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8463 · RP10; 8.2.10Required practical activitiesCore

RP10: Infrared Absorption and Emission

Objects emit infrared radiation according to their temperature and surface properties. A dull black surface is generally a good infrared emitter and absorber compared with a shiny metallic surface in the relevant school comparison. A shiny surface reflects strongly. Do not infer every infrared property from visible colour alone: visible light and infrared are different spectral regions.

35 min Video by Science Shorts Required practical activities
INFRA-RED RADIATION - Science GCSE Physics Required PracticalCurated by Kepler, not produced by us. Watch it before the notes below.
Open on YouTube

Before you start

  • Use your teacher’s tier checklist: this overview can include Higher-only material, and is not a tier-specific revision list.
  • AQA separate GCSE Physics 8463. Use the specified required-practical reference and the centre’s approved apparatus and supervised procedure.
  • Distinguish temperature from thermal-energy transfer and recognise controlled variables.
  • Use the centre’s approved supervised method and equipment limits. These notes prepare you to explain and evaluate practical work; they do not replace carrying it out and recording the evidence.

Watch, pause, try

Video by Science Shorts. This overview supports surface absorption/emission reasoning. Pearson’s required four-container cooling procedure is specified separately in the written guide; a Leslie-cube demonstration alone does not replace it.

Viewing focus: Distinguish emitting, absorbing and reflecting radiation

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Name two conditions that must be comparable when measuring infrared detector response from different hot surfaces.

Show answer and reasoning

Surface temperature and detector distance/orientation must be comparable; background response and settling time also need consistent treatment.

After watching 2: Why is “the black container lost more heat by radiation” stronger than the cooling data alone justify?

Show answer and reasoning

Temperature change records net effects. Conduction, convection, geometry and covering properties can also affect the result, so evaluate them before isolating radiation.

What you'll be able to do

  • Distinguish emitting, absorbing and reflecting radiation
  • Compare surfaces using matched geometry and temperature
  • Evaluate what a cooling experiment can and cannot isolate
1

Separate a surface property from its temperature

Objects emit infrared radiation according to their temperature and surface properties. A dull black surface is generally a good infrared emitter and absorber compared with a shiny metallic surface in the relevant school comparison. A shiny surface reflects strongly. Do not infer every infrared property from visible colour alone: visible light and infrared are different spectral regions.

In an emission comparison, keep the surfaces at comparable temperatures and place a suitable detector at the same distance and orientation. A Leslie cube provides different surfaces on a common heated container. Rotate or reposition consistently and allow the sensor to settle. Background readings and changing surface temperature can affect small differences.

2

Choose an absorption or cooling comparison deliberately

For absorption, comparable surfaces receive radiation from the same source at the same distance and orientation. Compare suitable temperature changes or detector evidence while controlling mass and geometry. A hotter starting sample or a closer source is an alternative explanation for a larger response. Distinguish energy absorbed from energy reflected away.

A covered-container cooling comparison records temperature at fixed times after equal hot-water volumes are added. Keep container geometry, initial temperature, lids, probe positions and surroundings comparable. Different surface coverings can affect radiation, but the water also transfers energy through conduction and convection. A temperature drop is evidence of net cooling, not a direct count of infrared energy emitted.

3

Use evidence within the apparatus limits

Plot temperature against time, or detector response against surface type, as appropriate. Compare samples over matched intervals and at comparable temperatures. A detector reading may be relative rather than calibrated in watts; do not attach energy units to an arbitrary signal. Repeat and reset the arrangement to assess variation and check for a systematic distance or timing bias.

A good emitter is also a good absorber at a given wavelength under the relevant conditions, but that does not mean its temperature must always be highest. Temperature depends on the balance of incoming and outgoing transfers. A surface can cool faster in one arrangement and warm faster when irradiated in another. State the experiment’s direction of transfer before explaining a result.

4

Worked applications

Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.

Example 1Two equal-volume covered containers both start at 75 °C. After 240 s, one is at 65 °C and the other at 69 °C. Compare the drops and the strength of the radiation inference.
1Drops are 10 K and 6 K.
2The first has the greater net cooling under the matched conditions.
3Conduction, convection and covering differences still need evaluation.
Answer
A 4 K difference in drop; the data alone do not isolate a pure radiated-energy difference.
Example 2At matched temperature and detector geometry, a dull surface gives a background-corrected signal of 48 arbitrary units and a shiny surface 12. Find the response ratio and state its limit.
148/12 = 4.
2This is a ratio of measured detector responses under the stated conditions.
3Without calibration it is not an absolute energy or power measurement.
Answer
A response ratio of 4; do not label it 4 J or 4 W.
Example 3A sample warms by 3.0 K during a 120 s irradiation period. Another starts hotter and warms by 2.0 K. Why is a surface ranking uncertain?
1Unequal starting temperatures change outgoing transfers.
2Net temperature rise depends on thermal capacity and both incoming and outgoing transfers.
3Match initial conditions, mass and geometry before isolating a surface effect.
Answer
A smaller rise alone does not establish poorer absorption.

Common mistakes to avoid

Calling all detector signals joules.
Check what the detector measures and whether it is calibrated.
Equating visible white with poor infrared absorption in every material.
Use the actual surface and spectral context.

Key takeaways

  • You can distinguish emitting, absorbing and reflecting radiation.
  • You can compare surfaces using matched geometry and temperature.
  • You can evaluate what a cooling experiment can and cannot isolate.

Apply it in an exam

Question 1: A sample warms by 3.0 K during a 120 s irradiation period. Another starts hotter and warms by 2.0 K. Why is a surface ranking uncertain?

Show answer and reasoning

A smaller rise alone does not establish poorer absorption.

  • Unequal starting temperatures change outgoing transfers.
  • Net temperature rise depends on thermal capacity and both incoming and outgoing transfers.
  • Match initial conditions, mass and geometry before isolating a surface effect.

Explore further

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

  • Does a cooling curve measure radiation alone?
  • Why can a dull surface both warm and cool faster in different setups?

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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