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Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)
Question
In a Leslie cube experiment a student records IR sensor readings (in millivolts): matt black 5.4 mV; matt white 3.8 mV; shiny silver 1.1 mV. Explain the pattern in the results.
Solution
All three faces are at the same temperature, so any difference must be due to the surface.
Matt black has the highest reading (5.4 mV) because matt-black surfaces are the best emitters of IR.
Shiny silver has the lowest reading (1.1 mV) because shiny silver surfaces are the worst emitters / best reflectors.
Answer
Matt black emits the most IR (best emitter); shiny silver emits the least (worst emitter / best reflector). All three faces are at the same temperature, so the differences are entirely due to surface colour and texture.
Question
A student finds that their Leslie cube readings vary unexpectedly. Suggest THREE ways the student could improve the experiment to make a fairer comparison.
Solution
Keep the distance from each face to the detector exactly the same. IR intensity falls with distance.
Wait for the cube to reach a steady temperature so all four faces are at the same temperature.
Carry out the experiment in a room at constant temperature with no draughts; use a heat-shield to block IR from other warm objects.
Answer
(1) Keep the detector at a constant distance from each face. (2) Allow the cube to reach a steady uniform temperature before taking readings. (3) Minimise background IR — constant room temperature, no draughts, shield from other warm objects.
Examiner note
AQA examiner reports specifically reward 'same distance' and 'constant temperature' as fair-test improvements.
Question
A vacuum flask keeps coffee hot for hours. The walls of the flask are silvered. Explain why.
Solution
Heat from the hot coffee would otherwise be lost by infrared radiation through the glass walls.
Silvered surfaces are poor emitters of IR — very little IR leaves the inner wall.
Silvered surfaces are also good reflectors — any IR that does leave the inner wall is reflected back from the outer silvered surface.
Answer
The silvered surfaces are poor emitters of IR and good reflectors. They minimise heat loss from the hot coffee by reducing both the radiation emitted and any radiation that escapes is reflected back inward.
Question
Explain why the heat-exchanger fins on the back of a fridge are painted matt black.
Solution
Heat absorbed from inside the fridge is transferred to the fins at the back.
Matt black surfaces are the best emitters of infrared radiation.
Painting the fins matt black maximises the rate at which the fins lose heat to the room — keeping the fridge cool inside.
Answer
Matt black is the best emitter of IR. Painting the fins matt black maximises the rate at which heat is radiated to the surrounding air, keeping the fridge cool.
Question
On a hot summer day, two identical garden sheds stand side by side. One is painted matt black and the other matt white. Predict which shed is warmer inside after several hours of bright sunshine and explain your reasoning.
Solution
Both sheds absorb IR from the sun, but a matt-black surface absorbs more than a matt-white surface (which reflects much of the sunlight).
More absorbed energy is transferred into the air inside the matt-black shed → temperature inside rises faster and reaches a higher steady value.
Both sheds emit IR back to the environment, but the matt-black absorbs more than it emits while the sun shines, so net heating is greater for the black shed.
Answer
The matt-black shed is warmer inside because matt-black surfaces absorb more sunlight (visible + IR) than matt-white surfaces, which reflect a large fraction.
Question
State two properties of a perfect black body. (2 marks)
Solution
Property 1: it absorbs all the radiation that falls on it — none is reflected or transmitted.
Property 2: it emits the maximum possible amount of radiation for its temperature (a perfect emitter).
Answer
(1) Absorbs all the incident radiation (no reflection). (2) Emits the maximum possible radiation at every wavelength for its temperature.
Question
An electric stove element glows dull red when first switched on, then brighter orange after a few minutes. Explain in terms of intensity and peak wavelength.
Solution
At first the element is just above room temperature — peak emission is in the IR (invisible) and the tail of the spectrum reaches into red.
As the element heats up, the intensity of emission rises sharply (more energy per second).
The peak wavelength shifts to shorter values, so more energy is emitted in the red and orange parts of the visible spectrum — the element looks brighter orange.
Answer
As T rises the total emission grows (brighter) and the peak shifts to shorter wavelengths (red → orange).
Question
Star A appears red. Star B appears blue-white. Both stars approximate perfect black bodies. Which star has the higher surface temperature, and why?
Solution
Star colour reflects the peak wavelength of its emission.
Blue light has a shorter wavelength than red light. The blue-white star's emission peak is at a shorter wavelength.
Shorter peak wavelength means higher temperature, so star B (blue-white) is hotter than star A (red).
Answer
Star B (blue-white) has the higher surface temperature. Shorter peak wavelength = higher T.
Question
Explain, in terms of absorption and emission of radiation, why Earth's average temperature stays roughly constant from year to year.
Solution
Earth absorbs solar radiation (mostly visible light) from the Sun.
The warm Earth emits infrared radiation back to space.
In steady state the rate of energy absorbed from the Sun equals the rate of energy emitted as IR; Earth's average temperature therefore stays roughly constant.
Answer
Absorbed solar energy ≈ emitted IR energy → temperature stays roughly constant. If absorption increases (e.g. more greenhouse gases trap IR) the balance is met at a higher T.
Question
Explain why a clear night in winter is usually colder than a cloudy night, even when the daytime temperatures were the same.
Solution
At night the ground emits IR to space.
Clear sky lets IR escape directly to space → ground cools rapidly.
Cloudy sky absorbs the outgoing IR and re-emits some of it back down, reducing the net loss of energy from the ground → less cooling.
Answer
Clouds absorb outgoing IR and re-radiate some back down, slowing the ground's cooling. Clear nights lose IR straight to space → colder.
Electromagnetic radiation with wavelengths longer than red light. Emitted by all objects above absolute zero; we feel it as heat.
The release of infrared radiation by an object. Faster for hotter objects and for matt-black surfaces.
The taking-in of incident infrared radiation by an object. A good absorber is also a good emitter (matt black); shiny surfaces are poor absorbers and poor emitters.
A hollow metal cube with four different surfaces (typically matt black, matt white, shiny black and shiny silver) used to compare the rate of IR emission from each surface at the same temperature.
The state in which an object emits and absorbs IR at equal rates, so its temperature stays constant.
An idealised object that absorbs ALL the radiation that falls on it (no reflection) and emits the maximum possible radiation for its temperature.
The rate at which energy is radiated per unit area of a surface (W/m²). Increases steeply as temperature rises.
The wavelength at which an object emits the most radiation. For a black body, peak wavelength decreases as temperature increases.
The warming of Earth's surface caused by greenhouse gases absorbing outgoing IR radiation and re-emitting some of it back to the surface.
The balance between radiation absorbed from the Sun and radiation emitted by Earth to space. A steady balance fixes Earth's average temperature.
Mistake
Saying matt black is only a good absorber but a poor emitter.
Why it happens
Students see 'good absorber' and assume the opposite for emission.
How to avoid it
A good absorber is also a good emitter (and vice-versa). Matt black is best at both. Shiny silver is worst at both.
Source: AQA Examiner Report Paper 2 2024.
Mistake
Saying 'silver metal is a bad emitter because it is a metal'.
Why it happens
Confusing material with surface.
How to avoid it
It is the SURFACE finish (colour + texture) that matters, not the underlying material. A copper face painted matt black emits like a matt-black surface.
Mistake
Moving the IR detector closer to one face than another in the Leslie cube experiment.
Why it happens
Trying to 'get a better reading' from one face.
How to avoid it
Always keep the detector at the same fixed distance from every face. IR intensity falls off with distance.
Mistake
Forgetting to say that the Leslie cube must be at a steady temperature.
Why it happens
Going straight to readings.
How to avoid it
Write 'wait for the cube to reach a steady temperature so all faces are at the same T, then take readings'. AQA awards a mark for this method point.
Mistake
Saying a perfect black body 'only absorbs' and does not emit radiation.
Why it happens
Confusion with the word 'black' = no light coming out.
How to avoid it
A perfect black body emits the maximum possible radiation at every wavelength for its temperature. Hot black bodies (stars, glowing iron) shine brilliantly.
Source: AQA Examiner Report Paper 2 2024.
Mistake
Mentioning intensity only when asked how spectrum changes with T.
Why it happens
Students forget the wavelength shift.
How to avoid it
Quote BOTH: intensity rises AND peak wavelength decreases. Two marks for two ideas.
Mistake
Saying greenhouse gases 'reflect' IR.
Why it happens
Mental analogy with a greenhouse roof.
How to avoid it
Greenhouse gases absorb outgoing IR and re-emit in all directions, sending some back down. They do not reflect.
Mistake
Saying 'doubling the temperature in °C gives 16× more emission'.
Why it happens
Stefan–Boltzmann is a T⁴ law in kelvin, not °C.
How to avoid it
The T⁴ relationship applies to absolute temperature in kelvin. Doubling °C is meaningless because °C has an arbitrary zero.