Know that for an object to be at a constant temperature it needs to transfer energy away from the object at the same rate that it receives energy

2.3.3 Radiation

Learning Objective

Explain why an object that stays at a constant temperature must lose energy at the same rate that it gains energy, and describe how emission, absorption and reflection of thermal radiation depend on temperature, surface area and surface properties.

Key Concepts

• Thermal radiation – transfer of energy by electromagnetic waves.

  • For everyday temperatures the radiation is mainly in the infra‑red (IR) part of the spectrum.
  • All objects with a temperature above absolute zero emit radiation (no lower limit).

• No material medium required – radiation can travel through vacuum (e.g. Sun → Earth).
• Surface properties affect three related processes:

  • Emission – how readily a surface radiates energy.
  • Absorption – how readily a surface takes up incident radiation.
  • Reflection – how much incident radiation is reflected away.
  • For a given wavelength, emissivity (ε) = absorptivity (α) (Kirchhoff’s law). Hence a good emitter is also a good absorber, and a poor emitter is a good reflector.

• Effect of colour and finish (typical classroom examples):

  • Black, matte surfaces – high ε (≈ 1); excellent emitters and absorbers, low reflectivity.
  • Shiny metal or polished surfaces – low ε (≈ 0.05 – 0.2); poor emitters/absorbers, high reflectivity.
  • White or highly reflective paints – low ε, high reflectivity in the IR.

Energy Balance for a Constant Temperature

For an object to remain at a steady temperature the total power it receives must equal the total power it loses:

\[

\boxed{P{\text{in}} = P{\text{out}}}

\]

If \(P{\text{in}} > P{\text{out}}\) the temperature rises; if \(P{\text{in}} < P{\text{out}}\) it falls.

Stefan–Boltzmann Law (Quantitative Form)

The power radiated by a surface is

\[

P_{\text{out}} = \varepsilon \,\sigma \,A \,T^{4}

\]

• \(P_{\text{out}}\) – radiated power (W)
• \(\varepsilon\) – emissivity (0 ≤ ε ≤ 1)
• \(\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}\) – Stefan–Boltzmann constant
• \(A\) – surface area (m²)
• \(T\) – absolute temperature (K)

Because ε = α, the same factor also tells how much incident radiation the surface will absorb.

How to Apply the Energy‑Balance Idea

1. Identify all sources of incoming energy (solar radiation, electrical heating, conduction from a warmer neighbour, absorbed thermal radiation, etc.).
2. Calculate the total incoming power, \(P_{\text{in}}\) (W).
3. Write the outgoing power using the Stefan–Boltzmann law: \(P_{\text{out}} = \varepsilon\sigma A T^{4}\).
4. Set \(P{\text{in}} = P{\text{out}}\) and solve for the unknown (usually the equilibrium temperature \(T\)).

Worked Example

Question: A black sphere (\(\varepsilon = 1\)) of radius 0.10 m is placed in a room at 293 K. An electric heater supplies 50 W of power to the sphere. Find the equilibrium temperature of the sphere.

1. Surface area of the sphere:

   \[

   A = 4\pi r^{2}=4\pi(0.10)^{2}=0.126\ \text{m}^{2}

   \]

2. At equilibrium \(P{\text{in}} = P{\text{out}}\):

   \[

   50\ \text{W}= \sigma A T^{4}

   \]

3. Solve for \(T\):

   \[

   T = \left(\frac{50}{\sigma A}\right)^{1/4}

   = \left(\frac{50}{5.67\times10^{-8}\times0.126}\right)^{1/4}

   \approx 3.31\times10^{2}\ \text{K}

   \]

4. Result:

   \[

   \boxed{T \approx 331\ \text{K}\;(58^{\circ}\text{C})}

   \]

The sphere heats until it radiates exactly 50 W, at which point its temperature is steady.

Practical Activities (IGCSE‑level experiments)

• Good vs. bad emitters: Place a black matte card and a white glossy card under an identical lamp. Use a thermometer or thermistor to record the temperature rise after a fixed time. The black card (high ε) will become hotter.
• Good vs. bad absorbers: Cover two identical thermometers with a black paper and a white aluminium foil respectively. Expose both to the same infrared source. The thermometer under the black cover (high α) registers a higher temperature, demonstrating the link between absorption and emissivity.
• Effect of a reflective cover: Put a thermometer inside a small aluminium‑foil “shroud”. Compare its reading with an uncovered thermometer when both are exposed to the same IR source. The foil‑shrouded thermometer stays cooler because most radiation is reflected away.
• Vacuum demonstration: Shine a lamp on a metal plate inside a sealed glass jar (or a vacuum chamber if available). The plate heats even though the air is thin, illustrating that radiation does not need a medium.

Common Misconceptions

• “Only hot objects radiate.” – In fact *all* objects emit radiation; hotter objects emit more.
• “If an object feels warm, it must be gaining energy.” – It may be losing energy at the same rate it gains, keeping its temperature constant.
• “Emissivity is always 1.” – Only a perfect black‑body has ε = 1; real surfaces have lower values.
• “Radiation needs air.” – Radiation can travel through vacuum; it is the only heat‑transfer method that does.
• “Emission and absorption are unrelated.” – Kirchhoff’s law states that for a given wavelength ε = α; a good emitter is also a good absorber.

Summary Table

Check Your Understanding

1. Why can an object at a constant temperature still have a non‑zero net energy flow?
2. How would the equilibrium temperature in the worked example change if the sphere’s emissivity were reduced to 0.5?
3. Explain in words why the Stefan–Boltzmann law gives a \(T^{4}\) dependence for radiated power.
4. Describe an experiment that shows colour (or surface finish) affects both absorption and emission.
5. What would happen to the temperature of a black‑coated thermometer placed in a sunny window compared with a white‑coated one? Explain using emissivity and absorptivity.