Answer on paper first, then reveal the mark scheme one mark at a time.
1Grade 1–3FoundationTriple only3 marks
(a) Complete the sentence. Choose the answer from the box.
gamma | infrared | ultraviolet
All objects emit and absorb __________ radiation, whatever their temperature. [1]
(b) Which two statements describe a perfect black body? Tick (✓) two boxes. [2]
☐ It absorbs all of the radiation that is incident on it. ☐ It reflects all of the radiation that is incident on it. ☐ It is the best possible emitter of radiation. ☐ It only emits radiation when it is very hot. ☐ It transmits all of the radiation that is incident on it.
(a) infrared(1)
(b) it absorbs all of the radiation that is incident on it(1)
(b) it is the best possible emitter of radiation if more than two boxes are ticked, deduct one mark for each extra tick (minimum 0)(1)
Answer: (a) infrared (b) absorbs all incident radiation; best possible emitter
Examiner tip: A perfect black body reflects and transmits nothing: it absorbs everything, and it is also the best emitter, which students often forget.
2Grade 1–3Triple only3 marks
Two identical metal mugs are filled with water. The water in mug P is at 85 °C. The water in mug Q is at 40 °C.
(a) Which mug emits more infrared radiation each second? Give a reason for your answer. [2]
(b) A student says: “An ice cube at 0 °C does not emit any infrared radiation because it is cold.”
Explain why the student is wrong. [1]
(a) mug P(1)
(a) (because) it is at a higher temperature / it is hotter allow the hotter an object, the more infrared it emits per second reason only scores if P is chosen(1)
(b) all objects emit (and absorb) infrared radiation (whatever their temperature) allow any object above absolute zero emits infrared ignore it emits less infrared than a hotter object, on its own(1)
Answer: (a) P, because it is hotter. (b) All objects emit infrared, whatever their temperature.
Examiner tip: “Cold” objects still emit infrared – they just emit less of it per second than hot objects.
3Grade 4–5Triple only6 marks
The graph shows how the intensity of the radiation emitted by three objects, A, B and C, varies with wavelength. Each object behaves as a perfect black body and each is at a different temperature.
(a) Which object, A, B or C, is at the highest temperature? Give a reason for your answer. [2]
(b) Determine the wavelength at which object B emits radiation with the greatest intensity. [1]
(c) Describe how the wavelength at which the intensity is greatest changes as the temperature increases. [1]
(d) Object A is cooled to a lower temperature. Describe how its curve would be different. [2]
(a) C(1)
(a) it has the highest peak / it emits the greatest intensity at every wavelength allow its peak is at the shortest wavelength reason only scores if C is chosen(1)
(b) 830 nm allow any value from 800 nm to 850 nm(1)
(c) it decreases / becomes shorter allow the peak moves to the left(1)
(d) the peak would be lower / intensity lower at every wavelength allow the whole curve is lower(1)
(d) the peak would be at a longer wavelength / further to the right(1)
Answer: (a) C – highest peak (greater intensity at every wavelength) (b) about 830 nm (c) gets shorter (d) lower peak, at a longer wavelength
Examiner tip: Two separate things change with temperature – the height of the peak and its position; describe both.
4Grade 4–5Triple only4 marks
A blacksmith heats a steel bar in a furnace. At room temperature the bar does not glow, even though it is emitting radiation. As it is heated, the bar starts to glow dull red. When it is heated further it glows orange, then yellow-white.
Explain these observations. Use ideas about the radiation emitted by the bar. [4]
at room temperature the bar emits (mainly) infrared, which cannot be seen / is not visible(1)
as the temperature increases the bar emits more radiation per second / the intensity increases (at every wavelength)(1)
the peak of the emitted radiation shifts to shorter wavelengths allow the distribution of wavelengths changes towards shorter wavelengths(1)
so more of the shorter visible wavelengths (orange, yellow, green, blue) are emitted, so the colour changes from red towards white allow red light has the longest visible wavelength so it is seen first(1)
Answer: Cool: mostly infrared, invisible. Hotter: more radiation emitted and the distribution shifts to shorter wavelengths, so first red (longest visible wavelength) then more of the shorter visible wavelengths are added, giving orange then yellow-white.
Examiner tip: Say what happens to the wavelength distribution, not just “it gets hotter so it glows more”.
5Grade 6–7HigherTriple only5 marks
A glass of water has been standing in a room for several hours. The temperature of the water is now constant at 20 °C.
(a) A student says: “The water is not emitting any infrared radiation, because its temperature is not changing.”
Explain why the student is wrong. [3]
(b) The glass of water is moved into bright sunlight.
Explain what happens to the temperature of the water when it is first moved. [2]
(a) the water (still) emits infrared radiation (all objects do)(1)
(a) the water also absorbs infrared radiation (from its surroundings)(1)
(a) the rate of absorption is equal to the rate of emission (so the temperature stays constant) allow it absorbs energy at the same rate as it emits energy allow there is no net energy transfer(1)
(b) the water absorbs radiation at a greater rate than it emits radiation allow absorbs more energy per second than it emits(1)
(b) so its temperature increases(1)
Answer: (a) It still emits IR, but absorbs IR at the same rate as it emits it, so its temperature does not change. (b) It absorbs radiation faster than it emits it, so its temperature rises.
Examiner tip: A constant temperature means emission and absorption are balanced – not that both have stopped.
6Grade 6–7HigherTriple only6 marks
A metal panel on a satellite in space is painted matt black. When the panel faces the Sun, it absorbs radiation at a rate of 450 W.
At one moment, the panel is emitting radiation at a rate of 310 W. Assume no energy is transferred from the panel in any other way.
(a) Calculate the net energy gained by the panel in 2.0 minutes. Assume the rates of absorption and emission stay constant. [3]
(b) The panel has a mass of 0.80 kg. The specific heat capacity of the metal is 900 J/kg °C.
Calculate the increase in temperature of the panel in this time.
Use the Physics Equations Sheet. [2]
(c) The actual temperature increase in 2.0 minutes is less than your answer to (b).
Suggest why. [1]
(a) net rate = 450 − 310 = 140 (W)(1)
(a) E = 140 × 120(1)
(a) E = 16 800 (J) allow 16 800 J with no working shown for 3 marks allow ecf for this mark from an incorrect net power, provided t = 120 s is used 280 J (t = 2 used) scores max 1(1)
(c) as the panel's temperature increases it emits radiation at a greater rate (so the net rate of energy gain decreases) allow the rate of emission does not stay at 310 W(1)
Answer: (a) 16 800 J (b) 23 °C (c) the hotter panel emits radiation at a greater rate, so the net power falls
Examiner tip: Use the net power (absorbed minus emitted) and convert minutes to seconds before using E = Pt.
7Grade 6–7HigherTriple only6 marks
The diagram shows some of the ways that energy is transferred by radiation between the Sun, the Earth's surface, the atmosphere and space.
Arrows A to E represent radiation.
Scientists predict that changes to the atmosphere will cause it to absorb a greater proportion of the infrared radiation emitted by the Earth's surface.
Explain how the Earth's surface can stay at a constant average temperature, and why this change to the atmosphere would cause the average temperature of the Earth's surface to increase.
You should refer to arrows A to E in your answer. [6]
radiation from the Sun (A) is absorbed by the Earth's surface (and transfers energy to it)(1)
some radiation from the Sun is reflected back into space (B) by the atmosphere / clouds (or by the surface, e.g. ice), so it is not absorbed(1)
the Earth's surface emits infrared radiation (C and D)(1)
some of this infrared passes through the atmosphere into space (C)(1)
some of this infrared is absorbed by the atmosphere (D)(1)
the atmosphere emits infrared, some of which goes back to the surface (E) and is absorbed(1)
the temperature is constant when the surface absorbs radiation at the same rate as it emits radiation(1)
if the atmosphere absorbs more infrared, D increases and C decreases / less escapes to space(1)
so more infrared is emitted back to the surface (E increases)(1)
the surface now absorbs radiation at a greater rate than it emits it, so its temperature increases(1)
as the surface gets hotter it emits more infrared per second, until absorption and emission balance again at a higher temperature(1)
Level
Marks
What it looks like
Level 3
5–6
A detailed, logically sequenced explanation. Uses the arrows correctly to explain that a constant temperature needs the rate of absorption to equal the rate of emission, and explains that the change increases E (and decreases C), so the surface absorbs faster than it emits and its temperature rises.
Level 2
3–4
A partial explanation. Describes absorption, reflection and emission using some of the arrows, and links the change to a rise in temperature, but the balance between rates of absorption and emission is incomplete or not clearly linked.
Level 1
1–2
Simple, mostly unconnected statements about radiation being absorbed, reflected or emitted, or a statement that the temperature rises with little or no explanation.
No relevant content
0
Nothing worthy of credit.
Answer: The surface absorbs A and E and emits C and D; its temperature is constant when the rate of absorption equals the rate of emission. If the atmosphere absorbs more (D up, C down) it re-emits more back (E up), so the surface absorbs faster than it emits and warms until a new balance is reached at a higher temperature.
Examiner tip: Top-band answers link the arrows to the balance between rate of absorption and rate of emission – just naming the greenhouse effect is not enough.
8Grade 4–5Triple only5 marks
A student investigated how the infrared radiation emitted by a can depends on the temperature of the can.
She filled a matt black metal can with hot water and placed an infrared detector 10 cm from the side of the can. She recorded the detector reading as the water cooled.
Temperature of water in °C
80
70
60
50
40
30
Detector reading in arbitrary units
52
41
37
22
14
7
(a) Identify the independent variable. [1]
(b) Give one control variable in this investigation. [1]
(c) Identify the anomalous result in the table. [1]
(d) Another student says: “The detector reading is directly proportional to the temperature of the water in °C.”
Use data from the table to show that this student is wrong. [2]
(a) temperature of the water allow temperature of the can(1)
(b) any one from: distance between the can and the detector / the surface of the can facing the detector / the same detector / the same room or surroundings temperature ignore amount of water, unless qualified(1)
(c) 37 (at 60 °C)(1)
(d) chooses a pair of values where the temperature doubles, e.g. 40 °C and 80 °C allow reading ÷ temperature calculated for two results, e.g. 52 ÷ 80 = 0.65 and 14 ÷ 40 = 0.35(1)
(d) shows the reading does not double, e.g. it increases from 14 to 52 (about 3.7 times) allow shows the ratios are not equal do not accept a comparison that uses the anomalous result(1)
Answer: (a) temperature of the water (b) e.g. distance from can to detector (c) 37 (d) temperature doubles from 40 °C to 80 °C but reading increases from 14 to 52 (×3.7), not ×2
Examiner tip: To test direct proportion, double one variable and check whether the other exactly doubles – or check that the ratio stays constant.
9Grade 8–9HigherTriple only6 marks
The filament of a lamp behaves like a black body. The table shows the percentage of the radiation emitted by a filament that is visible light, at different filament temperatures.
Filament temperature in °C
1700
2200
2700
3200
Percentage of emitted radiation that is visible light
0.7
3.2
7.8
14.0
(a) Explain why the percentage of the emitted radiation that is visible light increases as the filament temperature increases. [2]
(b) A lamp has a power input of 60 W. Its filament is at a constant temperature of 2700 °C.
Assume that all of the power input is emitted as radiation from the filament.
Calculate the power emitted as visible light. [2]
(c) The filament is made of tungsten, which melts at 3422 °C.
Suggest one reason why lamp filaments are not designed to operate at 3200 °C, even though this would increase the percentage that is visible light. [1]
(d) Give the reason why it is valid to assume that, at a constant temperature, the power emitted by the filament is equal to the power transferred to it. [1]
(a) as temperature increases, the peak of the emitted radiation shifts to shorter wavelength(s) allow the wavelength distribution moves towards shorter wavelengths(1)
(a) (the peak moves from the infrared towards the visible region) so a greater proportion of the radiation has wavelengths in the visible range ignore more radiation is emitted, on its own(1)
(b) 60 × 7.8 / 100(1)
(b) 4.7 (W) allow 4.68 (W) allow 4.7 W with no working shown for 2 marks(1)
(c) 3200 °C is close to the melting point so the filament might melt / break allow the filament would wear out (evaporate) faster, so the lamp would have a shorter life(1)
(d) at a constant temperature an object emits energy at the same rate as it absorbs / gains energy allow otherwise its temperature would change(1)
Answer: (a) The peak shifts to shorter wavelengths (from infrared towards the visible), so a larger proportion is visible. (b) 4.7 W (c) e.g. the filament could melt or would wear out quickly. (d) At constant temperature the rate of energy transfer in equals the rate of emission.
Examiner tip: “It gets hotter so it gives out more light” only explains the total – you need the shift of the peak to explain the percentage.
10Grade 8–9HigherTriple only7 marks
On average, each square metre at the top of the Earth's atmosphere receives 340 W of radiation from the Sun. 30% of this radiation is reflected back into space by clouds, ice and other surfaces. The rest is absorbed by the Earth and its atmosphere.
(a) Calculate the power absorbed by each square metre. [2]
(b) The average temperature of the Earth is constant.
State the power per square metre that the Earth emits into space. Give a reason for your answer. [2]
(c) Large areas of ice melt and are replaced by darker land and sea. The percentage of the Sun's radiation that is reflected falls to 28%.
Explain what happens to the average temperature of the Earth. Your answer should include a calculation. [3]
(a) 238 (W) allow 238 W with no working for 2 marks(1)
(b) 238 W (per m²) allow ecf from (a)(1)
(b) (because) at a constant temperature the rate of emission is equal to the rate of absorption(1)
(c) power absorbed increases to 340 × 0.72 = 245 W (per m²) allow 244.8 W, or an increase of 6.8 W per m²(1)
(c) the Earth now absorbs radiation at a greater rate than it emits radiation, so its temperature increases(1)
(c) as the temperature increases the Earth emits more radiation per second, until the rate of emission again equals the rate of absorption (at a higher temperature) allow the temperature rises until it reaches a new constant (higher) value(1)
Answer: (a) 238 W (b) 238 W per m² – at constant temperature the emission rate equals the absorption rate (c) absorbed power rises to 245 W per m²; absorption now exceeds emission so the temperature rises until emission increases to 245 W per m² at a new, higher temperature
Examiner tip: The temperature does not keep rising forever: a hotter Earth emits more, so it settles at a new, higher constant temperature.
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