Know what happens to an object if the rate at which it receives energy is less or more than the rate at which it transfers energy away from the object

2.3.3 Radiation

Learning objective

Explain what happens to an object when the rate at which it receives energy is less than, equal to, or greater than the rate at which it loses energy by radiation.

Key concepts

  • All objects with a temperature above absolute zero emit electromagnetic radiation, chiefly in the infrared region.
  • Radiation can travel through empty space – no material medium is required.
  • The temperature of an object changes only if there is a net gain or loss of energy:

    • Net gain → temperature rises.
    • Net loss → temperature falls.
    • Net zero → thermal equilibrium (temperature remains constant).

Qualitative dependence of radiated power

For the IGCSE syllabus it is sufficient to know that the power radiated by an object is proportional to its surface area, its emissivity and the fourth power of its absolute temperature:

Radiated power ∝ A · e · T⁴

Thus a small increase in temperature produces a large increase in the amount of energy emitted.

Link between emissivity, absorptivity, colour and texture

  • According to Kirchhoff’s law, a surface that is a good emitter (high emissivity e) is also a good absorber of radiation at the same wavelength. Consequently:

    • High‑emissivity surfaces (e ≈ 0.9–1) – typically black, matte, rough – have high absorptivity and therefore absorb most incoming infrared radiation.
    • Low‑emissivity surfaces (e ≈ 0.1–0.3) – shiny metals, polished or reflective paints – have low absorptivity and reflect most incoming infrared radiation.

  • Colour and texture therefore affect both emission and absorption (and consequently reflection) of infrared radiation.

Energy‑balance scenarios

ScenarioEnergy received (Pin)Energy lost (Pout)Net power (Pnet)Resulting behaviour
1. Pin < PoutLess than the power radiated awayGreater than the power receivedNegative (loss)Object cools – temperature falls until equilibrium is reached.
2. Pin = PoutExactly equal to the power radiated awayExactly equal to the power receivedZeroThermal equilibrium – temperature remains constant.
3. Pin > PoutGreater than the power radiated awayLess than the power receivedPositive (gain)Object heats – temperature rises until a new equilibrium is achieved.

Everyday examples

  1. Black car vs. white car in sunlight: A black, matte car has high emissivity and absorptivity, so Pin > Pout. Its temperature rises quickly. A white, reflective car has low emissivity/absorptivity, receives less energy and stays cooler.
  2. Heating a kitchen pan: While on a burner the pan receives a large amount of energy (Pin) and Pin > Pout. After the flame is turned off, the pan loses more energy than it receives and cools.
  3. Earth’s energy balance: Solar radiation is the incoming energy. The Earth emits infrared radiation to space. When the two rates are equal the global temperature is stable; an imbalance leads to climate change.
  4. Spacecraft thermal control: Surfaces are coated to give a chosen emissivity so that internal heat production plus solar input matches the radiated loss, keeping instruments at safe temperatures.

Factors that influence the rate of energy gain or loss

  • Emissivity (e): Determines both how strongly a surface emits infrared radiation and how strongly it absorbs incoming infrared radiation.
  • Colour and texture: Dark, rough surfaces → high e, high absorptivity; light, smooth surfaces → low e, low absorptivity.
  • Surface area (A): A larger area increases the total amount of energy received and emitted proportionally.
  • Temperature (T): Radiated power varies with T⁴, so even modest temperature changes cause large changes in energy loss.
  • Surrounding temperature (Tenv): A hot environment can supply radiation, reducing net loss or producing a net gain.

Supplementary material – quantitative extension

For students who wish to explore the quantitative side, the Stefan‑Boltzmann law gives the exact radiated power:

Prad = σ A e (T⁴ − Tenv⁴)

where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. This equation is not required for the core IGCSE assessment but is useful for higher‑level work.

Summary

  • Net gain (Pnet > 0) → temperature rises.
  • Net loss (Pnet < 0) → temperature falls.
  • Net zero (Pnet = 0) → thermal equilibrium.

Suggested diagram: Energy‑flow sketch showing incoming solar radiation, reflected/absorbed radiation, emitted infrared radiation, and the balance that determines heating, cooling, or equilibrium.