Pen colours: blue = objects and particles, red = laws, energy flows and key steps, green = answers and results, pencil = construction lines. Boxes: Why? = the reasoning, Mistake = what students get wrong, yellow note = exam tip.
In one line: everything is made of moving particles; temperature tells us how fast they move on average, and thermal energy always flows from hot to cold.
Thermal energy flows from the hotter body to the colder body until both reach the same temperature.
This topic has three parts. First, a particle model of matter: what temperature and internal energy mean for the molecules. Second, what thermal energy does to a body: it warms it up (Q = mcΔT) or changes its phase (Q = mL). Third, how thermal energy travels: conduction, convection and radiation. Radiation leads to stars: their power, their colour and how bright they look from Earth.
Data used in these notes: kB = 1.38 × 10⁻²³ J K⁻¹ · σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ · Wien constant 2.9 × 10⁻³ m K · c(water) = 4200 J kg⁻¹ K⁻¹ · c(ice) = 2100 J kg⁻¹ K⁻¹ · Lf(ice) = 334 kJ kg⁻¹ · Lv(water) = 2.26 MJ kg⁻¹.
The same particles in three phases. What changes is how far apart they are, how strongly they pull on each other and how they move.
Solid
Liquid
Gas
Arrangement
regular, close together
irregular, close together
far apart (about 10 times the spacing in a solid)
Motion
vibrate about fixed positions
move around and slide past each other
move fast and randomly in straight lines between collisions
Forces between particles
strong
weaker, but keep the particles close
almost zero (only during collisions)
Shape and volume
fixed shape and volume
takes the shape of the container; fixed volume
fills the whole container; easy to compress
Density:ρ = m / V · unit kg m⁻³ Solids and liquids have similar densities (iron: 7.9 × 10³ kg m⁻³ solid, 7.0 × 10³ kg m⁻³ liquid). A gas is about 1000 times less dense (air ≈ 1.2 kg m⁻³), because its particles are about 10 times further apart (10³ in volume).
Mistake “A big block of copper has a bigger density than a small one.” Density is mass per unit volume: it depends on the material (and its temperature), not on the size of the block.
The Celsius and Kelvin scales use the same size of degree; only the zero is different.
T (K) = θ (°C) + 273 A change of temperature is the same number on both scales: ΔT in K = Δθ in °C.
Learn it word for word. Temperature (in kelvin) is a measure of the average kinetic energy of the particles: E̅k = (3/2) kB T, where kB is the Boltzmann constant and T is in kelvin.
So absolute zero (0 K = −273 °C) is the temperature at which the particles have their lowest possible kinetic energy. Nothing can be colder. Doubling the kelvin temperature doubles the average kinetic energy; doubling the Celsius temperature does not.
Worked example 1The temperature of a cup of tea falls from 318 K to 295 K. (a) Give both temperatures in °C. (b) State the temperature drop in °C.
(a) 318 − 273 = 45 °C; 295 − 273 = 22 °C.
(b) Drop = 318 − 295 = 23 K, and a change in kelvin equals the change in °C: 23 °C. (Check: 45 − 22 = 23.)
Worked example 2Find the average kinetic energy of the molecules of a gas at 27 °C. What is it at 327 °C? Helium and neon are at the same temperature: compare their average kinetic energies.
Kelvin first: T = 27 + 273 = 300 K.
E̅k = (3/2) kB T = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
327 °C = 600 K: the kelvin temperature doubles, so E̅k doubles: 1.24 × 10⁻²⁰ J.
Same temperature → same average kinetic energy, even though a neon atom is heavier. The lighter helium atoms therefore move faster on average.
Mistake Using °C in E̅k = (3/2)kBT, or saying “twice the temperature in °C means twice the kinetic energy”. Always change to kelvin first.
The potential energy of two molecules against their separation. Moving them apart (melting, boiling) increases the potential energy.
Particles have two kinds of energy. They move randomly, so they have random kinetic energy. They pull on each other (like tiny springs between them), so they have intermolecular potential energy. In a solid this potential energy is negative; it rises towards zero as the particles move apart. For an ideal gas it is zero.
Learn it word for word. Internal energy of a system: 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.
Exam tip Always say whose energy it is: the energy of the molecules (or atoms, or particles). Then write total (not average), random kinetic energy and intermolecular potential energy. “Kinetic energy plus potential energy”, with no molecules, atoms or particles, scores 0.
Does a gas on a fast plane have more internal energy? No. Internal energy uses the random motion of the molecules relative to each other. The plane moves all the molecules together; that ordered motion is not part of the internal energy. Same gas, same temperature → same internal energy.
Word
What it is
Unit
Temperature
measures the average kinetic energy of one particle
K or °C
Internal energy
total random KE + total intermolecular PE of all the particles
J
Thermal energy (heat)
energy transferred because of a temperature difference
J
Mistake “A hot body has more internal energy than a cold body.” Not always: a bath of warm water has far more internal energy than a hot spark, because it has many more particles. Temperature is about the average per particle; internal energy is the total.
Thermal energy and thermal equilibrium
Put a hot body in contact with a cold one. Energy is transferred from the hot body to the cold body, never the other way, until they reach the same temperature. Then there is no more net transfer: the bodies are in thermal equilibrium.
Learn it word for word. Thermal energy (heat) is the energy transferred from one body to another because of a temperature difference. The temperature difference decides the direction of the net transfer: from higher to lower temperature.
The energy given out by the hot body equals the energy taken in by the cold body (if none is lost to the surroundings). This is the idea behind every “mixing” calculation: energy lost = energy gained.
Mistake “Heat is stored in a hot body.” A body has internal energy; heat is energy on the move between bodies. The word “capacity” in specific heat capacity does not mean the body holds heat like a bottle holds water.
Learn it word for word. Specific heat capacity c: the energy required to raise the temperature of 1 kg of a substance by 1 K.
Q = m c ΔT · Q in J, m in kg, ΔT in K (or °C), c in J kg⁻¹ K⁻¹ Water: 4200 · ice: 2100 · aluminium: 900 · iron: 440 · copper: 385 · lead: 130 J kg⁻¹ K⁻¹
A large c means the substance needs a lot of energy to warm up, and it gives out a lot of energy as it cools. Water has a very large c: this is why the sea warms and cools slowly, and why water is used to carry energy in radiators.
Worked example 3(a) 6600 J of thermal energy is given to a 0.40 kg block of aluminium. Find its temperature rise. (b) 6600 J is removed from 0.40 kg of water at 60 °C. Find its final temperature.
(a) ΔT = Q/(mc) = 6600/(0.40 × 900) = 18 K. (We do not need the starting temperature.)
(b) ΔT = 6600/(0.40 × 4200) = 3.9 K, so the water ends at 60 − 3.9 ≈ 56 °C.
Same energy, same mass, but water changes temperature about 4.7 times less: its c is 4.7 times bigger.
Worked example 4A 0.30 kg copper block at 250 °C is dropped into 0.80 kg of water at 18 °C. Find the final temperature. Ignore the container and any losses.
Let the final temperature be θ. Copper cools from 250 to θ; water warms from 18 to θ.
Energy lost by copper = energy gained by water: 0.30 × 385 × (250 − θ) = 0.80 × 4200 × (θ − 18).
Check: the final temperature is between 18 °C and 250 °C, and much closer to the water’s temperature, because the water has much more mass × c.
Method for every mixing problem 1. Write “energy lost = energy gained”. 2. For each body write mcΔT with ΔT as (higher − lower), so every term is positive. 3. Solve for the unknown. 4. Check the answer lies between the starting temperatures.
A heater of known power P warms the sample for a time t, so the energy supplied is Q = Pt (or IVt). Measure the temperature rise, then c = Pt/(mΔT). Stir, insulate the container, and include the energy taken by the container if it is not negligible.
Worked example 5: the heating curve of a 150 g sample heated by a heater of constant power.
Worked example 5A 150 g sample of a solid at 15 °C is heated by a heater of constant power. Its specific heat capacity as a solid is 2000 J kg⁻¹ K⁻¹. The graph shows how its temperature changes with time. Use the graph to find (a) the power of the heater, (b) the melting point, (c) the specific latent heat of fusion, (d) the specific heat capacity of the liquid.
Read the graph first: three parts. 0 → 100 s: the solid warms from 15 °C to 55 °C. 100 → 500 s: flat, the solid melts. 500 → 550 s: the liquid warms from 55 °C to 65 °C.
Key idea: the heater gives the same energy every second, so on every part of the graph the energy supplied is Q = P × (time of that part).
(a) Power (use the part where we know everything: the solid warming). Q = mcΔT = 0.150 × 2000 × (55 − 15) = 12 000 J in 100 s, so P = Q/t = 12 000/100 = 120 W.
(b) Melting point = the temperature of the flat part = 55 °C.
(c) Latent heat. Time on the flat part: 500 − 100 = 400 s. Energy: Q = Pt = 120 × 400 = 48 000 J. Then Q = mLf gives Lf = 48 000/0.150 = 3.2 × 10⁵ J kg⁻¹.
(d) c of the liquid. Time: 550 − 500 = 50 s, so Q = 120 × 50 = 6000 J. Temperature rise: 65 − 55 = 10 K. c = Q/(mΔT) = 6000/(0.150 × 10) = 4.0 × 10³ J kg⁻¹ K⁻¹.
Reading a heating curveFlat part → phase change: use Q = mL with Q = P × (length of the flat part). Sloping part → one phase warming: use Q = mcΔT with Q = P × time. The gradient of a sloping part is ΔT/Δt = P/(mc): the steeper the line, the smaller c. Here the liquid line (10 K in 50 s) is less steep than the solid line (40 K in 100 s), so the liquid has the larger c.
Worked example 6A 20 W heater runs for 4.0 minutes in 0.30 kg of a liquid held in a 0.10 kg copper cup (c = 385 J kg⁻¹ K⁻¹). The temperature of both rises by 6.0 K. Find the specific heat capacity of the liquid. Is the true value larger or smaller than your answer?
Energy supplied: Q = Pt = 20 × 240 = 4800 J.
Energy taken by the cup: 0.10 × 385 × 6.0 = 231 J. The rest goes to the liquid: 4800 − 231 = 4569 J.
c = 4569/(0.30 × 6.0) = 2.5 × 10³ J kg⁻¹ K⁻¹.
Some energy escapes to the room, so less than 4569 J really went into the liquid: the true c is smaller than calculated.
2. The method of mixtures
The method of mixtures. The metal must be moved quickly so it loses little energy on the way.
Heat a metal sample in boiling water (so its temperature is known), move it quickly into an insulated calorimeter of cold water, stir, and record the highest temperature reached. Then use energy lost = energy gained.
Worked example 7A 0.50 kg metal block at 95 °C is put into a 0.12 kg aluminium calorimeter holding 0.60 kg of water at 18 °C. The final temperature is 24 °C. Find c for the metal, and say how energy losses affect the answer.
Energy gained by the cup and the water (both warm by 6 K): 0.12 × 900 × 6 + 0.60 × 4200 × 6 = 648 + 15 120 = 15 768 J.
Energy lost by the metal (cools by 95 − 24 = 71 K): 0.50 × c × 71 = 35.5c.
35.5c = 15 768 → c = 444 J kg⁻¹ K⁻¹.
The block cools a little in the air on the way, so it really entered the water below 95 °C. Its true temperature drop is smaller, so the true c is larger than calculated.
liquid → gas, throughout the liquid, at the boiling point
taken in
condensing
gas → liquid
given out
evaporation
liquid → gas, only at the surface, at any temperature
taken in (the liquid cools)
Learn it word for word. A phase change happens at constant temperature. The energy supplied changes the intermolecular potential energy (the particles move further apart, bonds are broken), not the average random kinetic energy. So the temperature stays the same, but the internal energy increases.
Why does freezing give energy out? When a liquid freezes, bonds between the molecules form. Forming bonds releases energy: the intermolecular potential energy decreases and this energy is transferred to the surroundings. The average kinetic energy does not change, so the temperature stays at the freezing point.
Why does evaporation cool a liquid? Only the fastest molecules near the surface have enough energy to escape. They take a lot of kinetic energy with them, so the average kinetic energy of the molecules left behind falls, and so does the temperature. Evaporation is faster at a higher temperature, with a larger surface area, and when air blows the escaped molecules away.
Learn it word for word. Specific latent heat L: the energy required to change the phase of 1 kg of a substance at constant temperature. Lf (fusion) is for melting or freezing; Lv (vaporisation) is for boiling or condensing.
Q = m L · L in J kg⁻¹ · water: Lf = 334 kJ kg⁻¹, Lv = 2260 kJ kg⁻¹
Lv is about 7 times larger than Lf. In melting the molecules move only a little further apart; in boiling they are pulled completely apart, so much more work is done against the forces between them.
Worked example 8To measure Lf of ice: 20 g of ice at 0 °C is dropped into a 200 g aluminium cup holding 250 g of water at 26 °C. The lowest temperature reached is 19 °C. Find Lf.
Energy lost by the cup and water (cool by 7 K): (0.20 × 900 + 0.25 × 4200) × 7 = 1230 × 7 = 8610 J.
Energy gained by the ice: melting, 0.020 × Lf, then the melted water warms from 0 to 19 °C: 0.020 × 4200 × 19 = 1596 J.
Tips: the ice must be at 0 °C (take it from a mixture of ice and water) and dried before it goes in; find its mass by weighing the cup again at the end.
Worked example 9 on one temperature line: the ice takes in energy in three stages; the water gives out energy as it cools.
Worked example 9(a) An ice cube of mass 30 g at −12 °C is dropped into 250 g of water at 25 °C. Find the final temperature. (b) The same is done with 120 g of ice at −12 °C. Find the final temperature and the mass of ice that melts. Ignore the container. cice = 2100, cwater = 4200 J kg⁻¹ K⁻¹, Lf = 334 kJ kg⁻¹.
Masses in kg. Ice: 0.030 kg. Water: 0.250 kg. Call the final temperature θ.
Energy taken in by the ice, stage by stage. ① ice warms from −12 °C to 0 °C: 0.030 × 2100 × 12 = 756 J ② ice melts at 0 °C: 0.030 × 334 000 = 10 020 J ③ the melted ice (now water) warms from 0 °C to θ: 0.030 × 4200 × θ = 126θ
Check that all the ice melts. To get to ②, the ice needs 756 + 10 020 = 10 776 J. Cooling the water all the way to 0 °C would give 0.250 × 4200 × 25 = 26 250 J. 26 250 > 10 776, so yes: all the ice melts and θ is above 0 °C.
Energy given out by the water as it cools from 25 °C to θ: 0.250 × 4200 × (25 − θ) = 26 250 − 1050θ.
Energy lost = energy gained. 26 250 − 1050θ = 756 + 10 020 + 126θ.
Solve. Numbers on one side, θ on the other: 26 250 − 10 776 = 1050θ + 126θ → 15 474 = 1176θ → θ = 13 °C. Check: between 0 °C and 25 °C ✓.
(b) Do the check again with 120 g: warming the ice needs 0.120 × 2100 × 12 = 3024 J, and melting all of it needs 0.120 × 334 000 = 40 080 J: 43 104 J in total. The water can give only 26 250 J (by cooling to 0 °C). Not all the ice melts.
So the final temperature is 0 °C (ice and water together). Energy left for melting after warming the ice: 26 250 − 3024 = 23 226 J. Mass melted = 23 226/334 000 = 0.070 kg = 70 g (50 g of ice is left).
Stage
Equation
Energy in (a)
① ice warms to 0 °C
mice cice (0 − θice)
756 J
② ice melts
mice Lf
10 020 J
③ melted ice warms to θ
mice cwater (θ − 0)
126θ
water cools to θ
mwater cwater (θwater − θ)
26 250 − 1050θ
Always do the check first If the water cannot give enough energy to melt all the ice, the answer is 0 °C and some ice is left. Using the full equation anyway gives a negative θ, which is impossible for melted ice.
Worked example 10Steam at 100 °C is bubbled into 0.50 kg of water at 20 °C until the water reaches 60 °C. What mass of steam condenses? Why is a burn from steam worse than one from boiling water?
The steam condenses (gives out mLv), then the condensed water cools from 100 to 60 °C (gives out m × 4200 × 40).
The water gains 0.50 × 4200 × 40 = 84 000 J.
m(2.26 × 10⁶ + 4200 × 40) = 84 000 → m = 84 000/2 428 000 = 0.035 kg = 35 g.
Steam on the skin first condenses, giving out its large latent heat (35 g gives about 79 kJ) before it even starts to cool. Boiling water only gives out mcΔT.
Mistake Forgetting a step in a phase problem. Ice at −12 °C needs three terms (warm the ice, melt it, warm the water), and you must check that all of it melts.
Thermal energy moves through a material from the hot end to the cold end, with no movement of the material itself.
Learn it word for word. Conduction: the particles at the hot end have a higher average kinetic energy. They pass kinetic energy to their neighbours in collisions (and through vibrating bonds), so energy is transferred from the hot side to the cold side. In metals, free electrons move through the metal and carry energy quickly, so metals are good conductors.
Left: the quantities in the conduction equation. Right: two rods in series (Worked example 11b).
ΔQ/Δt = k A ΔT/Δx ΔQ/Δt: rate of energy transfer (W) · k: thermal conductivity (W m⁻¹ K⁻¹) · A: cross-sectional area · ΔT/Δx: temperature gradient (temperature difference ÷ thickness) Double the area → double the rate · double the thickness → half the rate · copper k ≈ 400, glass ≈ 0.8, air ≈ 0.025 W m⁻¹ K⁻¹
Worked example 11(a) A wall of area 15 m² and thickness 0.25 m has k = 0.80 W m⁻¹ K⁻¹. Inside is 21 °C, outside −4 °C. Find the rate of energy loss. (b) Two rods of the same length and area, with conductivities k and 4k, are joined end to end. The free ends are kept at 120 °C and 20 °C. Find the temperature at the joint.
(a) ΔT = 21 − (−4) = 25 K. ΔQ/Δt = 0.80 × 15 × 25/0.25 = 1200 W. A heater of 1.2 kW keeps the room warm.
(b) In the steady state the same rate flows through both rods (energy does not pile up at the joint): kA(120 − T)/L = 4kA(T − 20)/L.
120 − T = 4T − 80 → 5T = 200 → T = 40 °C.
The better conductor needs a smaller temperature difference (20 K) to carry the same rate as the poorer one (80 K).
Why does a tiled floor feel colder than a carpet at the same temperature? Both are at room temperature, but tile is a much better conductor. It carries energy away from your foot quickly, so your skin cools faster. Your skin senses the rate of energy loss, not the temperature of the floor.
Learn it word for word. Convection happens in fluids (liquids and gases). The fluid near the heater warms, expands and becomes less dense, so it rises. Cooler, denser fluid sinks to take its place. This sets up a convection current: the moving fluid carries the energy.
Examples: a radiator warms the whole room; sea breezes (by day the land warms faster, the air above it rises, and cooler air flows in from the sea); ocean currents and winds. Convection cannot happen in a solid, and it does not happen in space, because there is no fluid to move.
Exam tip Convection is due to density differences. Say “expands, less dense, rises” for the warm fluid and “denser, sinks” for the cool fluid. “Heat rises” scores nothing: it is the warm fluid that rises.
Every body above 0 K emits electromagnetic waves, mostly infrared at everyday temperatures. Radiation needs no material: this is how energy from the Sun crosses empty space to reach the Earth.
Learn it word for word. A black body is a perfect absorber: it absorbs all the radiation that falls on it, reflecting and transmitting none. It is also the best possible emitter at any temperature.
Stefan–Boltzmann law (black body): L = σ A T⁴ L: power emitted (luminosity, W) · A: surface area · T: surface temperature in kelvin · σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ Double T → 2⁴ = 16 times the power · double A → twice the power
Worked example 12(a) By what factor does the power emitted by a black body increase when it is heated from 50 °C to 250 °C? (b) A black sphere of radius 0.10 m is at 600 K. Find the power it emits.
(a) Kelvin: 323 K and 523 K. Power ∝ T⁴, so the factor is (523/323)⁴ = 6.9. (Using °C would give 5⁴ = 625: wrong.)
(b) A = 4πr² = 4π × 0.10² = 0.126 m².
L = σAT⁴ = 5.67 × 10⁻⁸ × 0.126 × 600⁴ = 920 W.
Mistake Using °C in σAT⁴. The law only works with the absolute temperature: always add 273 first.
Black-body spectra at three temperatures (calculated, not sketched). Hotter means more power at every wavelength and a peak at a shorter wavelength.
Wien’s displacement law:λmax T = 2.9 × 10⁻³ m K λmax: the wavelength at the peak of the spectrum · T in kelvin Double T → the peak moves to half the wavelength
A black body emits a continuous range of wavelengths. As it gets hotter, the curve gets higher everywhere (more power) and the peak moves to shorter wavelengths. A heated piece of iron goes from dull red to orange to yellow-white. Blue stars are hotter than red stars.
The area under the curve is the power emitted per square metre of surface, σT⁴. If the same body has a bigger area at the same temperature, it emits more power, but each square metre still gives the same curve: the graph of power per m² does not change.
Worked example 13Find the peak wavelength for (a) human skin at 33 °C, (b) a star with surface temperature 9700 K, (c) a lamp filament at 2900 K. Name the part of the spectrum each time.
(a) T = 306 K: λmax = 2.9 × 10⁻³/306 = 9.5 × 10⁻⁶ m: infrared (this is what thermal cameras detect).
(b) λmax = 2.9 × 10⁻³/9700 = 3.0 × 10⁻⁷ m = 300 nm: ultraviolet. The star looks blue-white.
(c) λmax = 2.9 × 10⁻³/2900 = 1.0 × 10⁻⁶ m: infrared. A filament lamp gives out most of its energy as infrared, not light.
Learn it word for word. Luminosity L: the total power radiated by a star (W). For a star modelled as a black body, L = σAT⁴ = σ4πR²T⁴. Apparent brightness b: the power received per unit area at the observer (W m⁻²).
The star’s power spreads out over bigger and bigger spheres, so the apparent brightness falls as 1/d².
b = L / (4πd²) · d: distance from the star Luminosity is a property of the star. Apparent brightness also depends on how far away we are. Luminosity can also be given in units of the Sun’s luminosity, L☉.
Worked example 14(a) A star 25 ly away has an apparent brightness of 2.0 × 10⁻⁹ W m⁻². Find its luminosity (1 ly = 9.46 × 10¹⁵ m). (b) Another star has L = 3.8 × 10²⁷ W and b = 1.2 × 10⁻⁹ W m⁻². Find its distance.
(a) d = 25 × 9.46 × 10¹⁵ = 2.37 × 10¹⁷ m.
L = 4πd²b = 4π × (2.37 × 10¹⁷)² × 2.0 × 10⁻⁹ = 1.4 × 10²⁷ W.
(b) d = √(L/(4πb)) = √(3.8 × 10²⁷/(4π × 1.2 × 10⁻⁹)) = 5.0 × 10¹⁷ m.
Worked example 15(a) Star A has 2.5 times the radius and 1.5 times the surface temperature of star B. Find LA/LB. (b) The two stars have the same apparent brightness. Find dA/dB. (c) A star has 0.6 times the Sun’s surface temperature and 250 times its luminosity. Find its radius in solar radii.
(b) Same b means L/d² is the same: dA/dB = √(LA/LB) = √31.6 = 5.6.
(c) 250 = (R/R☉)² × 0.6⁴ → R/R☉ = √250/0.6² = 15.8/0.36 = 44. A cool, very luminous star must be huge: a red giant.
Ratio problems Write L ∝ R²T⁴ and b ∝ L/d², then put in the ratios. The constants σ and 4π cancel, so you never need them.
Writing the explanations
“Define internal energy” (2 marks) ✔ the total random kinetic energy of the molecules (atoms / particles) ✔ plus the total intermolecular potential energy of the molecules (from the forces between them) The answer must mention molecules, atoms or particles: “KE + PE” on its own scores 0. Also not “average”.
“Explain why the temperature stays constant while a substance melts (or boils)” (2–3 marks) ✔ the energy supplied increases the intermolecular potential energy / breaks the bonds between the molecules / the molecules move further apart ✔ the average (random) kinetic energy of the molecules does not change ✔ temperature measures the average kinetic energy, so the temperature is constant (the internal energy still increases) For freezing: bonds form, potential energy decreases and energy is released; the kinetic energy does not change.
“Explain, in terms of molecules, why evaporation cools a liquid” (2 marks) ✔ the fastest (most energetic) molecules escape from the surface ✔ so the average kinetic energy of the remaining molecules decreases, and temperature is a measure of the average kinetic energy Use “average kinetic energy”, not “the liquid loses heat”.
“Explain how thermal energy is transferred through a metal by conduction” (2 marks) ✔ particles at the hot end have a higher average kinetic energy ✔ they pass energy to neighbouring particles in collisions; free electrons carry energy quickly through the metal Energy flows from high to low temperature; the material itself does not move.
“Explain why there is a temperature difference across a layer that energy is conducted through” (2 marks) ✔ energy is conducted through the layer at a rate ΔQ/Δt = kAΔT/Δx ✔ a temperature gradient (ΔT/Δx) is needed for energy to flow; the faster the energy flows (or the poorer the conductor), the bigger the temperature difference In the steady state the rate is the same through every layer in series.
“Describe how convection transfers energy” (2–3 marks) ✔ the fluid near the heat source warms, expands, becomes less dense and rises ✔ cooler, denser fluid sinks to replace it, forming a convection current Always mention density. Only fluids show convection.
“Determine the surface temperature of a star from its spectrum” (2 marks) ✔ read the peak wavelength λmax from the graph (check the unit: nm or μm) ✔ use Wien’s law: T = 2.9 × 10⁻³/λmax, with λmax in metres; the answer is in kelvin Peak wavelength, not the wavelength where the curve ends.
Common mistakes
Ten traps in B.1 1. Using °C in E̅k = (3/2)kBT, σAT⁴ or Wien’s law. All three need kelvin. 2. Converting a temperature change to kelvin by adding 273. A change of 10 °C is a change of 10 K. 3. Defining internal energy as “kinetic energy plus potential energy” without saying it is the energy of the molecules (atoms, particles), and without total, random and intermolecular. 4. Saying a hotter body always has more internal energy. Internal energy also depends on the number of particles (and the phase). 5. Saying the temperature rises during melting, or that the kinetic energy increases during a phase change. The potential energy changes; the temperature is constant. 6. Missing a step in a mixing problem with ice (warm the ice → melt it → warm the water), or forgetting the container. 7. In mixtures, writing ΔT as (final − initial) for the hot body, which gives a negative term. Use (higher − lower) on both sides of “lost = gained”. 8. Forgetting that the rate through rods in series is the same in each rod. 9. “Heat rises.” The warm, less dense fluid rises. 10. Mixing up luminosity (W, a property of the star) and apparent brightness (W m⁻², depends on the distance).
Practise next
You have read the notes. Now practise, in this order:
Levelled practice: first a ★ at Level 2 in every topic (the goal), then Levels 3 and 4 for a 6 or 7. Hints and worked answers help you as you go.
IB-format practice (Paper 1A · Paper 2): timed, one attempt, and your report goes to your teacher. Do these when the practice goal is reached.