, so red light shows the highest temperatures ('red-hot'); λ_max increases with T. In fact No. λ_max is inversely proportional to T. Hotter bodies peak at SHORTER wavelengths: blue-white stars are hotter than red stars.
Students often think Wien's law and the Stefan–Boltzmann law can be mixed: the peak wavelength depends on T⁴. In fact No. Wien's law is λ_max T = 2.9 × 10⁻³ m K, with T to the first power. The fourth power belongs to the Stefan–Boltzmann law, L = σAT⁴.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A pure substance is heated steadily from a solid until it has completely boiled. According to the molecular model, which statement about its molecules is correct?
Answer and reasoning
Their mean separation increases greatly on melting and increases again on boiling. — A student who pictures liquid molecules as widely spaced picks this. Liquids are almost incompressible and have densities close to their solids, so the separation hardly changes on melting; the large increase happens on boiling.
Their mean separation changes little on melting but increases greatly on boiling. — In a solid and a liquid the molecules are close together, which is why their densities are similar; melting frees them from fixed positions without separating them much. On boiling they move far apart: the gas is roughly a thousand times less dense than the liquid.
The molecules themselves get larger on melting, and larger again on boiling. — A student who gives particles the properties of the bulk material picks this. The molecules stay the same size throughout; what changes is how far apart they are, how they are arranged and how they move.
The molecules start to move only on melting, and they move faster still on boiling. — A student who thinks solid particles are motionless picks this. Molecules in a solid already vibrate about fixed positions; melting lets them slide past one another. And during boiling itself the temperature, and so the average kinetic energy, is constant.
2 The density of aluminium is 2.7 g cm⁻³. What is the volume, in m³, of an aluminium block of mass 0.081 kg?
Answer and reasoning
3.0 × 10⁻¹ m³ — A student who finds 30 cm³ correctly and then divides by 100 to convert to m³ gets 0.30 m³, a block the size of a large cupboard. A cubic metre is (100 cm)³ = 10⁶ cm³, so 30 cm³ = 3.0 × 10⁻⁵ m³.
3.0 × 10⁻² m³ — A student who divides the mass in kilograms by the density in grams per cubic centimetre gets 0.081/2.7 = 0.030 and calls it m³. The units must match: use 81 g with 2.7 g cm⁻³, or 0.081 kg with 2700 kg m⁻³.
3.0 × 10⁻⁵ m³ — ρ = m/V, so V = m/ρ = 81 g/2.7 g cm⁻³ = 30 cm³. Since 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³, V = 30 × 10⁻⁶ m³ = 3.0 × 10⁻⁵ m³. (In SI: V = 0.081 kg/2700 kg m⁻³ gives the same result.)
2.2 × 10⁻⁴ m³ — A student who treats density as volume divided by mass multiplies instead of dividing: 81 × 2.7 = 219 cm³ = 2.2 × 10⁻⁴ m³. Density is mass per unit volume, so V = m/ρ.
Working ρ = m/V ⇒ V = m/ρ. In SI: ρ = 2.7 g cm⁻³ = 2.7 × 10³ kg m⁻³; V = 0.081 kg/(2.7 × 10³ kg m⁻³) = 3.0 × 10⁻⁵ m³ (= 30 cm³).
3 Sample P is warmed from 10 °C to 30 °C. Sample Q is warmed from 290 K to 310 K. Which statement about the two temperature changes is correct?
Answer and reasoning
P's rise of 20 °C is equivalent to a rise of 293 K. — A student who adds 273 to every Celsius value picks this. The 273 cancels in a difference: (30 + 273) − (10 + 273) = 20 K. Only a temperature, never a change, has 273 added.
The changes differ, since kelvins and °C differ in size. — A student who assumes the two scales have different-sized degrees, as the Celsius and Fahrenheit scales do, picks this. The kelvin was defined to be the same size as the degree Celsius, so a change of 20 °C is a change of 20 K, the same as Q's.
Both undergo the same temperature change, 20 K. — P: ΔT = 30 − 10 = 20 °C, which is 20 K because the kelvin and the degree Celsius are the same size. Q: ΔT = 310 − 290 = 20 K. The changes are equal.
P's rise is larger, since P's temperature tripled. — A student who treats a ratio of Celsius temperatures as meaningful picks this. On the absolute scale P goes from 283 K to 303 K, a factor of only 1.07; the change, 20 K, is the same as Q's.
Working P: ΔT = 30 °C − 10 °C = 20 °C = 20 K. Q: ΔT = 310 K − 290 K = 20 K. Equal changes. (P in kelvin: 283 K to 303 K, ratio 1.07, not 3.)
4 Which of the following is the internal energy of a system?
Answer and reasoning
The total random kinetic energy of its molecules plus their total intermolecular potential energy — This is the guide's definition: the potential energy arising from the forces between molecules plus the kinetic energy of their random motion.
The total random kinetic energy of its molecules, which is the quantity its temperature measures — A student who leaves out the potential-energy part picks this. Internal energy also includes the intermolecular potential energy, which is what changes during melting and boiling. (Temperature measures the AVERAGE kinetic energy, not the total.)
The kinetic energy of the body's motion plus its gravitational potential energy and its molecules' energy — A student who reads 'internal' as 'all the energy the body has' picks this. The kinetic energy of the body as a whole and its gravitational potential energy are macroscopic stores, not part of internal energy.
The quantity of heat that is stored in the system, which it gives out again when it cools down — A student who thinks of heat as a substance held in a body picks this. Heat is energy transferred because of a temperature difference; what a body contains is internal energy, defined in terms of molecular kinetic and potential energy.
5 A 0.20 kg copper block at 80 °C is lowered into a large tank containing 200 kg of water at 20 °C. The water has far more internal energy than the block. Which statement describes the resultant thermal energy transfer?
Answer and reasoning
From the block to the water, until they reach the same temperature. — The direction of the resultant transfer is set by temperature difference alone: from the hotter block to the cooler water. It stops at thermal equilibrium, when the temperatures are equal, just above 20 °C here because the tank is so large.
From the water to the block, since the water has more internal energy. — A student who treats internal energy as if it were temperature picks this. The water's large internal energy comes from its large mass; each water molecule has less average kinetic energy than each copper particle, and energy flows from high to low temperature.
Coldness passes from the water into the block, so the block cools down. — A student who thinks of cold as something that flows picks this. Only energy is transferred: the block cools because it loses energy to the water, and nothing called 'cold' enters it.
From the block to the water, until they have equal internal energies. — A student who thinks energy levels out like water between tanks picks this. The direction is right, but transfer stops when the TEMPERATURES are equal; the 200 kg of water still has vastly more internal energy than the block.
6 An electric kettle heats 250 g of water from 18 °C to 90 °C. The specific heat capacity of water is 4.18 × 10³ J kg⁻¹ K⁻¹. How much energy is transferred to the water? Ignore energy transferred to the kettle and surroundings.
Answer and reasoning
7.5 × 10⁴ J — Convert the mass first: m = 250 g = 0.250 kg. Q = mcΔT = 0.250 × 4.18 × 10³ × (90 − 18) = 0.250 × 4180 × 72 = 7.5 × 10⁴ J. A temperature change of 72 °C is 72 K.
3.6 × 10⁵ J — A student who adds 273 to the temperature change uses ΔT = 345 K and gets 0.250 × 4180 × 345 = 3.6 × 10⁵ J. A change of 72 °C is a change of 72 K; the 273 cancels in the subtraction.
7.5 × 10⁷ J — A student who substitutes the mass in grams, as given, uses m = 250 and gets 250 × 4180 × 72 = 7.5 × 10⁷ J. Specific heat capacity is per kilogram, so convert first.
9.4 × 10⁴ J — A student who uses the final temperature, 90 °C, in place of the change gets 0.250 × 4180 × 90 = 9.4 × 10⁴ J. ΔT is the change, 90 − 18 = 72 K.
Working m = 250 g = 0.250 kg; ΔT = 90 − 18 = 72 K. Q = mcΔT = (0.250 kg)(4.18 × 10³ J kg⁻¹ K⁻¹)(72 K) = 0.250 × 4180 × 72 = 75 240 J ≈ 7.5 × 10⁴ J.
7 One end of a glass rod is held in a flame. How does the molecular model explain conduction of energy along the rod?
Answer and reasoning
Hot particles travel along the rod from the hot end to the cold end, carrying their energy with them. — A student who merges conduction with convection picks this. In a solid the particles only vibrate about fixed positions; energy is passed from particle to particle, but the particles do not travel.
Faster-vibrating particles at the hot end pass kinetic energy to their slower neighbours, and so on along the rod. — Particles at the hot end have a greater average kinetic energy. Through collisions and the forces between neighbours they pass energy to particles with less, and the energy spreads towards the cold end while the particles stay in place.
Heat flows along the rod like a fluid, filling the spaces between the particles as it goes. — A student who thinks of heat as a substance picks this. Nothing material flows: energy is passed on through the interactions of particles with different kinetic energies.
Particles at the hot end expand and push against their neighbours, passing the heat along. — A student who thinks particles swell when heated picks this. Particles do not change size; those at the hot end vibrate with more kinetic energy and pass energy on to their neighbours.
8 The rate of thermal energy transfer by conduction is given by ΔQ/Δt = −kA(ΔT/Δx). What does the minus sign show?
Answer and reasoning
The rate of energy transfer falls steadily with time and along the length of the conductor. — A student who reads every minus sign as 'decreasing' picks this. In steady state the rate is constant in time and the same through every cross-section; the sign shows the direction of flow.
Cold is conducted in the opposite direction to heat, from the cold end to the hot end. — A student who thinks cold is something that flows picks this. Only energy is transferred, and it flows from higher to lower temperature; nothing flows the other way.
Energy flows towards lower temperature: where T falls along +x, the net flow is along +x. — Where the temperature decreases along +x, ΔT/Δx is negative, so −kA(ΔT/Δx) is positive: energy flows in the +x direction, from hot to cold. The minus sign encodes the direction of flow.
The conductor is losing energy, so ΔQ/Δt is a negative quantity for every conducting material. — A student who links minus signs with losses picks this. In steady state a conductor gains energy at its hot face as fast as it loses it at its cold face, and ΔQ/Δt is positive in the direction of decreasing temperature.
9 A small metal sphere of radius 5.0 cm is heated to 527 °C. Treat it as a black body and take σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. What power does the sphere emit as thermal radiation?
Answer and reasoning
1.4 × 10² W — A student who uses the Celsius temperature, 527, in T⁴ gets 1.4 × 10² W. The Stefan–Boltzmann law needs the absolute temperature, 800 K; with a fourth power the error is large.
1.8 × 10² W — A student who uses πr², the area of the disc the sphere appears as, gets a quarter of the true value, 1.8 × 10² W. Radiation leaves the whole surface, area 4πr².
7.3 × 10² W — T = 527 + 273 = 800 K; A = 4πr² = 4π(0.050)² = 3.14 × 10⁻² m². L = σAT⁴ = 5.67 × 10⁻⁸ × 3.14 × 10⁻² × 800⁴ = 7.3 × 10² W.
7.3 × 10⁶ W — A student who substitutes r = 5.0 (centimetres) gets an area 10⁴ times too large and 7.3 × 10⁶ W. Convert to r = 0.050 m before using σ in W m⁻² K⁻⁴.
Working T = 527 + 273 = 800 K. A = 4πr² = 4π(0.050 m)² = 3.14 × 10⁻² m². L = σAT⁴ = (5.67 × 10⁻⁸)(3.14 × 10⁻²)(800)⁴ = 5.67 × 10⁻⁸ × 3.14 × 10⁻² × 4.10 × 10¹¹ = 730 W ≈ 7.3 × 10² W.
10 A star has a luminosity of 40 L⊙ and is 25 ly from the Earth. Take L⊙ = 3.83 × 10²⁶ W and 1 ly = 9.46 × 10¹⁵ m. What is the apparent brightness of the star at the Earth?
Answer and reasoning
8.7 × 10⁻⁸ W m⁻² — A student who divides by πd² instead of 4πd² gets a value four times too large, 8.7 × 10⁻⁸ W m⁻². The power spreads over the whole sphere of radius d, area 4πd².
2.2 × 10⁻⁸ W m⁻² — L = 40 × 3.83 × 10²⁶ = 1.53 × 10²⁸ W; d = 25 × 9.46 × 10¹⁵ = 2.37 × 10¹⁷ m. b = L/4πd² = 1.53 × 10²⁸/(4π × (2.37 × 10¹⁷)²) = 2.2 × 10⁻⁸ W m⁻².
2.7 × 10⁻⁷ W m⁻² — A student who uses b = L/d² gets 2.7 × 10⁻⁷ W m⁻², about 12.6 times too large. The area over which the power spreads is 4πd².
2.0 × 10²⁴ W m⁻² — A student who substitutes d = 25 (light-years) gets 2.0 × 10²⁴ W m⁻². The distance must be in metres: 25 ly = 2.37 × 10¹⁷ m.
Working L = 40 L⊙ = 40 × 3.83 × 10²⁶ W = 1.532 × 10²⁸ W. d = 25 × 9.46 × 10¹⁵ m = 2.365 × 10¹⁷ m. b = L/(4πd²) = 1.532 × 10²⁸/(4π × 5.593 × 10³⁴) = 1.532 × 10²⁸/7.029 × 10³⁵ = 2.2 × 10⁻⁸ W m⁻².
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27 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 At 100 °C and atmospheric pressure, liquid water has a density of 958 kg m⁻³ and steam has a density of 0.598 kg m⁻³. What do these data show about the molecules in steam compared with those in the liquid water?
Answer and reasoning
Their mean separation is about 12 times larger, as the volume per molecule is 1600 times larger. — Mass is conserved and the molecules are identical, so the density ratio 958/0.598 ≈ 1600 is the ratio of the volume available per molecule. Separation is a length, so it scales as the cube root of volume: ∛1600 ≈ 12.
Their mean separation is about 1600 times larger, because the density is 1600 times smaller. — A student who scales a length in the same ratio as a volume picks this. The volume per molecule is 1600 times larger, but separation is a length: it grows by ∛1600 ≈ 12.
Each molecule is about 1600 times larger, so each one takes up 1600 times the space. — A student who thinks molecules swell when a liquid becomes a gas picks this. A steam molecule is identical to a liquid-water molecule; each has more space around it because the molecules are further apart.
Each molecule has about 1/1600 of the mass, because a gas is much lighter than a liquid. — A student who thinks gases are nearly weightless picks this. Boiling 1 kg of water gives 1 kg of steam, so each molecule keeps its mass; the density is low because the same mass is spread through a larger volume.
Working Density ratio = 958/0.598 = 1602 ≈ 1600. For the same mass (same number of identical molecules), the volume per molecule is ≈ 1600 times larger in steam. Mean separation ∝ (volume per molecule)^(1/3), so the ratio of separations = 1602^(1/3) = 11.7 ≈ 12.
2 Liquefied natural gas is stored at 112 K, the boiling point of methane at atmospheric pressure. What is this temperature on the Celsius scale?
Answer and reasoning
+385 °C — A student who always adds 273 gets 112 + 273 = 385. Adding 273 converts Celsius to kelvin; to go from kelvin to Celsius, 273 is subtracted.
+112 °C — A student who thinks the two scales give the same number keeps 112. The kelvin and the degree Celsius are the same size, but the scales have different zeros: 0 °C = 273 K.
+161 °C — A student who computes 273 − 112 to avoid a negative number gets +161. The conversion is θ = T − 273, and a temperature well below that of melting ice must be negative in °C.
−161 °C — θ/°C = T/K − 273 = 112 − 273 = −161 °C. A temperature below 273 K is below the melting point of ice, so it must be negative on the Celsius scale.
Working θ/°C = T/K − 273 = 112 − 273 = −161, so θ = −161 °C.
3 The average kinetic energy of the molecules of a gas is 8.28 × 10⁻²¹ J. Take k_B = 1.38 × 10⁻²³ J K⁻¹. What is the temperature of the gas in degrees Celsius?
Answer and reasoning
400 °C — A student who finds T = 400 K correctly but thinks kelvin and Celsius values are the same stops here. 400 K is 400 − 273 = 127 °C.
127 °C — T = 2E̅_k/(3k_B) = 2 × 8.28 × 10⁻²¹/(3 × 1.38 × 10⁻²³) = 400 K, and θ = 400 − 273 = 127 °C.
673 °C — A student who finds 400 K and then adds 273 gets 673. Adding 273 converts Celsius to kelvin; converting kelvin to Celsius needs 273 subtracted.
327 °C — A student who takes E̅_k = k_B T gets T = 8.28 × 10⁻²¹/1.38 × 10⁻²³ = 600 K = 327 °C. The relationship is E̅_k = (3/2)k_B T, so T is two-thirds of that, 400 K.
Working E̅_k = (3/2)k_B T ⇒ T = 2E̅_k/(3k_B) = (2 × 8.28 × 10⁻²¹ J)/(3 × 1.38 × 10⁻²³ J K⁻¹) = 400 K. θ = 400 − 273 = 127 °C.
4 A sealed container of gas is heated from 20 °C to 40 °C. A student claims that the average kinetic energy of the gas molecules has doubled. Which evaluation of the claim is correct?
Answer and reasoning
Correct: E̅_k is proportional to temperature, and the temperature has doubled. — A student who uses the Celsius values in a proportional relationship picks this. E̅_k is proportional to the ABSOLUTE temperature, which rises only from 293 K to 313 K.
Incorrect: E̅_k cannot be judged unless the amount of gas is known. — A student who confuses the average energy per molecule with the total energy of the gas picks this. E̅_k = (3/2)k_B T depends only on T; the amount of gas affects the internal energy, not the average per molecule.
Incorrect: some energy goes into expanding the molecules, so E̅_k less than doubles. — A student who thinks heated molecules swell picks this. The molecules stay the same size; the energy supplied increases their average random kinetic energy, and E̅_k ∝ T in kelvin, so going from 293 K to 313 K raises E̅_k by only about 7%, for the reason in the correct option, not because molecules expand.
Incorrect: T rises from 293 K to 313 K, so E̅_k increases by only about 7%. — E̅_k = (3/2)k_B T is proportional to the kelvin temperature. 313/293 = 1.07, so the average kinetic energy rises by about 7%. The Celsius values double only because the Celsius zero is arbitrary.
Working T₁ = 20 + 273 = 293 K; T₂ = 40 + 273 = 313 K. E̅_k ∝ T, so E̅_k₂/E̅_k₁ = 313/293 = 1.07, an increase of about 7%, not a doubling.
5 A block of ice at 0 °C melts completely to give water at 0 °C. What happens to the internal energy of the H₂O, and why?
Answer and reasoning
It is unchanged: the temperature is still 0 °C, and internal energy depends on temperature alone. — A student who applies the ideal-gas result (internal energy depends only on T) to a phase change picks this. Energy is absorbed during melting at constant temperature and goes into intermolecular potential energy, so internal energy increases.
It is unchanged: the energy that is absorbed is balanced by the energy released as bonds break. — A student who believes breaking bonds releases energy picks this. Separating attracting molecules requires energy; no energy is released on melting, and the internal energy increases by the latent heat absorbed.
It increases: intermolecular potential energy rises; the average kinetic energy is unchanged. — The latent heat absorbed breaks the ordered structure of the ice, increasing the intermolecular potential energy. The temperature is constant, so the average random kinetic energy is unchanged; the internal energy (KE + PE) increases by mL_f.
It increases: the molecules move faster in the liquid than in the solid at the same temperature. — A student who links every energy input to faster molecules picks this. At the same temperature the average kinetic energy is the same in ice and water; the increase is entirely in the intermolecular potential energy.
6 Which statement correctly distinguishes evaporation from boiling for water at atmospheric pressure?
Answer and reasoning
Evaporation occurs only at the boiling point, and only at the surface; boiling occurs throughout the whole liquid. — A student who links all vapour formation to the boiling point picks this. Puddles and washing dry at everyday temperatures: evaporation happens at any temperature, at the surface.
Evaporation occurs only at the surface and at any temperature; boiling occurs throughout, at the boiling point. — Surface molecules with enough kinetic energy escape at any temperature, which is evaporation. Boiling happens once the liquid reaches its boiling point (100 °C at atmospheric pressure), with vapour forming in bubbles throughout the liquid.
Boiling produces bubbles of air or of hydrogen and oxygen; evaporation produces water vapour. — A student who thinks the bubbles in boiling water are air or its elements picks this. Boiling is a phase change: the bubbles are water vapour formed inside the liquid.
Boiling occurs throughout while the temperature keeps rising; evaporation occurs at the surface. — A student who expects continued heating always to raise the temperature picks this. While water boils at constant pressure its temperature stays at the boiling point; the energy supplied is latent heat.
7 A solid is heated by a heater of constant power. Its temperature rises steadily to 80 °C, stays at 80 °C for 5 minutes while the solid melts, and then rises again. Which statement about the 5 minutes at 80 °C is correct?
Answer and reasoning
Energy is absorbed and makes the molecules move faster, although the temperature reading does not change. — A student who assumes any energy input speeds molecules up picks this. The temperature is a direct measure of the average kinetic energy; a constant temperature means a constant average kinetic energy.
No energy is absorbed by the substance, because its temperature is not changing during this time. — A student who equates energy transfer with temperature change picks this. The heater supplies energy at the same rate throughout, and melting needs energy, Q = mL_f; this latent heat is absorbed at constant temperature.
Energy is released as bonds between molecules break, so the intermolecular potential energy decreases. — A student who believes breaking bonds releases energy picks this. Pulling attracting molecules out of their fixed arrangement requires energy, so the potential energy increases during melting.
Energy is absorbed and increases the intermolecular potential energy; the average kinetic energy is constant. — The heater still supplies energy and the substance absorbs it as latent heat of fusion. Because the temperature is constant, the average random kinetic energy is constant; the energy goes into intermolecular potential energy as the ordered solid structure breaks down.
8 0.40 kg of ice at 0 °C is melted and the water is then warmed to 20 °C. Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹; specific latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹; specific heat capacity of water = 4.18 × 10³ J kg⁻¹ K⁻¹. What is the total energy that must be supplied?
Answer and reasoning
3.3 × 10⁴ J — A student who thinks melting at constant temperature needs no energy calculates only the warming, 0.40 × 4180 × 20 = 3.3 × 10⁴ J. Melting absorbs latent heat, here 1.3 × 10⁵ J, four times the energy needed for the warming.
9.4 × 10⁵ J — A student who uses the latent heat of vaporization for the melting gets 0.40 × 2.26 × 10⁶ + 3.34 × 10⁴ = 9.4 × 10⁵ J. Melting is fusion; L_v applies to boiling, which does not happen here.
6.2 × 10⁵ J — A student who adds 273 to the temperature change uses ΔT = 293 K for the warming and gets 1.336 × 10⁵ + 0.40 × 4180 × 293 = 6.2 × 10⁵ J. The rise from 0 °C to 20 °C is 20 K.
9 A 0.50 kg block of aluminium and a 0.50 kg block of copper, both at 20 °C, are each supplied with 9.0 kJ of energy. Specific heat capacities: aluminium 900 J kg⁻¹ K⁻¹, copper 390 J kg⁻¹ K⁻¹. Assume no energy is lost. Which statement about their final temperatures is correct?
Answer and reasoning
The aluminium ends at about 66 °C, the copper at 40 °C. — A student who thinks a large specific heat capacity makes a material heat up faster picks this. ΔT = Q/(mc), so the larger c of aluminium gives the SMALLER temperature rise.
They end at the same temperature, as each gets 9.0 kJ. — A student who equates energy supplied with temperature picks this. Equal energy gives equal temperatures only if mc is equal; copper needs less energy per kelvin, so it gets hotter.
The copper ends at about 66 °C and the aluminium at 40 °C. — ΔT = Q/(mc). Aluminium: 9000/(0.50 × 900) = 20 K, so 20 + 20 = 40 °C. Copper: 9000/(0.50 × 390) = 46 K, so 20 + 46 = 66 °C. The smaller c gives the larger rise.
The copper ends at about 46 °C and the aluminium at only 20 °C. — A student who takes the calculated temperature change as the final temperature picks this. The rises are 46 K and 20 K; they must be added to the starting temperature of 20 °C.
Working ΔT = Q/(mc). Aluminium: 9000 J/(0.50 kg × 900 J kg⁻¹ K⁻¹) = 20 K, final 20 + 20 = 40 °C. Copper: 9000/(0.50 × 390) = 46.2 K, final 20 + 46 = 66 °C.
10 Which statement about the mechanisms of thermal energy transfer is correct?
Answer and reasoning
All three are directed upwards, because heat naturally rises from warm places. — A student who thinks heat itself rises picks this. It is hot FLUID that rises in convection; conduction and radiation carry energy in any direction, including downwards.
Convection can occur in solids as well as in liquids and gases, just like conduction. — A student who treats convection as any 'carrying of heat' picks this. Convection needs bulk movement of the material, which only fluids can undergo; in solids energy is transferred by conduction.
Thermal radiation is emitted only by very hot objects, such as the Sun or a fire. — A student who knows only glowing sources radiate noticeably picks this. Every body above absolute zero emits thermal radiation; cool bodies emit less power and mainly in the infrared.
Only thermal radiation can cross a vacuum; conduction and convection both need a medium. — Conduction passes energy between particles and convection moves the fluid itself, so both need matter. Thermal radiation is electromagnetic waves, which cross empty space; this is how energy reaches the Earth from the Sun.
11 A copper rod of length 0.50 m and diameter 2.0 cm is lagged along its sides. One end is kept at 100 °C and the other at 20 °C. The thermal conductivity of copper is 390 W m⁻¹ K⁻¹. What is the rate of thermal energy transfer along the rod in steady state?
Answer and reasoning
2.0 × 10¹ W — r = 1.0 cm = 0.010 m, so A = π(0.010)² = 3.14 × 10⁻⁴ m². |ΔT/Δx| = 80 K/0.50 m = 160 K m⁻¹. Rate = kA|ΔT/Δx| = 390 × 3.14 × 10⁻⁴ × 160 = 19.6 W ≈ 2.0 × 10¹ W, from the hot end to the cold end.
7.8 × 10¹ W — A student who uses the diameter, 0.020 m, as the radius gets an area of 1.26 × 10⁻³ m² and a rate of 78 W, four times too large. r = d/2 = 0.010 m.
8.7 × 10¹ W — A student who adds 273 to the temperature difference uses 353 K and gets 390 × 3.14 × 10⁻⁴ × 353/0.50 = 87 W. The difference of 80 °C is a difference of 80 K.
2.0 × 10⁵ W — A student who uses the radius as 1.0 (centimetres) in an equation with k in W m⁻¹ K⁻¹ gets an area 10⁴ times too large and a rate of 2.0 × 10⁵ W. Convert r = 1.0 cm to 0.010 m first.
Working r = 2.0 cm/2 = 0.010 m; A = πr² = π(0.010 m)² = 3.14 × 10⁻⁴ m². ΔT/Δx = (20 − 100) K/0.50 m = −160 K m⁻¹. ΔQ/Δt = −kA(ΔT/Δx) = −(390)(3.14 × 10⁻⁴)(−160) = +19.6 W ≈ 2.0 × 10¹ W, directed from the 100 °C end to the 20 °C end.
12 Warm air rises above a radiator in a cold room. Which is the correct explanation?
Answer and reasoning
Heat naturally rises, so the thermal energy from the radiator moves upwards and carries the air along with it. — A student who thinks heat itself rises picks this. It is the warm air that rises, because it is less dense than its surroundings; the energy is carried by the air, not the other way round.
Each air molecule expands when it is heated, so it becomes lighter than a cold molecule and floats upwards. — A student who thinks molecules swell when heated picks this. Molecules keep their size and mass; the air expands because its molecules move further apart, which lowers the density of the air.
Warm air has almost no weight, unlike the colder air around it, so nothing holds it down and it simply floats upwards. — A student who thinks hot gases are weightless picks this. Warm air has weight; it rises because a given volume of it weighs less than the same volume of the cooler air around it.
Air heated by the radiator expands, so it is less dense than the cooler air around it, which sinks and pushes it up. — Warmed air expands, so a given volume has less mass: its density falls. The denser, cooler air around it sinks beneath it and pushes it upwards, and a convection current forms.
13 Star X has 1.2 times the radius of star Y and 1.5 times its surface temperature (in kelvin). Both stars are modelled as black bodies. What is the ratio (luminosity of X)/(luminosity of Y)?
Answer and reasoning
5.1 — A student who thinks luminosity depends only on temperature gets 1.5⁴ = 5.1. X is also larger, with 1.2² = 1.44 times the surface area, and that must be included.
6.1 — A student who takes the surface area as proportional to the radius gets 1.2 × 1.5⁴ = 6.1. Area is 4πr², so the area ratio is 1.2² = 1.44, not 1.2.
2.2 — A student who treats luminosity as proportional to T instead of T⁴ gets 1.44 × 1.5 = 2.2. The power per unit area of a black body is σT⁴, so the temperature ratio must be raised to the fourth power.
7.3 — L = σ4πr²T⁴, so L_X/L_Y = (r_X/r_Y)²(T_X/T_Y)⁴ = 1.2² × 1.5⁴ = 1.44 × 5.06 = 7.3.
Working L = σAT⁴ with A = 4πr². L_X/L_Y = (r_X/r_Y)² × (T_X/T_Y)⁴ = (1.2)² × (1.5)⁴ = 1.44 × 5.0625 = 7.29 ≈ 7.3.
14 Two stars appear equally bright when viewed from the Earth. Which statement must be true?
Answer and reasoning
Equal power per unit area reaches Earth from each. — Apparent brightness is the power received per unit area at the observer. Equal apparent brightness means exactly this, and nothing more about the stars themselves.
Each star radiates the same total power out into space. — A student who equates brightness with luminosity picks this. Equal brightness can arise from a luminous distant star and a dim nearby one; b = L/4πd² depends on both L and d.
The two stars are at the same distance from the Earth. — A student who thinks brightness depends only on distance picks this. Stars differ enormously in luminosity, so equally bright stars can be at very different distances.
The two stars must have equal surface temperatures. — A student who thinks brightness is set by temperature alone picks this. Brightness depends on luminosity, which depends on surface area as well as temperature, and on distance.
15 Star P has a luminosity of 16 L⊙ and is at a distance 4d from the Earth. Star Q has a luminosity of 1 L⊙ and is at a distance d. How does the apparent brightness of P compare with that of Q?
Answer and reasoning
They appear equally bright: 16 times the luminosity offsets 4 times the distance. — b = L/4πd². b_P/b_Q = (L_P/L_Q) × (d_Q/d_P)² = 16 × (1/4)² = 16/16 = 1. P's sixteen-fold luminosity exactly compensates for its four-fold distance.
P appears 16 times as bright as Q, because it is 16 times as luminous. — A student who equates apparent brightness with luminosity picks this. P emits 16 times the power, but at four times the distance that power is spread over 16 times the area.
P appears 4 times as bright as Q, as brightness is proportional to 1/d. — A student who thinks brightness falls in proportion to distance gets 16/4 = 4. Brightness falls as 1/d², so four times the distance divides it by 16.
Q appears the brighter of the two, because Q is nearer to the Earth. — A student who thinks brightness depends only on distance picks this. Luminosity matters too: P is 16 times as luminous, which exactly compensates for its greater distance.
Working b = L/4πd². b_P/b_Q = (16 L⊙/(4π(4d)²))/(1 L⊙/(4πd²)) = 16/16 = 1, so the two stars appear equally bright.
16 A surface at 27 °C can be modelled as a black body. Wien's displacement law: λ_max T = 2.9 × 10⁻³ m K. At what wavelength is its emission most intense?
Answer and reasoning
1.1 × 10⁻⁴ m, in the far infrared — A student who uses the Celsius temperature gets 2.9 × 10⁻³/27 = 1.1 × 10⁻⁴ m. Wien's law needs the absolute temperature, 300 K.
8.7 × 10⁻¹ m, in the radio band — A student who thinks λ_max increases with temperature multiplies instead of dividing: 2.9 × 10⁻³ × 300 = 0.87 m. λ_max is inversely proportional to T.
3.6 × 10⁻¹³ m, in the gamma-ray band — A student who mixes Wien's law with the Stefan–Boltzmann law divides by T⁴ and gets 3.6 × 10⁻¹³ m. Wien's law uses T to the first power: λ_max = 2.9 × 10⁻³/T.
9.7 × 10⁻⁶ m, in the infrared — T = 27 + 273 = 300 K; λ_max = 2.9 × 10⁻³/300 = 9.7 × 10⁻⁶ m (about 10 μm), in the infrared. Everyday objects emit mainly infrared, which is why they do not glow visibly.
Working T = 27 + 273 = 300 K. λ_max = (2.9 × 10⁻³ m K)/(300 K) = 9.7 × 10⁻⁶ m, which is in the infrared (about 10 μm).
17 The black-body spectrum of star X peaks at 290 nm and that of star Y peaks at 870 nm. The two stars have the same radius. Which conclusion is correct?
Answer and reasoning
Y has three times the surface temperature of X and has 81 times the luminosity. — A student who thinks a longer peak wavelength means a hotter body picks this. λ_max is inversely proportional to T, so the star peaking at the shorter wavelength, X, is the hotter one.
X has three times the surface temperature of Y and three times the luminosity. — A student who treats luminosity as proportional to T picks this. The temperature ratio is right, but L = σAT⁴, so tripling T multiplies L by 3⁴ = 81.
X has three times the surface temperature of Y, and 81 times its luminosity. — λ_max T = constant, so T_X/T_Y = 870/290 = 3. With equal radii, L ∝ T⁴, so L_X/L_Y = 3⁴ = 81.
X and Y may be equally hot; the peak shows only what each star is made of. — A student who thinks star colour reveals composition picks this. The peak of a continuous black-body spectrum is fixed by temperature through Wien's law; composition shows up in spectral lines.
Working Wien: λ_max T = 2.9 × 10⁻³ m K, so T_X/T_Y = λ_Y/λ_X = 870/290 = 3.0 (T_X ≈ 1.0 × 10⁴ K, T_Y ≈ 3.3 × 10³ K). Equal radii: L_X/L_Y = (T_X/T_Y)⁴ = 3⁴ = 81.
18 A black body is heated to a higher temperature. How does its emission spectrum change?
Answer and reasoning
It emits more at every wavelength, and the peak moves to a longer wavelength. — A student who links higher temperature with longer (redder) wavelengths picks this. λ_max is inversely proportional to T: the peak moves to shorter wavelengths as T rises.
It emits more at every wavelength, and its peak moves to a shorter wavelength. — A hotter black body emits more intensely at every wavelength, so the whole curve rises, and by Wien's law λ_max = 2.9 × 10⁻³/T the peak moves to a shorter wavelength.
Its peak moves to shorter wavelengths, so it emits less at long wavelengths. — A student who pictures the curve sliding sideways picks this. The hotter curve lies above the cooler one at every wavelength, including long ones; it rises as the peak shifts.
It emits only one wavelength, λ_max, which gets shorter as it gets hotter. — A student who thinks a black body emits only its peak wavelength picks this. The spectrum is continuous over all wavelengths; λ_max is only where the intensity is greatest.
19 Water in an electric kettle is boiling at 100 °C. The 2.0 kW heater stays switched on for a further 2.5 minutes. Assume all the energy supplied is transferred to the water. Specific latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹; specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹; specific heat capacity of water = 4.18 × 10³ J kg⁻¹ K⁻¹. What mass of water boils away?
Answer and reasoning
9.0 × 10⁻¹ kg — A student who uses the latent heat of fusion gets 3.0 × 10⁵/3.34 × 10⁵ = 0.90 kg. Boiling is the liquid–gas change, so it needs L_v, about seven times larger than L_f.
7.2 × 10⁻¹ kg — A student who thinks boiling water keeps getting hotter treats the energy as raising its temperature, so uses Q = mcΔT with the only temperature given, 100 °C, as ΔT = 100 K: 3.0 × 10⁵/(4180 × 100) = 0.72 kg. While water boils its temperature stays at 100 °C, so ΔT = 0; the energy is latent heat of vaporization, Q = mL_v.
2.2 × 10⁻³ kg — A student who leaves the time in minutes uses t = 2.5 and gets 2000 × 2.5/2.26 × 10⁶ = 2.2 × 10⁻³ kg. The watt is a joule per second, so the time must be 150 s.
1.3 × 10⁻¹ kg — Q = Pt = 2000 × 150 = 3.0 × 10⁵ J. The temperature does not change, so all of this energy is latent heat of vaporization: m = Q/L_v = 3.0 × 10⁵/2.26 × 10⁶ = 0.13 kg.
Working Q = Pt = (2.0 × 10³ W)(2.5 × 60 s) = 3.0 × 10⁵ J. The water stays at 100 °C, so all of this is latent heat of vaporization: Q = mL_v, m = Q/L_v = (3.0 × 10⁵ J)/(2.26 × 10⁶ J kg⁻¹) = 0.133 kg ≈ 1.3 × 10⁻¹ kg.
20 Two solid cubes are cut from the same block of steel. Cube X has sides of length 2.0 cm and cube Y has sides of length 4.0 cm. Which statement correctly compares cube Y with cube X?
Answer and reasoning
Y has eight times the mass of X and the same density as X. — Density is a property of the material, so both cubes have the density of steel. Volume scales as the cube of the length ratio: V_Y = (4.0/2.0)³ V_X = 8 V_X, and since m = ρV the mass is also eight times larger.
Y has twice the mass of X and the same density as X. — A student who scales volume in the same ratio as length picks this. Doubling every side doubles the volume three times over: V = (2.0 cm)³ = 8.0 cm³ becomes (4.0 cm)³ = 64 cm³, eight times as much, so the mass is eight times as much.
Y has four times the mass of X and the same density as X. — A student who scales volume as the square of the length factor, as if it were an area, picks this. Volume is a product of three lengths, so doubling each side gives 2 × 2 × 2 = 8 times the volume and eight times the mass.
Y has a greater density than X, because it contains more steel. — A student who confuses density with the amount or heaviness of material picks this. Cube Y has eight times the mass but also eight times the volume, so m/V is unchanged: density describes the material, not how much of it there is.
Working Same material ⇒ same density ρ = m/V. Volume scales as length³: V_Y/V_X = (4.0/2.0)³ = 8, so m_Y = ρV_Y = 8 m_X. Density unchanged.
21 Which statement about the Kelvin and Celsius temperature scales is correct?
Answer and reasoning
A Kelvin temperature is converted to degrees Celsius by adding 273, since both directions use the same rule. — A student who remembers '273' but not which way it goes picks this. Kelvin values are the larger numbers (0 °C is 273 K), so going from kelvin to Celsius means subtracting 273: 112 K is −161 °C, not 385 °C.
A temperature has the same numerical value on both scales, as the kelvin is the SI name for the degree Celsius. — A student who treats the kelvin as a renamed degree picks this. The kelvin and the degree Celsius are the same size, but the scales start in different places: 0 K is −273 °C, so a given temperature has different values on the two scales.
A Celsius temperature is converted to kelvin by adding 273, whether the Celsius value is above or below zero. — T/K = t/°C + 273 for every temperature. The two scales have divisions of the same size and differ only in their zero, which is 273 °C lower on the Kelvin scale, so −78 °C is 195 K and 27 °C is 300 K.
A temperature below the freezing point of water is negative on the Kelvin scale, exactly as it is on the Celsius scale. — A student who thinks of the Kelvin scale as a relabelled Celsius scale picks this. 0 K is the lowest possible temperature, so Kelvin temperatures are never negative: water freezes at 273 K, and −78 °C is +195 K.
Working T/K = t/°C + 273 for every temperature, e.g. −78 °C → −78 + 273 = 195 K and 27 °C → 300 K; the reverse conversion subtracts 273 (112 K → −161 °C).
22 The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹. A student heats a copper block from 15 °C to 65 °C and, before using Q = mcΔT, adds 273 to the temperature change 'because c is given per kelvin'. Which evaluation of the student's method is correct?
Answer and reasoning
It is correct: every Celsius quantity, including a temperature change, must have 273 added to it before it is used in a physics equation. — A student who applies the conversion rule to a difference picks this. Adding 273 to both temperatures and subtracting cancels the 273, so ΔT = 50 K, not 323 K. Only an actual temperature, such as the T in E̅_k = (3/2)k_BT, needs converting.
It is wrong: a temperature change has the same numerical value in kelvin as in degrees Celsius, since the two scales differ only in where their zero lies. — Converting each temperature gives 288 K and 338 K, and 338 − 288 = 50 K, the same as 65 − 15 = 50 °C: the 273 cancels in a difference. ΔT is 50 K and the unit of c is consistent with it, so nothing should be added.
It is wrong only in the number used: the kelvin and the degree Celsius are different sizes, so ΔT should be multiplied by a factor rather than have 273 added. — A student who believes the two units differ in size picks this. They do not: one kelvin is exactly one degree Celsius in size, and the scales differ only in their zero. A change of 50 °C is a change of 50 K, with no factor and no 273.
It is wrong: the kelvin is simply the SI name for the degree Celsius, so a temperature and a temperature change are both identical on the two scales. — A student who thinks the scales are the same picks this. The conclusion about ΔT happens to be right, but the reason is not: a temperature is NOT the same on both scales (65 °C is 338 K). It is only a temperature change that has the same value in K and °C.
23 A metal sphere at 33 °C hangs from a thin thread inside a large evacuated chamber whose walls are at 18 °C. Energy is exchanged between the sphere and the walls only by thermal radiation. Which statement describes the resultant thermal energy transfer?
Answer and reasoning
Only the sphere emits thermal radiation, since the walls are too cool to radiate, so the whole transfer is from the sphere to the walls. — A student who thinks only hot objects radiate picks this. The direction is right, but walls at 18 °C (291 K) radiate strongly in the infrared; the sphere merely emits more than the walls send back, so the resultant transfer is from sphere to walls.
Both emit thermal radiation, but the sphere emits more than it absorbs, so the resultant transfer is from the sphere to the walls. — Every body above 0 K radiates. The sphere and the walls each emit and each absorb, and the resultant transfer is from the higher temperature (33 °C) to the lower (18 °C) because the hotter surface emits more power per unit area than it receives back.
The walls contain far more internal energy than the sphere does, so the resultant transfer of energy is from the walls to the sphere. — A student who lets the amount of internal energy set the direction picks this. The walls have more internal energy only because they have far more mass; the direction of the resultant transfer is decided by temperature difference alone, from 33 °C to 18 °C.
Coldness radiates from the cool walls towards the sphere, so the resultant transfer is from the walls to the sphere. — A student who thinks of cold as something that travels picks this. Nothing called cold is emitted; the sphere cools because it radiates more energy to the walls than the walls radiate back to it, so energy leaves the sphere.
24 On a still evening a person stands 2 m to one side of a bonfire, at the same height as the flames, and feels warmth on their face. By which mechanism does most of this energy reach their face, and why?
Answer and reasoning
Conduction: the hot air next to the flames passes energy from particle to particle across the 2 m of still air to the person's face. — A student who thinks air conducts well picks this. Still air is one of the poorest conductors there is, which is why trapped air is used as insulation; conduction through 2 m of air transfers a negligible amount of energy in the time the warmth is felt.
Convection: heat is a fluid that spreads out from the fire through the air in every direction, including sideways to the face. — A student who pictures heat as a substance that flows picks this. Convection is the movement of the heated fluid itself, and the air heated by the fire is less dense than its surroundings, so it moves upwards, not sideways at the height of the flames.
Radiation: still air is a poor conductor, and the hot air from the fire rises, so neither conduction nor convection carries much energy sideways. — Thermal radiation travels in straight lines from the flames through the air in every direction, so it reaches a face at the side. The convection current carries the hot air upwards, and air's thermal conductivity is so small that conduction across 2 m is negligible.
Convection: the heat from the fire rises, and as it spreads it warms the whole of the surrounding air, including the air by the person's face. — A student who thinks 'heat rises' picks this. It is hot, less dense air that rises, and it goes up rather than out; on a still evening the air at the side of a fire stays cool, and a card held in front of the face blocks the warmth at once, which shows it arrives as radiation.
25 One end of a steel rod is held at 80 °C and the other at 20 °C. After some time the temperature at every point along the rod stops changing. Which statement about the particles of the rod is correct in this steady state?
Answer and reasoning
Since the temperature at every point along the rod has stopped changing, no more energy is now being transferred from the hot end to the cold end. — A student who equates 'energy transferred' with 'temperature changing' picks this. A steady temperature at a point means energy arrives there at the same rate as it leaves; energy is still conducted along the rod, at a constant rate, as long as the ends are kept at different temperatures.
Particles near the hot end still have more kinetic energy on average than those near the cold end, and kinetic energy is still being passed along the rod towards the cold end. — Temperature measures the average kinetic energy of the particles, so a rod with a temperature difference along it has a kinetic-energy difference along it. Conduction continues: faster-vibrating particles keep passing energy to slower neighbours at a steady rate, which is why the hot end must be kept heated and the cold end keeps warming its surroundings.
The faster-moving particles are steadily travelling along the rod from the hot end to the cold end, carrying their kinetic energy with them. — A student who pictures conduction as particles migrating picks this. In a solid the particles stay in their positions and vibrate; energy is passed from a faster-vibrating particle to its slower neighbours through the forces between them, and no particle travels along the rod.
Every particle along the rod now vibrates with the same kinetic energy; what decreases towards the cold end is the amount of heat stored between the particles. — A student who thinks of heat as a fluid stored in the material picks this. There is no separate 'heat' between the particles: the hot end is hot because its particles have more kinetic energy on average, and that difference in kinetic energy is what drives conduction along the rod.
26 In a simple refrigerator the freezing compartment is placed at the top of the cabinet. Which statement explains why this cools the whole of the space below it?
Answer and reasoning
Heat naturally rises, so the thermal energy from all the food travels upwards to the compartment at the top, where it is removed from the cabinet. — A student who thinks 'heat rises' picks this. Thermal energy has no tendency to move upwards; it is heated fluid that rises because it is less dense. Here the flow is driven by the cooled air sinking, and the rising air simply fills the space it leaves.
Coldness is a heavy substance that is produced by the compartment and pours down through the cabinet, filling it from the bottom upwards like water. — A student who thinks of cold as something that flows picks this. Nothing called cold is produced or transferred; the compartment removes energy from the air touching it, and it is that denser air that sinks, while energy from the food is carried up by the rising warmer air.
Air cooled by the compartment becomes denser and sinks, and the warmer air below rises to replace it, so a convection current circulates. — Air in contact with the cold compartment contracts and its density rises above that of the air below, so it sinks; the warmer, less dense air below is displaced upwards, is cooled in its turn, and the circulation carries energy from the whole cabinet to the compartment.
The air molecules next to the compartment shrink and become heavier as they cool, so each one falls to the bottom of the cabinet under its own weight. — A student who thinks the molecules themselves change picks this. A molecule's mass and size do not depend on temperature; cooling makes the molecules move more slowly and pack closer together, so the AIR is denser, and it is this denser air that sinks as a body.
27 Star A has 100 times the luminosity of star B, yet from the Earth star A appears dimmer than star B. Which statement must be correct?
Answer and reasoning
We receive more power per unit area from A than from B. — A student who equates brightness with luminosity picks this. The power per unit area arriving at the Earth IS the apparent brightness, so if A appears dimmer the Earth receives less power per unit area from A, in spite of A emitting far more in total.
At equal distances, A and B would look equally bright. — A student who thinks brightness depends only on distance picks this. At equal distances the light from each star is spread over the same area, so A, which emits 100 times the power, would appear 100 times brighter than B.
B must have a much hotter surface than A has. — A student who reads brightness as a sign of temperature picks this. Nothing in the data fixes the temperatures: a star's luminosity depends on its surface area as well as its temperature, and how bright it appears depends on its distance as well. B's temperature could be higher or lower than A's.
A lies much further from the Earth than B does. — Apparent brightness is the power received per unit area at the observer, and it depends on both the star's luminosity and its distance. A star that emits 100 times as much power can only appear dimmer if its light is spread over a much larger sphere, so it must be much further away.
That was your twenty minutes. Real practice on B.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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