B.1 Thermal energy transfers flashcards: IB Physics SL and HL

All 75 B.1 Thermal energy transfers cards

1. DefinitionSL & HL

   Define density and state its SI unit and how it is measured for an irregular solid.

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   Density ρ is the mass per unit volume of a substance, ρ = m/V. SI unit kg m⁻³; 1 g cm⁻³ = 1.00 × 10³ kg m⁻³. Density is a scalar. For an irregular solid, find m on a balance and V by displacement of water in a measuring cylinder or eureka can. Exam tip: the mark is for "mass per unit volume", not "how heavy something is" or "mass over size". Quote a temperature for precise work because most substances expand on heating so ρ falls; water is anomalous, being densest near 4 °C. Density is an intensive property — cutting a block in half does not change ρ.

2. DefinitionSL & HL

   Define temperature in terms of molecular motion and state what thermal equilibrium means.

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   Temperature is a measure of the average random kinetic energy of the molecules of a substance; it determines the direction of thermal energy transfer. Two bodies are in thermal equilibrium when they are at the same temperature, so there is no net transfer of thermal energy between them. SI unit kelvin (K); temperature is a scalar. Exam tip: the mark scheme rejects "temperature is the heat in a body" and "total kinetic energy" — the word average is essential, and potential energy of the molecules is excluded. Thermal energy always flows spontaneously from higher to lower temperature, never from a body with more internal energy to one with less.

3. DefinitionSL & HL

   Define absolute zero and outline the basis of the Kelvin (thermodynamic) temperature scale.

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   Absolute zero (0 K = −273 °C) is the temperature at which molecules have minimum kinetic energy, so no further thermal energy can be extracted from a substance. The Kelvin scale is an absolute scale whose zero is absolute zero and whose degree size equals the Celsius degree, so T/K = θ/°C + 273. Exam tip: never write "°K"; write K. State "minimum", not "zero", kinetic energy — a fully rigorous answer allows for residual quantum zero-point energy. Kelvin must be used in every equation containing an absolute temperature (P = eσAT⁴, λ_max T = 2.9 × 10⁻³ m K, pV = nRT); only temperature differences may be left in °C.

4. DefinitionSL & HL

   Define internal energy and state precisely which two contributions make it up.

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   The internal energy of a substance is the total of the random kinetic energy of its molecules plus the potential energy arising from the intermolecular forces between them. SI unit joule (J); it is a scalar and a property (state function) of the body. Exam tip: the two-part answer is required — "total kinetic energy" alone loses a mark, and adding the kinetic energy of the body as a whole or the energy stored in the nuclei is wrong. During a phase change, temperature and hence average KE are constant while PE increases as bonds are broken, so internal energy rises with no temperature change. Internal energy is not "heat"; heat is energy in transit.

5. DefinitionSL & HL

   Define thermal energy transfer (heat) and distinguish it from temperature and from internal energy.

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   Thermal energy (heat) is energy transferred between two bodies as a direct result of a temperature difference between them. SI unit joule (J); scalar. Exam tip: heat is energy in transit — a body does not "contain heat", it contains internal energy. Temperature decides the direction of transfer; internal energy is the total molecular KE + PE stored. A cup of tea at 80 °C has a higher temperature than a swimming pool at 20 °C, but the pool has far greater internal energy because it has vastly more molecules. Thermal energy transfer occurs by conduction, convection and radiation, and always from hot to cold in an isolated system.

6. DefinitionSL & HL

   Define specific heat capacity and explain the physical meaning of a large value.

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   The specific heat capacity c of a substance is the energy required to raise the temperature of unit mass of the substance by one kelvin (one degree) without a change of phase: Q = mcΔT. SI unit J kg⁻¹ K⁻¹. Exam tip: "unit mass" and "one kelvin" must both appear; omitting "per unit mass" gives heat capacity instead and loses the mark. A large c (water, 4180 J kg⁻¹ K⁻¹) means a lot of energy is needed for a small temperature rise, so water is a good coolant and moderates coastal climates. Because ΔT is a difference, K and °C are interchangeable in Q = mcΔT only, never in T⁴ or Wien's law.

7. DefinitionSL & HL

   Define heat capacity (thermal capacity) and state how it relates to specific heat capacity.

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   The heat capacity C of a body is the energy required to raise the temperature of that whole body by one kelvin: Q = CΔT. SI unit J K⁻¹. It relates to specific heat capacity by C = mc, so it depends on the mass and material of the particular object, whereas c depends only on the material. Exam tip: students lose marks by quoting the unit as J kg⁻¹ K⁻¹ — heat capacity has no "per kilogram". Heat capacity is useful for a composite object such as a calorimeter with its stirrer and thermometer, where a single value C_cal replaces separate m and c for each part.

8. DefinitionSL & HL

   Define specific latent heat of fusion and state what happens to the molecules during melting.

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   The specific latent heat of fusion L_f is the energy required to change unit mass of a substance from solid to liquid at constant temperature (its melting point): Q = mL. SI unit J kg⁻¹. For ice L_f = 3.34 × 10⁵ J kg⁻¹. Exam tip: "at constant temperature" (or "without a change in temperature") is a required mark point; so is "unit mass". The supplied energy increases the potential energy of the molecules by partially breaking the bonds of the rigid lattice, so average KE and therefore temperature are unchanged. Freezing releases exactly the same energy per kilogram, which is why fruit growers spray water on crops to prevent frost damage.

9. DefinitionSL & HL

   Define specific latent heat of vaporisation and explain why it is much larger than that of fusion.

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   The specific latent heat of vaporisation L_v is the energy required to change unit mass of a substance from liquid to vapour at constant temperature (its boiling point): Q = mL. SI unit J kg⁻¹; for water L_v = 2.26 × 10⁶ J kg⁻¹. Exam tip: it exceeds L_f (3.34 × 10⁵ J kg⁻¹) because vaporisation must break essentially all the intermolecular bonds rather than merely loosening the lattice, and the vapour must also do work pushing back the atmosphere as its volume expands enormously. This is why steam at 100 °C causes far worse burns than water at 100 °C — it delivers 2.26 MJ kg⁻¹ on condensing before it even begins to cool.

10. DefinitionSL & HL

    Distinguish evaporation from boiling, and explain the cooling effect of evaporation.

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    Evaporation occurs only at the exposed surface of a liquid, at any temperature below the boiling point, and its rate rises with temperature, surface area, draught and dryness of the air. Boiling occurs throughout the bulk of the liquid, only at one fixed temperature for a given pressure, with bubbles of vapour forming inside. Exam tip: for the cooling effect, state that the fastest (most energetic) molecules escape, so the average kinetic energy of those remaining falls and hence the temperature falls. Answers saying "the cold air cools it" score zero. Sweating, panting dogs and refrigerant evaporation in a fridge's cold pipes are the standard real-world links.

11. DefinitionSL & HL

    Describe the arrangement, separation and motion of molecules in a solid, a liquid and a gas.

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    Solid: molecules closely packed in a regular lattice, separation about one molecular diameter, strong intermolecular forces, vibration about fixed positions — fixed shape and volume. Liquid: molecules still close together but irregularly arranged and free to slide past one another, forces slightly weaker — fixed volume, takes the shape of its container. Gas: separation about ten molecular diameters or more, negligible intermolecular forces except during collisions, rapid random translational motion — no fixed shape or volume, highly compressible. Exam tip: answers must compare all three of arrangement, separation and motion; writing only "solids are hard, gases spread out" earns nothing. Density falls sharply from solid/liquid to gas, typically by a factor of about 10³.

12. DefinitionSL & HL

    Define thermal conduction and thermal conductivity, and explain the molecular mechanism.

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    Conduction is the transfer of thermal energy through a substance without any bulk movement of the substance itself. Molecules at the hotter end vibrate with greater amplitude and pass energy on by collisions with neighbours; in metals, free (delocalised) electrons diffuse and carry energy far faster, which is why metals are the best conductors. Thermal conductivity k is the constant in ΔQ/Δt = kAΔT/Δx, numerically the rate of energy transfer per unit area per unit temperature gradient, SI unit W m⁻¹ K⁻¹. Exam tip: "without movement of the material" is a mark point; also state that good electrical conductors are good thermal conductors because both rely on free electrons.

13. DefinitionSL & HL

    Define convection and describe how a convection current is set up.

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    Convection is the transfer of thermal energy through a fluid (liquid or gas) by the bulk movement of the fluid itself. Fluid in contact with the hot surface is heated, expands, so its density falls; the surrounding denser, cooler fluid sinks and displaces it, pushing the warm fluid upward, and a continuous convection current is established as the risen fluid cools and sinks again. Exam tip: the essential chain is heated → expands → less dense → rises, and answers that stop at "hot air rises" lose marks. Convection cannot occur in a solid or a vacuum. Real-world links: sea and land breezes, domestic radiators placed low, and cooling of the atmosphere and oceans.

14. DefinitionSL & HL

    Define thermal radiation and state the factors that determine the power radiated by a surface.

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    Thermal radiation is the emission of energy as electromagnetic waves (chiefly infrared for everyday temperatures) by any body at a temperature above absolute zero; it requires no medium and travels at c = 3.00 × 10⁸ m s⁻¹ through a vacuum. The power radiated is given by P = eσAT⁴ and so depends on the surface area A, the absolute temperature T (very strongly, to the fourth power), and the nature of the surface through the emissivity e. Exam tip: state "absolute temperature in kelvin" — using °C in T⁴ is the single commonest error. Dark, matt surfaces are the best emitters and best absorbers; shiny, silvered surfaces are poor at both, which is why vacuum flasks are silvered.

15. DefinitionSL & HL

    Define a black body and state the properties of black-body radiation.

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    A black body is a (theoretical) body that absorbs all the electromagnetic radiation incident on it at every wavelength, reflecting and transmitting none, and which is therefore also the best possible emitter at every wavelength for a given temperature. Its emitted spectrum depends only on its absolute temperature, not on its material or surface. Exam tip: a full answer must include the emitter half — "a perfect absorber" alone is often only worth one of two marks. A small hole in a large cavity is the standard practical approximation; stars are treated as black bodies. The peak wavelength obeys λ_max T = 2.9 × 10⁻³ m K and the total power obeys P = σAT⁴ with e = 1.

16. DefinitionSL & HL

    Define emissivity and state its range of values with examples.

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    Emissivity e is 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: e = P_surface/P_black body. It is a dimensionless ratio with no unit, and 0 ≤ e ≤ 1, with e = 1 for a perfect black body. Typical values: matt black soot ≈ 0.95, ordinary Earth surface and vegetation ≈ 0.9, polished silver ≈ 0.02, Earth as a planet ≈ 0.6 when the atmosphere is included. Exam tip: quoting a unit for e loses the mark. A good emitter is an equally good absorber at the same wavelength, so e also equals the absorptivity.

17. DefinitionSL & HL

    Define luminosity and apparent brightness, and state the units of each.

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    Luminosity L is the total power radiated by a body (a star) in all directions, SI unit watt (W). Apparent brightness b is the power received per unit area at the observer, measured perpendicular to the direction of propagation, SI unit W m⁻². They are linked by b = L/(4πd²) where d is the distance to the star. Exam tip: students routinely confuse the two — luminosity is an intrinsic property of the star, whereas apparent brightness depends on how far away we are. State that b is an intensity, so its unit contains "per square metre"; giving b in watts is a guaranteed lost mark. The inverse-square law assumes no absorption by interstellar dust.

18. EquationSL & HLData booklet: Yes

    State the density equation, define each symbol with its SI unit, and give a sanity check.

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    ρ = m/V (data booklet). ρ = density in kg m⁻³; m = mass in kg; V = volume in m³. Valid for a homogeneous substance at a stated temperature; for a mixture it gives only the mean density. Rearranged forms: m = ρV and V = m/ρ. Common misuse: mixing units — a volume in cm³ with a mass in kg gives a value 10⁶ times too large. Sanity check: 1.0 litre of water has m = 1.0 kg and V = 1.0 × 10⁻³ m³, so ρ = 1.0 × 10³ kg m⁻³, which is the standard value. Air at room conditions is about 1.2 kg m⁻³, roughly 1/800 of water.

19. EquationSL & HLData booklet: No – memorise

    State the relation between Kelvin and Celsius temperature and state exactly when each may be used.

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    T/K = θ/°C + 273 (more precisely 273.15). T = absolute (thermodynamic) temperature in kelvin, K; θ = Celsius temperature in °C. A temperature difference is numerically the same on both scales, so ΔT/K = Δθ/°C. Not printed in the data booklet — memorise it. Common misuse: substituting °C into P = eσAT⁴, λ_max T = 2.9 × 10⁻³ m K, pV = nRT or E_k = (3/2)k_B T; all of these require kelvin. Only Q = mcΔT and ΔQ/Δt = kAΔT/Δx contain a difference and so tolerate °C. Sanity check: 27 °C = 300 K, 0 °C = 273 K, −273 °C = 0 K = absolute zero.

20. EquationSL & HLData booklet: Yes

    State the equation for energy transferred in a temperature change and list its conditions of validity.

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    Q = mcΔT (data booklet). Q = thermal energy transferred in J; m = mass in kg; c = specific heat capacity in J kg⁻¹ K⁻¹; ΔT = temperature change in K (or °C, since it is a difference). Valid only when there is no change of phase and c is effectively constant over the range. Rearranged forms to know: c = Q/(mΔT), m = Q/(cΔT), ΔT = Q/(mc), and the rate form P = mcΔT/Δt. Common misuse: using the final temperature rather than the change, or applying it across a plateau on a heating curve. Sanity check: heating 0.50 kg of water by 40 K needs 0.50 × 4180 × 40 ≈ 8.4 × 10⁴ J.

21. EquationSL & HLData booklet: Yes

    State the equation for energy transferred during a phase change and define each symbol.

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    Q = mL (data booklet). Q = thermal energy transferred in J; m = mass changing phase in kg; L = specific latent heat in J kg⁻¹ (L_f for fusion, L_v for vaporisation). Valid only at the melting or boiling point, where the temperature stays constant, so no ΔT appears. Rearranged: L = Q/m and m = Q/L. The rate form is P = mL/Δt, or ṁ = P/L for a steady boiler. Common misuse: writing Q = mLΔT, or using the total mass when only part has melted. Sanity check: boiling away 0.10 kg of water needs 0.10 × 2.26 × 10⁶ = 2.3 × 10⁵ J, about 54 times the 4.2 × 10³ J needed to heat it from 90 °C to 100 °C.

22. EquationSL & HLData booklet: No – derive

    State how to calculate the total energy for a multi-stage heating process such as ice at −20 °C to steam at 120 °C.

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    Add the energy of each stage separately: Q_total = mc_ice(20) + mL_f + mc_water(100) + mL_v + mc_steam(20). This combination is not printed in the booklet — you must construct it. Symbols: m in kg, c in J kg⁻¹ K⁻¹, L in J kg⁻¹, temperature intervals in K. Use c_ice ≈ 2100, c_water = 4180, c_steam ≈ 2000 J kg⁻¹ K⁻¹, L_f = 3.34 × 10⁵, L_v = 2.26 × 10⁶ J kg⁻¹. Common misuse: forgetting a latent-heat term, or using c_water for the ice stage. Sanity check for 1.0 kg: 4.2 × 10⁴ + 3.34 × 10⁵ + 4.18 × 10⁵ + 2.26 × 10⁶ + 4.0 × 10⁴ ≈ 3.1 × 10⁶ J, dominated by vaporisation.

23. EquationSL & HLData booklet: Yes

    State the equation for the rate of thermal conduction and explain every symbol and condition.

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    ΔQ/Δt = kAΔT/Δx (data booklet). ΔQ/Δt = rate of thermal energy transfer (power) in W; k = thermal conductivity in W m⁻¹ K⁻¹; A = cross-sectional area perpendicular to the flow in m²; ΔT = temperature difference across the slab in K; Δx = thickness in the direction of flow in m. ΔT/Δx is the temperature gradient in K m⁻¹. Valid for steady state, uniform k and a lagged (laterally insulated) sample. Common misuse: using the surface area of the whole object instead of the cross-section, or ΔT in the wrong place. Sanity check: a 4.0 mm window pane, A = 2.0 m², k = 1.0, ΔT = 15 K gives 1.0 × 2.0 × 15/0.0040 = 7.5 kW — hence double glazing.

24. EquationSL & HLData booklet: Yes

    State the Stefan–Boltzmann law in its emissivity form and give the conditions and pitfalls.

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    P = eσAT⁴ (data booklet). P = power radiated in W; e = emissivity, dimensionless, 0 ≤ e ≤ 1; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (Stefan–Boltzmann constant); A = surface area emitting, in m²; T = absolute surface temperature in K. For a black body e = 1. The net power for a body at T in surroundings at T_s is P_net = eσA(T⁴ − T_s⁴). Common misuse: T in °C, or forgetting that doubling T multiplies P by 16. Sanity check: human skin, A ≈ 1.8 m², e ≈ 0.97, T = 306 K gives P ≈ 0.97 × 5.67 × 10⁻⁸ × 1.8 × 306⁴ ≈ 8.7 × 10² W gross, with net loss eσA(306⁴ − 293⁴) ≈ 1.4 × 10² W into a 293 K room.

25. EquationSL & HLData booklet: Yes

    State the luminosity equation for a star and show how the surface area is obtained.

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    L = σAT⁴ with A = 4πR² for a sphere, so L = 4πR²σT⁴ (booklet gives L = σAT⁴ and treats stars as black bodies, e = 1). L = luminosity in W; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; A = surface area in m²; R = stellar radius in m; T = surface (photosphere) temperature in K. Useful ratio form: L/L_Sun = (R/R_Sun)²(T/T_Sun)⁴, which avoids ever substituting σ. Common misuse: using πR² (the cross-section) instead of 4πR², or forgetting to square the radius. Sanity check: the Sun, R = 7.0 × 10⁸ m, T = 5800 K, gives L ≈ 3.8 × 10²⁶ W.

26. EquationSL & HLData booklet: Yes

    State Wien's displacement law, define its symbols, and give a worked sanity check.

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    λ_max T = 2.9 × 10⁻³ m K (data booklet). λ_max = wavelength at which the emitted intensity is a maximum, in m; T = absolute surface temperature in K; the constant 2.9 × 10⁻³ m K is Wien's constant. Valid for a black body (or a good approximation to one, such as a star). Rearranged: λ_max = 2.9 × 10⁻³/T and T = 2.9 × 10⁻³/λ_max. Common misuse: substituting λ in nm without converting to m — an error of 10⁹ — or using °C. Sanity check: the Sun at 5800 K gives λ_max = 2.9 × 10⁻³/5800 = 5.0 × 10⁻⁷ m = 500 nm, in the green-yellow visible region; the Earth at 288 K gives ≈ 1.0 × 10⁻⁵ m, in the infrared.

27. EquationSL & HLData booklet: Yes

    State the apparent brightness equation and explain the assumptions behind the inverse-square law.

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    b = L/(4πd²) (data booklet). b = apparent brightness (intensity received) in W m⁻²; L = luminosity in W; d = distance from the star to the observer in m. The 4πd² is the surface area of the sphere over which the emitted power has spread. Assumes the source radiates isotropically and no radiation is absorbed or scattered by interstellar matter. Rearranged: L = 4πd²b, d = √(L/(4πb)). Ratio form: b₁/b₂ = (d₂/d₁)². Common misuse: forgetting to square d, or mixing light-years/parsecs with metres (1 ly = 9.46 × 10¹⁵ m, 1 pc = 3.09 × 10¹⁶ m). Sanity check: the Sun at d = 1.5 × 10¹¹ m gives b = 3.8 × 10²⁶/(4π(1.5 × 10¹¹)²) ≈ 1.4 × 10³ W m⁻².

28. EquationSL & HLData booklet: No – derive

    State the energy-balance equation used in the method of mixtures and its key assumption.

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    Energy lost by the hot body = energy gained by the cold body (and the calorimeter): m_h c_h (T_h − T_f) = m_c c_c (T_f − T_c) + C_cal(T_f − T_c). Symbols: masses m in kg, specific heat capacities c in J kg⁻¹ K⁻¹, C_cal the heat capacity of the calorimeter in J K⁻¹, temperatures in °C or K, T_f = final common (equilibrium) temperature. Built from Q = mcΔT, so it is not printed as such in the booklet. Assumes no thermal energy is exchanged with the surroundings. Common misuse: writing (T_h − T_c) for one of the changes instead of using T_f, or dropping the calorimeter term. Sanity check: T_f must always lie between T_c and T_h.

29. EquationSL & HLData booklet: No – derive

    State the equations for the electrical determination of specific heat capacity and specific latent heat.

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    Electrical energy supplied = VIt, so for heating VIt = mcΔT and for a phase change VIt = mL (or Pt = mL). V = potential difference in V; I = current in A; t = time in s; hence c = VIt/(mΔT) and L = VIt/m, with m the mass melted or vaporised. The booklet gives P = VI, Q = mcΔT and Q = mL; the combination is yours to make. Assumes all electrical energy enters the sample, with no losses to the surroundings and none absorbed by the heater or container. Common misuse: using the total mass rather than the mass that changed phase. Sanity check: a 50 W heater run for 200 s delivers 1.0 × 10⁴ J, enough to melt 1.0 × 10⁴/3.34 × 10⁵ ≈ 30 g of ice.

30. EquationSL & HLData booklet: No – derive

    State the rate (power) forms of the thermal-energy equations and when each is used.

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    Steady heating: P = ΔQ/Δt = mcΔT/Δt, so the temperature rises at dT/dt = P/(mc). Steady phase change: P = mL/Δt, so the mass changes phase at ṁ = P/L in kg s⁻¹. Steady conduction: P = kAΔT/Δx. Steady radiation: P = eσAT⁴, with net P = eσA(T⁴ − T_s⁴). All powers in W, energies in J, times in s. These are the standard forms for "a heater of power P is switched on for time t" questions and for equilibrium problems where power in = power out. Common misuse: forgetting to convert minutes to seconds. Sanity check: a 2.0 kW kettle holding 1.5 kg of water raises it by 80 K in mcΔT/P = 1.5 × 4180 × 80/2000 ≈ 251 s ≈ 4.2 min.

31. Graph/diagramSL & HL

    Describe the heating curve of a substance taken from solid below its melting point to vapour, and state what each feature represents.

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    Axes: temperature T/°C (y) against time t/s (x) for constant power input, or against energy supplied Q/J. Shape: a rising slope (solid), a horizontal plateau at the melting point, a rising slope (liquid), a longer horizontal plateau at the boiling point, then a rising slope (vapour). Sloping sections mean rising average KE, so temperature rises; plateaus mean the supplied energy raises molecular PE by breaking bonds at constant KE, so temperature is constant. Gradient of a sloping section = P/(mc), so a steeper slope means a smaller specific heat capacity. Plateau length ∝ mL/P, so the vaporisation plateau is far longer than the fusion one. Doubling the heater power halves every time interval but leaves the plateau temperatures unchanged.

32. Graph/diagramSL & HL

    Describe the cooling curve of a molten substance solidifying, and how it is used to find a melting point.

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    Axes: temperature T/°C (y) against time t/s (x). Shape: temperature falls with a decreasing rate (curve concave upwards) while the liquid cools, then a horizontal plateau while it freezes as latent heat is released at constant temperature, then a further falling curve as the solid cools. The plateau temperature gives the melting/freezing point of the substance — this is how the purity of a substance is checked, since impurities lower and broaden it. The falling sections curve rather than being straight because the rate of loss to the surroundings falls as the excess temperature falls. Some substances show a small dip below the plateau before it (supercooling). Plateau duration ∝ mL_f/(rate of energy loss), so a larger sample gives a longer plateau.

33. Graph/diagramSL & HL

    Describe the graph used to determine specific heat capacity by the electrical method and how c is extracted.

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    Axes: temperature rise ΔT/K (y) against time t/s (x) for a lagged block heated at constant electrical power P = VI. Shape: a straight line through the origin during the steady heating region, curving and flattening at long times as losses to the surroundings grow. Gradient = P/(mc), so c = P/(m × gradient) — take the gradient of the initial straight part only. Alternatively plot ΔT against energy supplied VIt/J, giving a straight line of gradient 1/(mc) and intercept zero. Increasing m or c reduces the gradient; increasing P raises it. Lagging the block and taking the gradient near t = 0 minimise the systematic overestimate of c caused by heat lost to the room.

34. Graph/diagramSL & HL

    Describe the black-body spectrum and how the curves change with temperature.

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    Axes: spectral intensity (power emitted per unit area per unit wavelength, W m⁻² m⁻¹ or arbitrary units) on y against wavelength λ/m (or nm) on x. Shape: a smooth asymmetric curve rising steeply from zero at short λ to a single peak at λ_max, then falling more gradually with a long tail at long wavelengths; the curve never touches the axis at large λ. As T increases, the peak moves to shorter wavelength (λ_max = 2.9 × 10⁻³/T) and the whole curve rises, with the total area under the curve equal to the power per unit area σT⁴, so the area grows as T⁴. Curves for different temperatures never cross. Reading λ_max off the axis gives T; the area gives P/A and hence, with A, the luminosity.

35. Graph/diagramSL & HL

    Describe the linearised graph used to verify Wien's displacement law and what its gradient gives.

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    Wien's law λ_max T = 2.9 × 10⁻³ m K is a reciprocal relation, so plot λ_max/m (y) against 1/T (K⁻¹) (x). Shape: a straight line through the origin. Gradient = Wien's constant, 2.9 × 10⁻³ m K; a zero intercept confirms the inverse proportionality and shows there is no systematic offset in the temperature measurement. Equivalently, plot ln λ_max against ln T: a straight line of gradient −1 and intercept ln(2.9 × 10⁻³). Plotting λ_max directly against T would give a curve from which no constant could be read reliably. Draw maximum and minimum gradient lines through the error bars to obtain the uncertainty in the constant as (gradient_max − gradient_min)/2.

36. Graph/diagramSL & HL

    Describe the temperature profile along a lagged uniform bar conducting heat in the steady state, and along an unlagged bar.

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    Axes: temperature T/°C (y) against distance x/m along the bar from the hot end (x). Lagged bar in the steady state: a straight line of constant negative gradient, because ΔQ/Δt is the same through every cross-section and ΔQ/Δt = kA(ΔT/Δx) with k and A constant. The magnitude of the gradient, ΔT/Δx in K m⁻¹, is the temperature gradient; combined with the measured power and area it gives k = (ΔQ/Δt)/(A × gradient). Unlagged bar: a curve that is steep near the hot end and flattens further along, because energy leaks from the sides so less flows through the more distant sections. Halving A or k for the same power doubles the gradient; a composite bar shows a kink at the junction, steeper in the poorer conductor.

37. Graph/diagramSL & HL

    Describe the graph of rate of cooling against excess temperature for a body cooling in a room, and its use.

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    Axes: rate of temperature fall |dT/dt| in K s⁻¹ (y) against excess temperature (T − T_surroundings) in K (x), the values of dT/dt being tangent gradients taken from a T–t cooling curve. Shape: for small excess temperatures, an approximately straight line through the origin, showing the rate of loss is proportional to the excess temperature (Newton's law of cooling); at large excess the points curve upward above the line because radiation grows as T⁴ rather than linearly. The gradient equals (rate of energy loss per kelvin)/(mc), so multiplying by mc gives the heat-loss coefficient in W K⁻¹. Increasing the surface area or blackening the surface steepens the line; lagging flattens it. The graph justifies the cooling correction applied in method-of-mixtures experiments.

38. Graph/diagramSL & HL

    Describe the log-log graph used to confirm the fourth-power dependence in the Stefan-Boltzmann law.

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    Taking logs of P = eσAT⁴ gives ln P = 4 ln T + ln(eσA), so plot ln(P/W) on y against ln(T/K) on x. Shape: a straight line of gradient 4, which is the evidence for the fourth-power law; the y-intercept is ln(eσA), from which e can be found if A is known. Equivalently plot P against T⁴, giving a straight line through the origin of gradient eσA. Common practical version: a filament lamp, with P = VI and T deduced from the filament resistance. Points at low temperature fall below the line because the surroundings also radiate back, so plot the net power eσA(T⁴ − T_s⁴) instead. A measured gradient near 4.0 within its uncertainty confirms the law.

39. Concept/explainSL & HL

    Explain why absolute temperature measures the average kinetic energy of molecules, and why two bodies at the same temperature need not have the same internal energy.

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    • Molecules move randomly with a distribution of speeds
    • Absolute temperature T is directly proportional to the average random kinetic energy per molecule, so T is intensive and independent of how many molecules are present
    • Internal energy is the TOTAL random kinetic energy plus the TOTAL intermolecular potential energy, so it is extensive and depends on mass and phase
    • A bathful of water at 300 K therefore holds far more internal energy than a cupful at 300 K
    • Thermal energy flows because of a temperature difference, never because one body has more internal energy. Exam tip: the incomplete answer is "temperature is the heat in the body" — heat is energy in transit, not a stored quantity.
40. Concept/explainSL & HL

    Outline, in terms of molecular spacing, intermolecular forces and molecular motion, how the solid, liquid and gaseous states differ.

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    • Solid: molecules at fixed lattice positions, separation ≈ one molecular diameter, strong intermolecular forces, vibrating about fixed points; fixed shape and volume
    • Liquid: similar separation but weaker forces, so molecules slide past one another; fixed volume, takes the container's shape
    • Gas: separation ≈ 10 molecular diameters (volume ≈ 10³ × the solid), forces negligible except during collisions, rapid random translational motion; fills the container
    • Intermolecular potential energy rises solid → liquid → gas, while density ρ = m/V falls. Exam tip: "the particles vibrate faster" scores nothing — the marks are for spacing AND relative force strength AND type of motion.
41. Concept/explainSL & HL

    Explain what is meant by internal energy, and identify which contribution changes when a solid is warmed below its melting point and when it then melts.

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    • Internal energy = total random kinetic energy of the molecules + total intermolecular potential energy
    • Warming a solid below its melting point increases the vibration, so the kinetic part rises and the temperature rises
    • The potential part barely changes because the mean separation barely changes
    • During melting the temperature, and hence the average kinetic energy, stays constant
    • All supplied energy becomes potential energy: work is done against intermolecular forces to break up the lattice and increase separation
    • Internal energy still rises at constant temperature. Exam tip: "the energy breaks bonds", without linking to intermolecular POTENTIAL energy or stating that kinetic energy is unchanged, loses the mark.
42. Concept/explainSL & HL

    Explain why the temperature of a pure substance stays constant while it changes phase, even though thermal energy is continuously supplied.

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    • The supplied energy does not increase the average molecular kinetic energy
    • It does work against the intermolecular forces, raising the intermolecular potential energy
    • The mean molecular separation therefore increases
    • Since T ∝ average kinetic energy, T stays constant
    • Internal energy nevertheless increases, so this is not "no energy change"
    • The energy needed per unit mass is the specific latent heat, Q = mL
    • On a heating curve this is a horizontal plateau whose length ∝ L. Exam tip: the commonest lost mark is "the kinetic energy is used to break the bonds" — kinetic energy is unchanged; it is the SUPPLIED energy that becomes potential energy.
43. Concept/explainSL & HL

    Explain why the specific latent heat of vaporisation of a substance is always much larger than its specific latent heat of fusion.

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    • On melting, molecules leave fixed lattice sites but stay in contact, so only some intermolecular bonds break and the separation hardly changes
    • On boiling essentially all intermolecular attraction must be overcome, so far more work is done against these forces
    • Molecules must also be separated to about ten times their diameter, a much larger potential-energy rise
    • The vapour additionally expands, doing work pushing back the atmosphere
    • For water L_f = 3.34 × 10⁵ J kg⁻¹ but L_v = 2.26 × 10⁶ J kg⁻¹, a factor ≈ 6.8. Exam tip: "boiling needs more energy" merely restates the question; marks need work against intermolecular forces PLUS expansion against the atmosphere.
44. Concept/explainSL & HL

    Explain the mechanism of thermal conduction, and why metals conduct far better than non-metals.

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    • Molecules at the hot end vibrate with greater amplitude and kinetic energy
    • They collide with neighbours and pass kinetic energy along, so energy moves without net transfer of matter
    • This lattice-vibration process is slow
    • A metal also contains delocalised free electrons moving rapidly and randomly through the whole lattice
    • Electrons gain kinetic energy at the hot end and travel far before colliding with ions at the cool end, so transfer is much faster
    • Hence k(Cu) ≈ 400 W m⁻¹ K⁻¹ while glass ≈ 0.8 and air ≈ 0.025. Exam tip: say "free/delocalised electrons" explicitly — "metals have more particles" earns nothing.
45. Concept/explainSL & HL

    Explain how convection transfers thermal energy in a fluid, and state why convection cannot occur in a solid.

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    • Fluid touching the hot surface gains energy by conduction and its molecules move further apart
    • The fluid expands, so its density falls (ρ = m/V)
    • Denser surrounding cooler fluid sinks and forces the less dense warm fluid upward
    • The warm fluid loses energy higher up, contracts, becomes denser and sinks, so a convection current circulates
    • Energy is transferred by BULK MOVEMENT of matter, unlike conduction and radiation
    • In a solid the molecules are locked in a lattice and cannot flow, so no convection occurs. Exam tip: "hot air rises" is not an explanation — the mark scheme wants expansion → density decrease → circulating current.
46. Concept/explainSL & HL

    Explain what is meant by a black body and by emissivity, and outline why every object emits thermal radiation.

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    • Every body above 0 K contains accelerating charges, which emit electromagnetic radiation, so all objects radiate continuously
    • A black body is a perfect emitter and absorber: it absorbs all incident radiation at every wavelength and emits the maximum possible power at every wavelength for its temperature
    • Its spectrum depends only on absolute temperature, not on material
    • Emissivity e is the ratio of the power radiated by a surface to that from a black body of equal area and temperature, so 0 ≤ e ≤ 1 and it has no unit
    • P = eσAT⁴, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
    • A good emitter is an equally good absorber. Exam tip: "black body" describes behaviour, not colour.
47. Concept/explainSL & HL

    Explain how the black-body emission spectrum changes as the absolute temperature of the body is increased.

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    • The curve is continuous with a single peak, rising steeply then falling gradually towards long wavelengths
    • The peak shifts to shorter wavelength: λ_max T = 2.9 × 10⁻³ m K, so λ_max ∝ 1/T
    • Intensity increases at EVERY wavelength, so the curves never cross
    • The area under the curve is the power per unit area, which rises as T⁴, so doubling T gives 16 times the power
    • A body at ≈ 800 K glows dull red; at ≈ 5800 K the peak lies in the visible. Exam tip: sketches lose marks for crossing curves, a symmetrical peak, or a curve touching the λ axis at λ = 0 instead of approaching it.
48. Concept/explainSL & HL

    Explain the physical origin of the inverse-square relationship b = L/(4πd²) for the apparent brightness of a star, and state the assumptions made.

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    • Luminosity L is the total power radiated by the star in all directions (W)
    • At distance d this power is spread over a sphere of surface area 4πd²
    • Apparent brightness b is the power received per unit area at Earth, so b = L/(4πd²) and b ∝ 1/d²
    • Doubling the distance quarters the measured brightness
    • Assumptions: emission is isotropic; there is no absorption or scattering by interstellar dust or the atmosphere; d ≫ the star's radius, so it acts as a point source. Exam tip: b is an intensity in W m⁻² and L a power in W — writing b = L/4πd, or quoting b in watts, is a standard lost mark.
49. Concept/explainSL & HL

    Compare conduction, convection and radiation as mechanisms of thermal energy transfer.

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    • Conduction: energy passed by molecular collisions and, in metals, free electrons; no bulk movement of matter; significant mainly in solids; ΔQ/Δt = kAΔT/Δx
    • Convection: energy carried by bulk movement of fluid driven by density differences; only in liquids and gases, never in a solid or a vacuum
    • Radiation: energy carried by electromagnetic waves, mainly infrared at everyday temperatures; needs no medium, so it alone operates through a vacuum; P = eσAT⁴
    • Only radiation depends on T⁴; the others depend on ΔT
    • A vacuum flask defeats all three: the vacuum stops conduction and convection, the silvering (low e) cuts radiation. Exam tip: name the mechanism AND its medium requirement.
50. Concept/explainSL & HL

    Explain, in molecular terms, why evaporation from the surface of a liquid cools the remaining liquid, and how it differs from boiling.

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    • Molecules in a liquid have a range of speeds and hence a distribution of kinetic energies
    • Only the most energetic molecules near the surface can overcome the intermolecular attractions and escape
    • Their loss lowers the average kinetic energy of the molecules left behind
    • Since T ∝ average kinetic energy, the remaining liquid cools
    • Evaporation occurs at any temperature and only at the surface; boiling occurs throughout the liquid at one fixed temperature for a given pressure
    • Evaporation rate rises with temperature, surface area and draught, and falls with humidity. Exam tip: "the hot molecules leave" is not enough — say the AVERAGE kinetic energy of those remaining decreases.
51. Concept/explainSL & HL

    A sample of ice is heated at a constant rate until it becomes steam. Explain the shape of the resulting temperature–time (heating) curve.

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    • Sloping sections are single-phase, where Q = mcΔT and the temperature rises steadily
    • Horizontal plateaus are the phase changes at the melting and boiling points, where Q = mL and T is constant
    • With constant power P, energy supplied ∝ time, so a sloping gradient equals P/(mc), i.e. gradient ∝ 1/c
    • The ice section is steeper than the water section because c_ice ≈ 2100 < c_water ≈ 4200 J kg⁻¹ K⁻¹
    • The boiling plateau is far longer than the melting one because L_v ≫ L_f
    • Curvature at the ends comes from energy exchange with the surroundings. Exam tip: the gradient is INVERSELY proportional to c — "steeper because c is bigger" is a frequent error.
52. Concept/explainSL & HL

    Explain how measurements of a star's radiation allow astronomers to determine its surface temperature and its radius (a Nature of Science link).

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    • A star's spectrum is close to a black-body spectrum, whose shape depends only on absolute temperature
    • Measuring the peak wavelength and applying λ_max T = 2.9 × 10⁻³ m K gives the surface temperature
    • Apparent brightness b is measured directly and the distance d is found independently, for example by parallax
    • Luminosity then follows from L = 4πd²b
    • Stefan–Boltzmann L = σAT⁴ = 4πr²σT⁴ gives the radius r
    • Laws established in the laboratory are thus extrapolated to objects that can never be sampled, and later confirmed by independent methods. Exam tip: quote the chain b → L → r and state the assumption that the star radiates as a black body with e ≈ 1.
53. Worked problemSL & HLData booklet: Yes

    A 0.50 kg block of aluminium is heated from 20 °C to 65 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. Calculate the thermal energy supplied.

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    Principle: with no phase change, Q = mcΔT (data booklet). ΔT = 65 − 20 = 45 °C = 45 K. Q = 0.50 kg × 900 J kg⁻¹ K⁻¹ × 45 K. Q = 0.50 × 900 = 450 J K⁻¹ (the heat capacity of the block); 450 × 45 = 20250 J. Answer: Q = 2.0 × 10⁴ J (2 s.f., matching the 2 s.f. data). Check/Trap: a TEMPERATURE DIFFERENCE has the same numerical value in K and in °C, so there is no need to add 273 here — but never substitute a Celsius temperature into T⁴ or pV = nRT.

54. Worked problemSL & HLData booklet: No – memorise

    The temperature of an ideal gas is raised from 27 °C to 327 °C. Determine the factor by which the average kinetic energy of a molecule increases.

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    Principle: average random kinetic energy of a molecule ∝ absolute temperature T (in kelvin). Convert: T₁ = 27 + 273 = 300 K; T₂ = 327 + 273 = 600 K. Ratio of average kinetic energies = T₂/T₁ = 600/300 = 2.0. Answer: the average kinetic energy doubles (factor 2.0). The r.m.s. speed rises only by √2 ≈ 1.4, since E_k ∝ v². Check/Trap: the classic Paper 1 distractor is 327/27 = 12, obtained by using Celsius. Any proportionality involving temperature — average kinetic energy, gas laws, P = eσAT⁴, Wien's law — demands kelvin. A second distractor is 1.4, which is the speed ratio, not the energy ratio.

55. Worked problemSL & HLData booklet: Yes

    A copper block of mass 0.150 kg at 98.0 °C is dropped into 0.200 kg of water at 18.0 °C in a calorimeter of negligible heat capacity. The final steady temperature is 23.2 °C. Determine the specific heat capacity of copper. (c_water = 4200 J kg⁻¹ K⁻¹)

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    Principle (method of mixtures, energy conservation, no losses): energy lost by copper = energy gained by water, m_Cu c_Cu ΔT_Cu = m_w c_w ΔT_w. Water: ΔT_w = 23.2 − 18.0 = 5.2 K, so Q = 0.200 × 4200 × 5.2 = 4368 J. Copper: ΔT_Cu = 98.0 − 23.2 = 74.8 K, so m_Cu ΔT_Cu = 0.150 × 74.8 = 11.22 kg K. c_Cu = 4368 / 11.22 = 389.3 J kg⁻¹ K⁻¹. Answer: c_Cu = 3.9 × 10² J kg⁻¹ K⁻¹ (2 s.f.; accepted value 385). Check/Trap: use the block's own fall (98.0 → 23.2), not 98.0 − 18.0. Thermal energy lost while transferring the block makes the measured c too SMALL.

56. Worked problemSL & HLData booklet: Yes

    Calculate the total energy required to convert 0.0200 kg of ice at −10.0 °C into steam at 100 °C. (c_ice = 2100, c_water = 4200 J kg⁻¹ K⁻¹; L_f = 3.34 × 10⁵, L_v = 2.26 × 10⁶ J kg⁻¹)

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    Principle: split into four stages, Q = mcΔT within a phase and Q = mL at each change. Stage 1, ice −10.0 → 0 °C: Q₁ = 0.0200 × 2100 × 10.0 = 420 J. Stage 2, melting at 0 °C: Q₂ = 0.0200 × 3.34 × 10⁵ = 6680 J. Stage 3, water 0 → 100 °C: Q₃ = 0.0200 × 4200 × 100 = 8400 J. Stage 4, vaporising at 100 °C: Q₄ = 0.0200 × 2.26 × 10⁶ = 45200 J. Total Q = 420 + 6680 + 8400 + 45200 = 60700 J. Answer: Q = 6.07 × 10⁴ J (3 s.f.). Check/Trap: never use ΔT = 110 in one step — that ignores both latent heats. Q₄ alone is 74 % of the total, which is why the boiling plateau dominates the heating curve.

57. Worked problemSL & HLData booklet: Yes

    An immersion heater of constant power 50.0 W is placed in a well-lagged aluminium block of mass 0.800 kg. In 300 s the temperature rises from 19.0 °C to 39.8 °C. Determine the specific heat capacity of aluminium and comment on the result.

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    Principle: electrical energy supplied = Pt = thermal energy gained = mcΔT. Energy: E = 50.0 W × 300 s = 1.50 × 10⁴ J. Temperature rise: ΔT = 39.8 − 19.0 = 20.8 K. Rearranged: c = Pt/(mΔT) = 15000 / (0.800 × 20.8) = 15000 / 16.64 = 901.4 J kg⁻¹ K⁻¹. Answer: c = 9.0 × 10² J kg⁻¹ K⁻¹ (2 s.f.); accepted value 900 J kg⁻¹ K⁻¹. Comment: because energy is lost to the surroundings and some warms the heater and thermometer, this method normally OVERestimates c. Check/Trap: quote c per KELVIN, and ensure good thermal contact — a delayed reading of the peak temperature makes ΔT too small and c too large.

58. Worked problemSL & HLData booklet: Yes

    Water is boiled by a 60.0 W heater in a vessel standing on an electronic balance. Once steady conditions are reached the mass falls by 15.4 g in 600 s. Determine the specific latent heat of vaporisation of water.

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    Principle: in the steady state all the electrical energy supplied vaporises water (temperature constant at 100 °C, so no mcΔT term), Q = mL. Energy supplied: E = Pt = 60.0 × 600 = 3.60 × 10⁴ J. Mass vaporised: m = 15.4 g = 1.54 × 10⁻² kg. L_v = E/m = 36000 / 0.0154 = 2.338 × 10⁶ J kg⁻¹. Answer: L_v = 2.34 × 10⁶ J kg⁻¹ (3 s.f.; accepted 2.26 × 10⁶). Check/Trap: take readings only AFTER boiling is steady, or part of the energy is still raising the water's temperature. Losses to the surroundings make L too large; repeating at two powers, L = (P₁ − P₂)/(Δm₁/Δt − Δm₂/Δt), cancels the constant loss.

59. Worked problemSL & HLData booklet: Yes

    A single glass window pane has area 1.80 m² and thickness 4.00 mm. Its inner surface is at 18.0 °C and its outer surface at 13.0 °C. The thermal conductivity of glass is 0.80 W m⁻¹ K⁻¹. Determine the rate of thermal energy transfer through the pane.

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    Principle: steady-state conduction, ΔQ/Δt = kAΔT/Δx (data booklet). Substitute in SI units: k = 0.80 W m⁻¹ K⁻¹, A = 1.80 m², ΔT = 18.0 − 13.0 = 5.0 K, Δx = 4.00 mm = 4.00 × 10⁻³ m. Numerator: 0.80 × 1.80 × 5.0 = 7.2 W m. Divide: 7.2 / (4.00 × 10⁻³) = 1800 W. Answer: ΔQ/Δt = 1.8 × 10³ W (2 s.f.), i.e. 1.8 kW. Check/Trap: convert millimetres to metres — leaving Δx = 4.00 gives 1.8 W, out by 10³. ΔT is a difference, so kelvin and Celsius are interchangeable. Use the SURFACE temperatures, not the room and outdoor air temperatures.

60. Worked problemSL & HLData booklet: Yes

    The window above is replaced by double glazing: two 4.00 mm glass panes separated by a 6.00 mm layer of still air (k_air = 0.025 W m⁻¹ K⁻¹). Estimate the rate of energy transfer, assuming almost all of the 5.0 K temperature difference now appears across the air layer.

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    Principle: the layers conduct in series, so the layer with the smallest k/Δx (largest thermal resistance) controls the rate. Using the air gap: ΔQ/Δt = kAΔT/Δx = 0.025 × 1.80 × 5.0 / (6.00 × 10⁻³). Numerator: 0.025 × 1.80 = 0.045; × 5.0 = 0.225 W m. Divide: 0.225 / (6.00 × 10⁻³) = 37.5 W. Answer: ≈ 38 W (2 s.f.), about 48 times less than the 1.8 kW single pane. Check/Trap: the assumption holds because k_air is ≈ 32 times smaller than k_glass, so nearly all the temperature drop is across the air. The gap must be only a few mm so that CONVECTION is suppressed, or the insulation is far worse than this estimate.

61. Worked problemSL & HLData booklet: Yes

    A person has exposed skin of area 1.60 m² at 33 °C and emissivity 0.90, and stands in a room whose walls are at 19 °C. Determine the net rate at which the person loses energy by radiation.

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    Principle: net radiated power P = eσA(T_body⁴ − T_surroundings⁴), from P = eσAT⁴ (booklet). Convert to kelvin: T_body = 33 + 273 = 306 K; T_surr = 19 + 273 = 292 K. Fourth powers: 306⁴ = 8.768 × 10⁹ K⁴; 292⁴ = 7.270 × 10⁹ K⁴; difference = 1.498 × 10⁹ K⁴. Prefactor: eσA = 0.90 × 5.67 × 10⁻⁸ × 1.60 = 8.165 × 10⁻⁸ W K⁻⁴. P = 8.165 × 10⁻⁸ × 1.498 × 10⁹ = 122 W. Answer: P ≈ 1.2 × 10² W (2 s.f.). Check/Trap: kelvin is essential — using 33⁴ − 19⁴ is wrong by a factor ≈ 10⁶. Also (306 − 292)⁴ is NOT 306⁴ − 292⁴; take the difference of the fourth powers.

62. Worked problemSL & HLData booklet: Yes

    The Sun's radiation spectrum peaks at a wavelength of 5.0 × 10⁻⁷ m. Assuming the Sun behaves as a black body, determine its surface temperature.

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    Principle: Wien's displacement law, λ_max T = 2.9 × 10⁻³ m K (data booklet). Rearrange: T = 2.9 × 10⁻³ / λ_max. Substitute: T = 2.9 × 10⁻³ m K / (5.0 × 10⁻⁷ m). Mantissas: 2.9/5.0 = 0.58; exponents: 10⁻³/10⁻⁷ = 10⁴. T = 0.58 × 10⁴ = 5.8 × 10³ K. Answer: T = 5.8 × 10³ K (5800 K, 2 s.f.). Check/Trap: the constant is in m K, so λ_max must be in METRES — entering 500 nm as 500 gives 5.8 × 10⁻⁶ K. Sanity check: a peak in the middle of the visible range should give a few thousand kelvin; hotter stars peak in the blue/UV (smaller λ_max), cooler stars in the red/infrared.

63. Worked problemSL & HLData booklet: Yes

    A star has luminosity 4.0 × 10²⁸ W and surface temperature 4500 K. Assuming it radiates as a black body, determine its radius.

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    Principle: L = σAT⁴ with A = 4πr² for a sphere, so L = 4πr²σT⁴ (booklet, e = 1). Fourth power: 4500² = 2.025 × 10⁷, so 4500⁴ = (2.025 × 10⁷)² = 4.101 × 10¹⁴ K⁴. Power per unit area: σT⁴ = 5.67 × 10⁻⁸ × 4.101 × 10¹⁴ = 2.325 × 10⁷ W m⁻². Surface area: A = L/σT⁴ = 4.0 × 10²⁸ / (2.325 × 10⁷) = 1.720 × 10²¹ m². Radius: r² = A/4π = 1.720 × 10²¹ / 12.57 = 1.369 × 10²⁰ m², so r = 1.17 × 10¹⁰ m. Answer: r = 1.2 × 10¹⁰ m (2 s.f.), about 17 solar radii. Check/Trap: A = 4πr² is the whole surface — using πr², the cross-section used for ABSORPTION, doubles r.

64. Worked problemSL & HLData booklet: Yes

    A star of luminosity 2.4 × 10²⁸ W lies 3.2 × 10¹⁸ m from Earth. Determine its apparent brightness as measured above the Earth's atmosphere.

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    Principle: the power spreads over a sphere of radius d, so b = L/(4πd²) (data booklet). Square the distance: d² = (3.2 × 10¹⁸)² = 1.024 × 10³⁷ m². Sphere area: 4πd² = 4π × 1.024 × 10³⁷ = 1.287 × 10³⁸ m². Divide: b = 2.4 × 10²⁸ / (1.287 × 10³⁸) = 1.865 × 10⁻¹⁰ W m⁻². Answer: b = 1.9 × 10⁻¹⁰ W m⁻² (2 s.f.). Check/Trap: b is an intensity in W m⁻², never in W. Because b ∝ 1/d², a star twice as far away with the same luminosity appears one QUARTER as bright, not half. The result assumes isotropic emission and no absorption by interstellar dust, which would reduce the measured b.

65. Worked problemSL & HLData booklet: Yes

    Steam at 100 °C has density 0.60 kg m⁻³ and molar mass 0.018 kg mol⁻¹. Estimate the average separation of its molecules and compare it with the value ≈ 3.1 × 10⁻¹⁰ m for liquid water.

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    Principle: find the number of molecules per unit volume, then treat each as occupying a cube of side d, so n = 1/d³. Moles per cubic metre: ρ/M = 0.60 / 0.018 = 33.3 mol m⁻³. Molecules per cubic metre: n = 33.3 × 6.02 × 10²³ = 2.01 × 10²⁵ m⁻³. Volume per molecule: 1/n = 4.98 × 10⁻²⁶ m³. Separation: d = (4.98 × 10⁻²⁶)^(1/3) = 3.7 × 10⁻⁹ m. Answer: d ≈ 3.7 × 10⁻⁹ m, about 12 times the liquid separation, so the vapour occupies roughly 12³ ≈ 1.7 × 10³ times the volume. Check/Trap: take the CUBE root of the volume per molecule, not of n. This factor of ≈ 10 in spacing justifies neglecting intermolecular forces in a gas.

66. Worked problemSL & HLData booklet: Yes

    0.30 kg of water at 80 °C is mixed with 0.50 kg of water at 20 °C in an insulated container of negligible heat capacity. Determine the final temperature of the mixture.

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    Principle: energy conservation — energy lost by the hot water = energy gained by the cold water. Since both are water, c cancels: m₁(θ₁ − θ) = m₂(θ − θ₂). Substitute: 0.30(80 − θ) = 0.50(θ − 20). Expand: 24 − 0.30θ = 0.50θ − 10. Collect: 24 + 10 = 0.80θ, so 34 = 0.80θ and θ = 34/0.80 = 42.5 °C. Answer: θ = 43 °C (2 s.f.). Check/Trap: the answer must lie between 20 °C and 80 °C, and nearer 20 °C because the cold mass is larger — the mid-point 50 °C is the standard distractor from ignoring the masses. Because c cancels, its value is not needed; a container of heat capacity C would lower θ.

67. Exam technique/trapSL & HL

    Distinguish what the mark scheme expects from the command terms state, outline, explain, determine, deduce, sketch, suggest and compare in B.1 thermal questions.

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    Trap: writing the same answer whatever the verb. State = a bare fact, no reasoning. Outline = a brief account of the main points. Explain = give REASONS, so every marking point needs a "because/therefore". Determine = obtain a value, showing equation, working and unit. Deduce = reach a stated conclusion from the given data plus reasoning. Sketch = a labelled qualitative shape with correct intercepts, asymptotes and relative positions, axes labelled with quantity and unit. Suggest = propose a plausible reason in an unfamiliar situation. Compare = give similarities AND differences using linking words such as "whereas"; two separate descriptions score nothing. Exam tip: the number of marks equals the number of distinct points needed, so match points to marks.

68. Exam technique/trapSL & HLData booklet: No – memorise

    Identify the temperature-unit traps in B.1 and state the rule for deciding when a Celsius value must be converted to kelvin.

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    Trap: students convert everywhere or nowhere. Rule: a temperature DIFFERENCE has the same numerical value in K and °C, so Q = mcΔT and ΔQ/Δt = kAΔT/Δx work with either. Any expression containing an ABSOLUTE temperature needs kelvin: P = eσAT⁴, λ_max T = 2.9 × 10⁻³ m K, average kinetic energy ∝ T, and pV = nRT. Why students fall for it: ΔT questions are taught first, so the habit of ignoring the 273 carries over. Correct approach: apply T(K) = θ(°C) + 273 the moment a lone T appears. A 33 °C body radiating as 33⁴ rather than 306⁴ is wrong by ≈ 10⁶, so sanity-check every radiation answer.

69. Exam technique/trapSL & HL

    Describe an experiment to determine the specific heat capacity of a metal block by the electrical method, including the variables, the main limitation and two improvements.

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    Apparatus: metal block of known mass with two drilled holes, immersion heater, temperature probe and datalogger, joulemeter (or ammeter, voltmeter and stopwatch), lagging, balance. Method: measure m; insert heater and thermometer with a drop of oil for thermal contact; record θ₀, switch on, record V, I and t. Independent variable time (or energy supplied); dependent variable temperature; controlled: power, mass, lagging, room temperature. Analysis: c = VIt/(mΔT), or plot energy against θ and take c = 1/(m × gradient). Limitation: energy lost to the surroundings and stored in the heater makes ΔT too small and c too LARGE. Improvements: lag thoroughly and start as far below room temperature as the finish is above, so gains and losses cancel; extrapolate the cooling curve back to switch-off.

70. Exam technique/trapSL & HL

    Describe how to determine the specific latent heat of fusion of ice by the method of mixtures, and identify the systematic errors and how to remove them.

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    Method: measure the mass and initial temperature of warm water in a lagged calorimeter, starting ≈ 10 K above room temperature. Dry small pieces of melting ice at 0 °C on filter paper, add them, stir until all have melted, record the lowest steady temperature, then reweigh to find the ice mass m_i. Analysis: m_i L_f + m_i c_w θ_f = (m_w c_w + C_cal)(θ_i − θ_f). Controlled: same calorimeter and lagging, ice always at 0 °C. Systematic errors: wet ice overstates m_i, making L_f too small; ice below 0 °C raises L_f; energy gained from the room after the minimum lowers L_f. Removals: dry the ice, run a control, and start above and finish below room temperature so heat exchange cancels.

71. Exam technique/trapSL & HL

    Explain how to handle uncertainties when a specific heat capacity is calculated from c = Pt/(mΔT), including which measurement usually dominates.

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    Rule: for a quantity formed only by multiplication and division, ADD the fractional (percentage) uncertainties: Δc/c = ΔP/P + Δt/t + Δm/m + Δ(ΔT)/ΔT. Trap: ΔT is a DIFFERENCE of two readings, so their ABSOLUTE uncertainties add first — two readings each ±0.5 °C give ΔT = 20.8 ± 1.0 K, i.e. 4.8 %. This term usually dominates, since m (±0.001 kg on 0.800 kg = 0.1 %) and t are far more precise. Correct approach: quote the absolute uncertainty to 1 significant figure and round the value to the same decimal place: c = (9.0 ± 0.5) × 10² J kg⁻¹ K⁻¹. Improvement: heat for longer to raise ΔT and cut the dominant percentage uncertainty.

72. Exam technique/trapSL & HL

    Distinguish systematic from random errors, and precision from accuracy, in calorimetry experiments, and state how each is detected and reduced.

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    Random error: unpredictable scatter — reading a thermometer at slightly different times, parallax, uneven stirring. It shows as scatter about the best-fit line and is reduced by repeating and averaging or by using a datalogger. Systematic error: a fixed one-way bias — heat loss to the surroundings, a thermometer zero offset, the neglected heat capacity of the calorimeter. It shows as a non-zero intercept where the line should pass through the origin; repeating does NOT reduce it, so calibrate, lag, or use the two-power or cooling-correction technique. Precision = small random uncertainty; accuracy = closeness to the true value. Exam tip: tightly clustered readings from a badly lagged calorimeter are precise but inaccurate — never infer accuracy from consistency.

73. Exam technique/trapSL & HL

    State the common errors made when sketching or interpreting heating curves, cooling curves and black-body spectra in Paper 2, and how to avoid them.

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    Heating curve: label both axes with quantity and unit; plateaus must be exactly horizontal at the melting and boiling points; the boiling plateau must be clearly longer than the melting one because L_v ≫ L_f; the water section must be LESS steep than the ice section, since gradient = P/(mc) and c_water > c_ice. Trap: a sloping plateau, which would mean the temperature rises during a phase change. Black-body curves: continuous, single maximum, never crossing, with the hotter curve above the cooler at every wavelength and its peak at shorter λ. Cooling curve: an exponential-type decay approaching room temperature asymptotically, never reaching or crossing it. Exam tip: unlabelled axes forfeit an easy mark.

74. Exam technique/trapSL & HL

    Explain how to linearise thermal-physics data so that a straight-line graph can be used, giving two B.1 examples and what the gradient yields.

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    Approach: rearrange into y = mx + c so that the gradient gives the unknown. Example 1 — electrical determination of c: plot energy supplied E = VIt against temperature θ. Gradient = mc, so c = gradient/m, and a non-zero intercept exposes energy lost before timing began. Example 2 — Stefan–Boltzmann: since P ∝ T⁴, plot P against T⁴ (T in kelvin) for a straight line through the origin of gradient eσA; alternatively plot ln P against ln T, whose gradient should be 4.00, testing the power law itself. Trap: plotting P against T gives a curve from which nothing can be extracted. Always add error bars and use maximum and minimum gradients to obtain the uncertainty.

75. Exam technique/trapSL & HL

    Explain the distinction the mark scheme insists on between heat, temperature and internal energy, and how to phrase an answer that scores full marks.

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    Trap: using the three words interchangeably, e.g. "the ice absorbs heat so its internal temperature rises". Use: temperature is a measure of the average random kinetic energy of the molecules (K, intensive); internal energy is the total kinetic plus potential energy of all the molecules (J, extensive); thermal energy (heat) is energy transferred because of a temperature difference — energy in transit, so a body does not "contain heat". Full-mark phrasing for a phase change: "thermal energy is supplied, internal energy increases because molecular potential energy increases, but temperature is constant because the average kinetic energy is unchanged." Why students fall for it: everyday language equates heat with temperature. Watch for it in Paper 1 items comparing a hot pin with a warm bath.

Practise this topic with exam-style questions: B.1 Thermal energy transfers questions (SL) · B.1 Thermal energy transfers questions (HL) · all flashcards

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