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

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Radiation

2.3.3 Radiation

Learning Objective

Understand that an object at a constant temperature must transfer energy away at the same rate that it receives energy.

Key Concepts

  • Radiation is the transfer of energy by electromagnetic waves.

  • All objects with a temperature above absolute zero emit radiation.

  • The amount of energy radiated depends on temperature, surface area, and emissivity.

Energy Balance for Constant Temperature

For an object to remain at a steady temperature:

\$\text{Energy received per unit time} = \text{Energy emitted per unit time}\$

If the rates differ, the object's temperature will change.

Stefan‑Boltzmann Law

The power radiated by a surface is given by:

\$P = \varepsilon \sigma A T^{4}\$

  • \$P\$ – power radiated (W)

  • \$\varepsilon\$ – emissivity (0 ≤ \$\varepsilon\$ ≤ 1)

  • \$\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}\$ – Stefan‑Boltzmann constant

  • \$A\$ – surface area (m²)

  • \$T\$ – absolute temperature (K)

Applying the Energy Balance

  1. Identify all sources of energy input (e.g., solar radiation, conduction, convection).

  2. Calculate the total power received, \$P_{\text{in}}\$.

  3. Calculate the power radiated 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 temperature \$T\$ if required.

Example Problem

A black sphere (\$\varepsilon = 1\$) with a radius of 0.10 m is placed in a room at 293 K. The sphere absorbs 50 W of electrical power. Determine the equilibrium temperature of the sphere.

  1. Surface area: \$A = 4\pi r^{2} = 4\pi (0.10)^{2} = 0.126\ \text{m}^{2}\$.

  2. At equilibrium, \$P{\text{in}} = P{\text{out}}\$:

    3. \$50\ \text{W} = \sigma A T^{4}\$

  3. Solve for \$T\$:

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

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

    \approx 331\ \text{K}\$\$

The sphere will settle at about \$331\ \text{K}\$ (58 °C). If the room temperature is lower, the sphere loses heat by radiation until this balance is reached.

Common Misconceptions

  • “Radiation only occurs from hot objects.” – All objects emit radiation; hotter objects emit more.

  • “If an object is warm, it must be gaining energy.” – It may be losing energy at the same rate it gains, staying at constant temperature.

  • “Emissivity is always 1.” – Real surfaces have emissivities less than 1, affecting the radiated power.

Summary Table

QuantitySymbolUnitRelevant Equation
Power radiated\$P\$W\$P = \varepsilon \sigma A T^{4}\$
Surface area of sphere\$A\$m²\$A = 4\pi r^{2}\$
Stefan‑Boltzmann constant\$\sigma\$W m⁻² K⁻⁴\$5.67 \times 10^{-8}\$
Emissivity\$\varepsilon\$–0 ≤ \$\varepsilon\$ ≤ 1

Suggested diagram: A sphere receiving electrical power and radiating energy to its surroundings, showing incoming and outgoing energy arrows.

Check Your Understanding

  1. Why does an object at constant temperature not necessarily have zero net energy flow?

  2. How would the equilibrium temperature change if the emissivity of the sphere were reduced to 0.5?

  3. Explain how the Stefan‑Boltzmann law leads to the \$T^{4}\$ dependence of radiated power.