Describe the effect of surface colour (black or white) and texture (dull or shiny) on the emission, absorption and reflection of infrared radiation

2.3.3 Radiation – Effect of Surface Colour and Texture

Learning objective

Describe how the colour (black or white) and the texture (dull or shiny) of a surface influence its:

• Emission of infrared (IR) radiation
• Absorption of IR radiation
• Reflection of IR radiation

Key concepts

• All bodies with a temperature above absolute zero emit electromagnetic radiation.

  
  Stefan‑Boltzmann law (ideal black‑body):

  
  P = σ A T⁴ where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, A = surface area, T = absolute temperature (K).

• Real surfaces emit less than a perfect black‑body.

  
  P = ε σ A T⁴ where ε (0 ≤ ε ≤ 1) is the *emissivity* in the infrared region.

• Kirchhoff’s law for opaque surfaces:

  
  Emissivity equals absorptivity at each wavelength, i.e. ε = A.

  Consequently, a surface that is a good emitter is also a good absorber.

• Energy balance for a single wavelength (or a narrow band such as the IR):

  
  A + R + ε = 1 where R is reflectivity.

  Hence R = 1 − (A + ε).

• Infrared radiation (≈ 0.7–1000 µm) is the dominant part of the spectrum for everyday heat transfer.

Colour effects (black vs. white)

Colour describes which part of the visible spectrum is reflected, but it also gives a good indication of behaviour in the IR band.

• Black surfaces

  • High emissivity (ε ≈ 0.9–1) → strong IR emission.
  • Because ε = A, absorptivity is also high – most incident IR is taken up.
  • Reflectivity is low (R ≈ 0–0.1).

• White surfaces

  • Low emissivity (ε ≈ 0.05–0.2) → weak IR emission.
  • Absorptivity is equally low (A ≈ ε) – little IR is absorbed.
  • Reflectivity is high (R ≈ 0.8–0.95).

Texture effects (dull vs. shiny)

Texture determines the *type* of reflection, not the intrinsic values of ε or A for a given colour.

• Dull (matte) surfaces

  • Reflection is diffuse – the reflected beam is scattered over a wide solid angle.
  • R is spread out, so less IR returns in the direction of the source.
  • For a given colour, ε and A remain essentially unchanged.

• Shiny (polished) surfaces

  • Reflection is specular – a narrow, well‑defined beam.
  • In the direction of the incident beam the effective reflectivity is higher, reducing the net absorbed IR compared with a dull surface of the same colour.
  • ε and A are still set by the colour; the texture only modifies the *direction* of the reflected component.

Combined influence of colour and texture

Physical reasoning

1. Absorption (A): incident IR energy is taken up, raising the material’s temperature.
2. Emission (ε): the heated material radiates energy away according to P = εσAT⁴.
3. Reflection (R): part of the incident energy is sent away without being absorbed; texture decides whether this is diffuse or specular.

Because A + R + ε = 1, a surface that reflects a large fraction of IR must necessarily absorb and emit only a small fraction. Colour primarily sets ε (and therefore A via Kirchhoff’s law); texture mainly alters the angular distribution of R.

Numerical example (AO2 style)

Two identical square plates (area = 0.10 m²) are at 60 °C (T = 333 K). One is black‑matte (ε = 0.95) and the other white‑shiny (ε = 0.10). Using the Stefan‑Boltzmann expression for real surfaces:

Pblack = εblack σ A T⁴ = 0.95 × 5.67×10⁻⁸ × 0.10 × (333)⁴ ≈ 1.23 W
Pwhite = εwhite σ A T⁴ = 0.10 × 5.67×10⁻⁸ × 0.10 × (333)⁴ ≈ 0.13 W

Ratio = Pblack / Pwhite ≈ 9.5.

Thus the black‑matte plate radiates almost ten times more IR power than the white‑shiny plate at the same temperature.

Practical implications

• Roofing: Dull white paint (high R, low ε) reflects solar IR, keeping interiors cooler.
• Radiators & heat‑sinks: Black, matte finishes (high ε, low R) maximise heat loss by radiation.
• Solar collectors: Black, polished surfaces concentrate IR onto a receiver while reflecting visible light to reduce glare.
• Clothing: Dark, matte fabrics feel warmer in sunlight because they absorb more IR than light, shiny fabrics.

Sample examination question

Q: Two identical metal plates are heated to 60 °C in a room at 20 °C. Plate A is black and matte; Plate B is white and shiny. Which plate will cool faster and why?

Answer outline:

• Plate A: ε ≈ 0.95, R ≈ 0.05 → high emission, low reflection → loses heat rapidly.
• Plate B: ε ≈ 0.10, R ≈ 0.85 (mostly specular) → low emission, high reflection → loses heat slowly.
• Therefore Plate A (black‑matte) cools faster because it radiates much more IR.

Suggested diagram