Thermal Radiation
This topic covers thermal radiation, a method of heat transfer via infrared electromagnetic waves. Understanding how an object's properties (like colour, texture, and temperature) affect its ability to absorb and emit this radiation is key to solving problems about temperature change and energy balance.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Thermal radiation is part of the electromagnetic spectrum (infrared) and can travel through a vacuum at the speed of light.
- All objects with a temperature above absolute zero (0 K) both emit and absorb thermal radiation continuously.
- An object cools down if its rate of emission is greater than its rate of absorption, and warms up if its rate of absorption is greater.
- Dark, matt surfaces are both good absorbers and good emitters of thermal radiation.
- Light-coloured, shiny surfaces are poor absorbers and poor emitters, meaning they are good reflectors.
- The rate of thermal radiation emission increases with both the surface area and the temperature of the object.
› Why does this happen?
Where does thermal radiation come from?
The atoms inside any object are constantly vibrating because of their thermal energy. These atoms contain charged particles. The vibration and movement of these charged particles creates electromagnetic waves, which we call thermal radiation. These waves travel outwards, carrying energy away from the object. The hotter an object is, the more its atoms vibrate, and so the more energy it radiates per second.
Why are good absorbers also good emitters?
Imagine a matt black object and a shiny silver object are placed inside a sealed, warm room. After a long time, both objects will reach the same temperature as the room and stay there (a state called thermal equilibrium). The black object is a good absorber, so it takes in a lot of thermal radiation from the room's walls each second. To stop its temperature from rising further, it must radiate away the exact same amount of energy that it absorbs each second. Therefore, it must also be a good emitter. The shiny object absorbs very little radiation, so to stay at the same temperature, it only needs to emit that same small amount of energy. This shows that being a good absorber and a good emitter are linked.
Formulae
Pnet = Pabsorbed - Pemitted Use this principle to find the net power (rate of energy change) for an object. A positive result means the object's thermal energy is increasing (it's getting hotter), and a negative result means it's decreasing (getting colder).
Pincident = Pabsorbed + Preflected + Ptransmitted Use this conservation of energy principle when radiation strikes a surface. It helps you find the amount of power absorbed if you know the power reflected and transmitted.
Definitions
- Thermal Radiation
- The transfer of heat energy through infrared electromagnetic waves. It does not require a medium to travel through.
- Thermal Equilibrium
- The state where an object is absorbing and emitting thermal radiation at the same rate, resulting in a constant temperature.
Worked example
An insulated black cube is at a higher temperature than its surroundings. It emits thermal radiation at a rate of 1.2 W and absorbs radiation at a rate of 0.2 W. An identical cube, except it is painted with a shiny silver coating that makes it 4 times worse as an emitter and an absorber, is placed in the same surroundings. What is the net rate of energy loss from the shiny cube?
- 1
First, identify the emission and absorption rates for the shiny cube.
Since it is 4 times worse at both, we divide the rates of the black cube by 4.
- 2
Calculate the shiny cube's emission rate:
Pemittedshiny = 1.2 W / 4 = 0.3 W.
- 3
Calculate the shiny cube's absorption rate:
Pabsorbedshiny = 0.2 W / 4 = 0.05 W.
- 4
Calculate the net power transfer for the shiny cube using Pnet = Pabsorbed - Pemitted.
- 5 Pnet = 0.05 W - 0.3 W = -0.25 W
- 6
The negative sign indicates a net loss of energy.
The rate of energy loss is 0.25 W.
Answer: 0.25 W
Common mistakes
- ×Forgetting to calculate the NET energy change. An object absorbs and emits radiation at the same time. You must find the difference between these two rates, not just focus on one of them.
- ×Incorrectly relating surface properties. Remember the simple rule: good absorbers are good emitters (dull, dark) and poor absorbers are poor emitters (shiny, light).
- ×Misinterpreting energy flow diagrams. When radiation is incident on an object, carefully account for reflected and transmitted parts to find the true absorbed amount before calculating any net change.
No-calculator tips
- ✓Look for simple ratios. Problems often state one surface is 'twice as good' or '4 times worse' an emitter. This signals a simple multiplication or division is all that's needed.
- ✓Use logic to check your answer's sign. If an object is hotter than its room, it must be cooling, so its net energy change must be negative (emission > absorption).
- ✓Breakdown calculations. For a question like '1000 W is incident, 150 W is reflected, 600 W is transmitted', find the absorbed power by doing 1000 - 150 = 850, then 850 - 600 = 250 W. Don't try to do it all in one go.