Topic 3 of 4
Thermal radiation
Energy can transfer by electromagnetic radiation across a vacuum. No material medium is required between the source and receiver.
Electromagnetic waves can transfer energy. This pathway differs from conduction through particles and convection with moving fluid.
Ordinary warm objects emit thermal radiation, mainly infrared at everyday temperatures. They need not glow visibly. Radiation from the Sun reaches Earth through space, showing that air is not required to carry it.
Objects emit and absorb
An object can emit radiation and absorb radiation from its surroundings at the same time. Its net energy change depends on the difference between what it receives and what it emits.
A hot object in cooler surroundings usually loses energy overall. At thermal equilibrium with surroundings at the same temperature, emission and absorption balance, so there is no net radiative transfer even though both continue.
Radiation transfers energy across a vacuum
Both bodies emit radiation. The arrows show directions, with no numerical rates implied.
Surface colour and texture
Otherwise identical bodies at the same surface temperature, with the same exposed area and surroundings.
For this usual comparison, dull black is a better emitter. It is also a better absorber of the same incident radiation.
Surface temperature
Same surface finish, exposed area and surroundings; only the surface temperature differs.
A hotter surface emits radiation at a greater rate.
Exposed surface area
Same surface finish, surface temperature and surroundings; the exposed surface area differs.
A larger exposed area gives a greater total rate of emission under these conditions.
Net radiative transfer also depends on what is absorbed from the surroundings. Real cooling can include conduction and convection as well.
Three factors to compare
- Surface colour and texture
- For the usual comparison of similar opaque surfaces, a dull black surface is a better absorber and emitter of thermal radiation than a shiny metallic surface. The shiny surface reflects a greater fraction of incident radiation. Compare at the same temperature, exposed area and surroundings. Visible colour alone is not a universal rule for every material and wavelength.
- Surface temperature
- Keeping the surface finish and area unchanged, a hotter surface emits energy at a greater rate. To compare net loss, also keep the surroundings unchanged: an object's temperature alone does not tell you how much radiation it receives.
- Surface area
- At the same temperature and with the same finish, a larger exposed surface emits energy at a greater total rate. For absorption under equal illumination and orientation, a larger exposed area also intercepts more radiation. Use the area actually exposed, not a hidden part of the surface.
A comparison should state what is held fixed. Changing both the finish and temperature does not isolate the effect of finish; changing the surroundings can alter net transfer even when the object's surface is unchanged.
Investigate absorption with a fair comparison
Use two otherwise identical metal cans containing the same mass of water. Give one a dull black outer finish and leave the other shiny. Treat the thin finish as having a negligible effect on the total mass and thermal response apart from radiation.
- Keep the samples comparable: use the same can material, mass and geometry, the same water mass, and the same initial temperature.
- Keep the source comparable: expose both to the same lamp at equal distances and orientations for the same duration. Keep the surroundings and air movement the same.
- Measure consistently: place matching thermometers at the same immersion depth in the water, clear of the can walls and base. Use the same lids and any stirring procedure.
- Record the change: read temperature at the same elapsed times, allowing for the sensors' response. Repeat the comparison from matched starting conditions.
| Time / min | Dull black can / °C | Shiny can / °C |
|---|---|---|
| 0 | 20 | 20 |
| 2 | 24 | 22 |
| 4 | 28 | 24 |
| 6 | 31 | 25 |
Interpret the readings
Compare temperature rises over the same interval
After 6 min, the dull black can's water has risen by 31 - 20 = 11°C. The shiny can's water has risen by 25 - 20 = 5°C.
With the matched samples, the greater rise is evidence of greater net energy gain under these conditions. It is consistent with the dull black surface absorbing radiation more effectively.
The readings do not show that every joule received was retained. Both cans also transfer energy to their surroundings, and conduction carries energy from each warmed can into its water. The temperature-rise ratio is not automatically a ratio of radiation absorbed from the lamp.
A can closer to the lamp may receive more radiation for that reason alone. A thermometer touching the illuminated metal may read a locally hotter region rather than the water. Repeating readings does not fix unequal distances or misplaced sensors.
A cooling comparison needs similar care: match material, mass, shape, initial temperature, sensor placement and surroundings. Dull and shiny surfaces differ in radiative emission, but a simple cooling experiment in air also includes conduction and convection. Do not claim those routes have vanished.
Emission is not the same as net loss. Keep both incoming and outgoing radiation in the explanation. A shiny surface is a weaker emitter and absorber in the stated comparison, not a surface that transfers no energy at all.