Lesson Explainer: Infrared Radiation | Nagwa Lesson Explainer: Infrared Radiation | Nagwa

Lesson Explainer: Infrared Radiation Physics • Third Year of Secondary School

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In this explainer, we will learn how to describe how the temperature of an object and its surface features affect the object’s emission and absorption of infrared radiation.

The electromagnetic spectrum is the range of all types of electromagnetic radiation, more commonly known as light. What type of light it is called depends on its wavelength, as seen in the chart below.

This explainer will focus on the region between visible light and microwaves, infrared radiation (hereafter IR). The wavelength of light where IR begins is at the lowest wavelength portion of visible light, which is the color red. This is approximately 700 nm and continues until it reaches microwaves at around wavelengths of 1 mm.

IR is interesting because of how it interacts with heat. When an object has heat, which is to say, a temperature, it emits IR. This means that every object you see is emitting light, but just at wavelengths humans cannot detect. This is the technology behind night vision cameras: they are able to detect IR from the surrounding environment.

Hotter objects emit more IR, and colder objects emit less. The only time an object does not emit any IR is when it has no temperature at all, at absolute zero. When you hold your hand near a hot pan, the heat you feel is coming from invisible light in the infrared spectrum, which heats your hand. The diagram below shows the same pan in a dark room, comparing the visible light and infrared light it is emitting. Brighter colors means more light.

The pan is not emitting any visible light, only infrared.

The rate at which an object is able to emit IR is related to how well it is also able to absorb it. Objects that can absorb IR and heat up quickly are able to also emit it quickly. When light interacts with an object though, it is not just absorption. There are three main ways light can interact with an object, as shown below.

Reflection is when light reflects off of a surface, such as a mirror, or off of polished or wet surfaces. If a surface is reflective, then IR bounces off of it, making it more relatively cool. It is unable to absorb as much light.

Transmission is when light passes through an object, like sunlight through a window. If IR passes through an object, this means it did not transfer heat as effectively.

Absorption is when the light is absorbed by the object, like how black paint makes a room darker. Objects that absorb more light, and thus are better at absorbing and emitting IR and thus heat, are darker colors.

This means color is an important indicator of how much IR is being absorbed and emitted.

Let’s look at an example.

Example 1: Recalling the Effect of Color on the Thermal Absorption and Emission Properties

Which color of a surface makes it better at emitting and absorbing infrared radiation, black or white?

1. White
2. Black

Think of how, during days when the sun is out, black asphalt rapidly heats up and is uncomfortable to touch. When touching the sidewalk, which is made of stone but a lighter color, it is much more bearable. The darker asphalt is better at absorbing the IR from sunlight, and emitting it.

Similarly, at night time, the asphalt is much cooler compared to its surroundings. It is very good at emitting its IR more rapidly, until it is no longer so hot.

An object that is better at emitting and absorbing IR will tend to be darker, and one that is worse will tend to be lighter. Wearing white on hot days can keep you a little cooler than wearing black.

The correct answer is B, black.

Besides color, the second factor affecting how much IR is absorbed or emitted is the reflectivity of a surface. A highly reflective or glossy material, such as cut obsidian, will poorly absorb and emit IR, even though it may be a deep black color.

In the other direction, rough surfaces such as fabric or cloth will not reflect IR well, instead absorbing it nicely. They will also emit it more easily, making it an ideal material for warm beds.

Let’s look at an example.

Example 2: Thermal Radiation and Reflectivity

An object that is reflective absorbs infrared radiation better than if it had any other reflectivity. An object that is reflective emits radiation better than if it had any other reflectivity.

1. strongly, weakly
2. weakly, weakly
3. weakly, strongly
4. strongly, strongly

An object with strong reflectivity will reflect more IR, meaning it must be weakly reflecting to absorb more IR.

An object with strong reflectivity will not be able to emit IR as well as one that is weakly reflective.

The correct answer is B, weakly, weakly.

The third factor is the temperature of an object. Temperature does not affect the absorption of IR, but rather the emission of it. Any object with a temperature, as long as it is above absolute zero, will emit IR. Hotter objects emit more IR, shown by the light coming off the objects. A brighter color means more IR.

Hotter objects emit more IR along with wider wavelengths too. For example, here’s a graph that shows intensity (amount of light) on the -axis and wavelength (type of light) on the -axis. The same object is shown on the graph at three different temperatures: A, B, and C.

Temperature A is hotter than temperature B, and temperature B is hotter than temperature C. Infrared covers a wide range of wavelengths from 700 nm to 1 mm, meaning that hotter temperatures are emitting more IR at various wavelengths.

The fourth factor that affects the absorption and emission of IR is the surface area. Simply put, IR can only be emitted and absorbed at the surface of an object: its insides do not matter. So an object that is larger is better able to emit IR and also has a larger area for absorbing incoming IR.

The diagram above shows a small object, a larger object, and then a larger object with holes in it, all with the same temperature. The larger objects are better able to emit and absorb IR than the smaller one, and the one with holes in it is able to do so the best thanks to its larger surface area provided by the holes.

Many reptiles have frills that expand and contract, allowing them to rapidly warm themselves under the sun or cool themselves off as needed. An extreme example is the extinct animal Dimetrodon, which had a huge frill that allowed it to warm up or cool itself very quickly as it needed, giving it an advantage against other animals during the Paleozoic period.

Example 3: Properties Affecting Thermal Radiation

Which of the following properties of objects does not directly affect the amount of infrared radiation the object emits and absorbs?

1. Reflectiveness
2. Color
3. Temperature
4. Surface area
5. Mass

Reflectiveness definitely determines how much IR is emitted and absorbed: a mirror reflects light off, turning it away rather than absorbing it.

The color also affects infrared radiation. A white shirt will be cooler in direct sunlight than a black one.

Temperature does not affect the absorption but rather the emission of IR. Hotter objects emit more IR.

A higher surface area means more IR is able to hit the object in the first place, allowing it to absorb more. IR is also only emitted through the surface of an object, so larger surface areas mean larger IR emissions.

The mass of an object does not determine how much IR is absorbed or emitted. If two objects of the same shape, color, temperature, and reflectivity are compared to each other, they will be found to emit the same amount of IR. It does not matter how much more thick or dense one is than the other, it will be the same.

The correct answer is thus E, mass.

Now that we know the factors that affect IR absorption and emission, we can begin looking at everyday objects in a variety of environments and generally determine how well they absorb or emit IR.

A rough, darkly colored fabric, such as dyed wool, will absorb and emit IR better than a smoother lighter colored fabric such as nylon. The reflective interior of a thermos will not absorb very much IR, keeping the heat in the liquid instead of transferring it outside the container.

Let’s look at some examples.

Example 4: Water Bottle Thermal Radiation

A bottle containing water is placed in constant-intensity sunlight and absorbs infrared radiation. During the bottle’s exposure to sunlight, the temperature of the water remains constant. Which of the following statements is correct?

1. The water absorbs more energy from infrared radiation than the energy it loses by cooling.
2. The water does not receive any energy from the sunlight.
3. The water loses more energy by cooling than the energy of the infrared radiation that it absorbs.
4. The water loses exactly as much energy by cooling as the energy of the infrared radiation that it absorbs.
5. Infrared radiation absorption and emission have no effect on temperature.

If the water absorbed more energy from the radiation than it lost by cooling, it would begin warming up, but its temperature is constant. It is not A.

The water is absorbing infrared radiation, so it is receiving energy from the sunlight, so it is not B.

If the water loses more energy via cooling than it gains from the sun, it would cool down, but its temperature is constant. It is not C.

Infrared radiation absorption and emission definitely have an effect on the temperature, so it is not E.

For the temperature of the water bottle to remain constant, despite continuously gaining energy from the sun, it would have to cool at the same rate it absorbed IR. The answer is thus D.

Example 5: Plastic Cube Thermal Radiation

A plastic block is placed in constant-intensity sunlight and absorbs infrared radiation. During the block’s exposure to sunlight, its temperature decreases. Which of the following statements is correct?

1. The plastic does not receive any energy from the sunlight.
2. Infrared radiation absorption and emission have no effect on temperature.
3. The plastic loses exactly as much energy by cooling as the energy of the infrared radiation that it absorbs.
4. The plastic loses more energy by cooling than the energy of the infrared radiation that it absorbs.
5. The plastic absorbs more energy from infrared radiation than the energy it loses by cooling.

The plastic is definitely receiving energy from the sunlight, as it is stated to be absorbing IR, so it is not A.

The IR absorption and emission do have an effect on temperature, so it is not B.

If the plastic block lost as much energy by cooling as the energy it gained, then its temperature would remain constant, but it is stated to be decreasing. It is not C.

If the plastic absorbed more energy from the IR than it lost, then it would be increasing in temperature; but it is decreasing, so it is not E.

The temperature of the block is decreasing, so it must be losing more energy than it is gaining. Since it is only gaining it via absorbing IR, it must be losing more energy from cooling than it is by absorbing IR. So the answer is D.

Let’s summarize what we have learned in this explainer.

Key Points

• Infrared radiation (IR) is light with a wavelength between 700 nm and 1 mm.
• IR is responsible for heat felt from hot objects.
• All objects with a temperature above absolute zero emit IR.
• The factors affecting the absorption and emission of IR are color, reflectivity, temperature, and surface area.

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