Describe how the rate of emission of radiation depends on the surface temperature and surface area of an object

2.3.3 Radiation

Objective

Describe how the rate of emission of radiation depends on the surface temperature and surface area of an object.

Key Concepts

• All objects with a temperature above absolute zero emit thermal radiation, which is electromagnetic radiation in the infrared region.
• Thermal radiation can travel through a vacuum; it does not need a material medium.
• The energy emitted per unit time is the radiative power (or rate of emission).
• For an ideal black‑body the radiative power is given by the Stefan‑Boltzmann law.

Stefan‑Boltzmann Law (Ideal Black‑Body)

The total power \(P\) radiated by a black‑body is proportional to its surface area \(A\) and to the fourth power of its absolute temperature \(T\):

\[

P = \sigma A T^{4}

\]

where \(\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}\) is the Stefan‑Boltzmann constant.

Real Surfaces – Emissivity

Real objects are not perfect black‑bodies. Their ability to emit radiation is described by the emissivity \(e\) (dimensionless, \(0 \le e \le 1\)).

\[

P = e\,\sigma A T^{4}

\]

Typical emissivity values:

Dependence on Surface Area

With temperature \(T\) and emissivity \(e\) constant:

\[

P \propto A

\]

• Doubling the exposed area doubles the radiative power.

Dependence on Surface Temperature

The temperature appears to the fourth power, giving a very strong dependence:

\[

P \propto T^{4}

\]

• Doubling the temperature (e.g. 300 K → 600 K) increases the power by \(2^{4}=16\) times.
• A 10 % rise in temperature (\(T\to1.10T\)) gives \((1.10)^{4}\approx1.46\) → a 46 % increase in power.

Combined Effect (Changing More Than One Parameter)

\[

\frac{P{2}}{P{1}} = \frac{e{2}}{e{1}}\;\frac{A{2}}{A{1}}\;\left(\frac{T{2}}{T{1}}\right)^{4}

\]

This ratio is useful for exam‑style calculations involving changes in area, temperature, or surface coating.

Summary Table

Worked Example

Problem: Calculate the radiative power emitted by a square black plate (emissivity \(e=0.95\)) that is 0.5 m × 0.5 m and has a temperature of 400 K.

1. Surface area: \(A = 0.5 \times 0.5 = 0.25\ \text{m}^{2}\).
2. Insert values into the Stefan‑Boltzmann equation:

   \[

   P = 0.95 \times 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4} \times 0.25\ \text{m}^{2} \times (400\ \text{K})^{4}

   \]

3. \((400)^{4} = 2.56 \times 10^{10}\).

4. \[

   P \approx 0.95 \times 5.67 \times 10^{-8} \times 0.25 \times 2.56 \times 10^{10}

   \approx 3.45 \times 10^{2}\ \text{W}

   \]

   \(\displaystyle P \approx 345\ \text{W}\).

The plate radiates roughly 345 W of power.

Real‑World Connections

• Clothing colour: Black fabric (high \(e\)) absorbs and re‑emits more infrared radiation than white fabric, so it feels hotter in sunlight.
• Radiators: Increasing the exposed surface area of a metal radiator raises the heat it can emit to warm a room.
• Earth’s energy balance: The planet’s surface emissivity (~0.95) determines how efficiently absorbed solar energy is radiated back to space, influencing climate.
• Polished metal spoon: Low emissivity (\(e\approx0.05\)) means it emits less infrared radiation, so it feels cooler to the touch than a matte black spoon at the same temperature.

Simple Classroom Investigation

Objective: Demonstrate the dependence of radiative power on surface area and temperature.

1. Obtain two metal blocks of the same material but different exposed surface areas (e.g., a small cube and a larger plate).
2. Heat each block with a Bunsen burner until they reach the same colour (approximate temperature).
3. Place an infrared thermometer a fixed distance from each block and record the temperature rise of a nearby water container.
4. Compare the temperature changes: the block with the larger area should cause a greater rise, confirming \(P \propto A\).
5. Repeat with a single block, first at a lower flame (lower \(T\)) then at a higher flame (higher \(T\)). Observe that the temperature rise scales roughly with the fourth power of the measured surface temperature, illustrating \(P \propto T^{4}\).

Safety note: Use tongs for hot objects and wear protective goggles.

Suggested Diagram (for the teacher)

• Sketch a sphere labelled “black‑body”. Mark its surface area \(A\) on the outline and write the temperature \(T\) inside.
• Draw arrows radiating outward, labelled “thermal (infra‑red) radiation”.
• Beside it, sketch a polished metal sphere with fewer, thinner arrows to visualise low emissivity.

Common Exam‑Style Questions

• Predict how the radiative power changes when the temperature of a filament is increased from 1500 K to 1800 K.
• Compare the radiative power of two objects at the same temperature but with surface areas of 0.02 m² and 0.08 m².
• Explain why a polished metal spoon feels cooler than a matte black spoon when both are at the same temperature.
• Calculate the new power emitted if the emissivity of a surface is increased from 0.30 to 0.90 while \(A\) and \(T\) remain constant.

Key Take‑aways

• The rate of emission of thermal (infra‑red) radiation follows \(P = e\,\sigma A T^{4}\).
• Doubling the surface area doubles the power; doubling the temperature increases the power by a factor of 16.
• Emissivity (\(e\)) measures how closely a real surface behaves like an ideal black‑body.
• Thermal radiation travels through vacuum – no medium is required.