Hello, welcome to this lesson on emitting infrared and black body radiation.

Most of the lesson will be focusing on some practical work, looking at how different objects or different surfaces might emit different amounts of infrared radiation.

And we'll finish the lesson by looking at a theoretical perfect emitter of radiation.

The outcome of this lesson is hopefully by the end of the lesson you'll be able to describe how to investigate the emission of infrared radiation by different surfaces.

You'll be able to state what is meant by a black body radiator and explain how radiation affects an objects temperature.

Here are some keywords which will come up this lesson.

Infrared radiation, emit, temperature, Leslie cube and black body.

Each word will be explained as it comes up in the lesson.

However, if you'd find it helpful to pause the video now to review these definitions for each of those keywords, then do that before the lesson begins.

This lesson has three sections.

In the first section we'll look at one way of investigating the emission of infrared.

In the second section, we'll use a piece of equipment called a Leslie cube to effectively do the same thing.

And in the third part of the lesson we'll draw together some conclusions and we'll look at black body radiators and how emitting and absorbing radiation will affect an object's temperature.

Let's get going with the first section now.

So you will know that a hot object can warm your hand without contact.

You don't have to touch a hot surface to be able to feel that it's hot, and that's because your hand will absorb the invisible infrared radiation that's emitted from hot objects.

So what is that infrared radiation? Well, it just means electromagnetic waves with frequencies in a range just lower than red light.

So here is the full spectrum of possible electromagnetic wave frequencies and you can see the frequencies and wavelengths, which are called infrared radiation.

They're the frequencies which are just lower than the frequencies of red visible light.

And all objects, not just hot ones.

Any object with a temperature is constantly emitting or radiating electromagnetic waves into the surroundings.

And that's what a thermal camera picks up.

Now the animation, which is gonna appear in just a moment, is gonna show a model of how the range of wavelengths emitted and their intensities depend on the temperature of an object between room temperature and 400 degrees C.

So it starts with room temperature and we're going up to 400 degrees C.

And what you're seeing is the shape of the graph changing between room temperature and 400 degrees C.

You can see the intensity of all wavelengths increases at higher temperatures.

We've moved towards 400 degrees C and the range of wavelengths emitted changes as well.

The range of wavelengths, as the object got hotter, the range of wavelengths got closer and closer to including some visible wavelengths, which is shorter wavelengths.

But at 400 degrees C, something is still not hot enough to be glowing and giving out any visible light as you can see in this graph.

The intensity of visible wavelengths is zero on this graph showing the distribution the the range of wavelengths given out, and that's just what I've said there temperatures up to about, it's actually about 600 degrees C.

That's when objects start to emit visible light.

So temperatures up to 600 degrees C, they might be hot, but they're not so hot that they're glowing until they get to about 600 degrees C.

So below 600 degrees C, the wavelengths which objects are emitting are mainly infrared wavelengths.

And that's what a thermal camera detects as we said earlier.

So above 600 degrees C objects will emit visible wavelengths as well as infrared.

Now that's not an exact temperature and sometimes it's difficult to detect, but it's thought to be around that temperature where objects get so hot that they start to emit visible wavelengths.

So for example, all of the objects in this picture are in the region of between 600, 1,200 degrees C, and they're all glowing.

So they're all giving out some visible wavelengths as well as infrared.

A light bulb filament in an old fashioned light bulb would actually reach temperature temperatures of about 3,000 degrees C so hot that it glows kind of more of a white hot than a red hot.

And the sun surface is around 6,000 degrees C.

So what I'm gonna show you again now is an extension of the animation I showed you before.

The previous animation went from room temperature to 400 degrees C for about 400, 450.

This animation has different scales because it's gonna start showing the distribution of wavelengths emitted when an object is 1500 degrees C, and it's going to increase in temperature to show the distribution of wavelengths given out by an object as it's heated towards 4,500 degrees C.

So temperature is increasing on the animation now.

We saw 3000 degrees C for a bulb filament, and that's 4,500 degrees C.

So you can see in all of these temperatures the distribution of wavelengths emitted includes some of the visible light wavelengths, which is shown by the visible spectrum band.

There you can also see as the temperature increases, the more intense all wavelengths emitted get.

Okay, time for a check on what we've just said.

Choose words from the below to fill the gaps.

All of these statements are about conclusions that can be made from that animation, from that graph that I've shown you about what happens to the infrared emitted, the range of wavelengths when you increase the temperature.

So choose from the words in the list and each word be may be used once more than once or not at all.

And then there's five gaps to try and fill in.

Just using choosing from those words.

Pause the video now and have a good go at that.

Hello, welcome back from your effort on that task.

We'll go through the answers.

So at higher temperatures the radiation emitted every second increases.

That should have been pretty straightforward.

The intensity of every wavelength emitted also increases the range of wavelengths emitted also increases, gets wider to include shorter wavelengths that weren't emitted before the higher temperature an object gets.

And the wavelength of the most intense radiation emitted decreases the most intense radiation emitted becomes a shorter wavelength are higher temperatures.

Very well done if you've got most or all of those right.

Okay, onto the practical for this part of the lesson, you are going to investigate whether the nature of the surface affects the infrared emitted from a test tube of hot water.

You're gonna need to use identical beakers or test tubes, which can be covered in different colours of paper.

You could also test kitchen foil with either the shiny side of the foil or the dull side facing outwards like in that second picture.

So the hot water will cool down due to conduction and evaporation as well as radiation.

We're gonna have to think about that.

Let's just go over those keywords first.

Match each energy transfer process, conduction evaporation and radiation to its correct definition.

Pause the video now and have a good go at that, right? I'll give you the answers.

Conduction was definition A that's energy transfer through a material due to a temperature difference and evaporation was definition C.

Particles of a liquid are gaining sufficient energy from collisions to escape from the surface of the liquid.

And radiation was definition B.

Well done if you got those.

So in this experiment, because we want to investigate the cooling by radiation energy transfer by radiation from different services, we're gonna have to think carefully about what's going on with conduction and evaporation.

Both of those have to be controlled so that the different temperature drops in a set time that we might get if we get differences that can only be due to a difference in the energy transfer by radiation.

We can't let conduction and evaporation affect our results.

We've gotta make sure it's the same for all of them.

It's a control variable.

They are control variables.

So firstly, using the same type and amount of paper for every tube should ensure that energy transfer by conduction out through the size of the tube and the paper is the same for each.

Now the temperature drop for evaporation should already be equal for each tube because of the identical volumes and identical shapes and identical starting temperatures of each tube.

However, we should still use a cotton wall plug or a lid or a bung to reduce the evaporation.

And using the same for each tube means whatever we do will be the same for every tube.

And if we use that cotton wall plug or lid or burn, that will help make any differences in temperature drop produce the differences in infrared emission more noticeable.

So if we reduce the cooling by evaporation, then the cooling because of infrared radiation will be more noticeable if we get rid of that background cooling by evaporation.

Another control variable is that each container should have the same starting temperature because temperature, as we've seen in the animations earlier in the lesson, will affect the intensity of the infrared emitted.

We should gonna need to start a timer.

When the water reaches a set starting temperature that you can decide when you do your test about 80 degrees will would be fine and the temperature should be recorded at regular intervals.

For example, every minute for a set time, for example, six or eight minutes or something like that.

Another control variable we'll have to think about is the mass of water 'cause that would affect the temperature drop for the same amount of energy transferred.

So you've gotta keep that the same for all the tubes that you test.

The best way of controlling the massive water in this case is by using the same volume each time.

However, using a measuring cylinder, if you boil water in a kettle and then pour into our measuring cylinder to get the volume right and then fill around getting the volume exactly right and then pour it into a test tube, chances are that hot water will have cooled down too much in the process.

In this experiment, perhaps just marking a line at the same height on each tube might be a more practical method for controlling the volume of water with sufficient precision.

Have a look at what Sofia's saying in this prediction.

Sofia's saying that she thinks the darker or duller surfaces will be the better emitters of infrared than the lighter or shinier surfaces.

Let's have a look at why she thinks that.

She's saying that darker and duller surfaces are better absorbers of infrared and that means they interact well with infrared wavelengths, darker and duller surfaces.

So if they interact well with infrared wavelengths, that makes them good absorbers and she thinks that should apply to emitting as as well as absorbing.

So she thinks the darkened surfaces will be good emitters because they're good absorbers.

Lucas has got a different prediction here.

Lucas is saying, I'm not sure how darker surfaces could be good emitters of infrared at the same time as being good absorbers.

Isn't emitting an absorbing opposite? That's his thinking, I think.

So he is saying if darker surfaces are better absorbers, wouldn't lighter surfaces be better emitters? What do you think and why? Pause video now just to consider which prediction you agree with or whether you'd like to make a different prediction to both of Sofia and Lucas.

Let's do a check on what we've said about this experiment.

So you're going to investigate whether the nature of a surface affects the infrared emitted from a container of hot water, which the following are control variables in this investigation? Pause the video and chooses which options you think.

Okay, I'll give you some feedback on which are control variables in this investigation.

The material of the outer layer should be a control variable.

That's gotta be the same for each one 'cause we're not testing different materials, we've gotta use the same material.

So the energy transfer by conduction cools the water in the same way every time.

The only thing we're changing is the colour or the finish of the outer layer.

We're gonna measure the temperature fall in a set time.

So that's B and C are independent and dependent variables.

And in fact all the others are control variables.

We're gonna keep the same thickness of that layer, the same volume of water, the same shape or size container, the same starting temperature of water and the same starting temperature of the container.

All of those could affect the cooling of the water when the only thing we want to allow to affect the cooling of the water is the amount of infrared it emits because of its different colour or finish.

Okay, time for you to do this practical now then.

So here your task, as we've said, is to investigate whether the nature of a surface layer wrapped around a container of hot water affects the infrared emitted.

Use the method that we've talked about.

Choose different colours or finishes of the same material for example, paper or foil like I described.

Ensure that other variables that could affect the temperature drop in a set time are controlled or we've just been through those.

Cover the top of the container to minimise evaporation and record oil results in a results table.

Results such as one like is on the screen now.

So pause this video now and have a go at doing this investigation.

Well done for your effort on that task.

Let's look at some example results for your feedback now.

So here are my example results.

When I trialled this experiment, I tested yellow paper, blue paper and black paper, and I used a starting temperature at 78 degrees.

But you can see my results actually don't show any difference between the different colours of paper used.

And that's fine because sometimes you test out an idea and actually that idea didn't affect the results at all in some experiments.

So that's just what I got.

I wonder if your results showed a difference or if your results were like mine and didn't really show any difference between the different colours of paper wrapped around different colour containers.

In the next learning cycle, we'll look at possible reasons why.

This takes us to the second section of the lesson where we are going to investigate the emission of infrared radiation by different surfaces, but using a Leslie cube.

So a different method of investigating the same question.

So data from the experiment we did in learning cycle one could be plotted on a line graph with a line for each colour or finish of the surface used.

You can see we've got a graph here of temperature against time.

And we've got three different lines, three different sets of data all plotted on the same graph.

However, there is another way you could plot the data or present the data.

And you could plot the temperature drop for each colour.

Just the temperature drop from the start point to the finish point could be plotted on a bar chart.

And you'll recognise that this isn't my data because this shows a significant difference between each colour of paper tested.

So these graphs are showing example or possible data.

So here's Aisha's results from these experiments, which is very similar to my data.

Now, Aisha's results do not show a difference between the different colours of paper used.

And I want you to decide which of the following could possibly explain why there was no difference between the different colours of paper used and the cooling.

Pause video and choose which you think.

Okay, I'll give you some feedback now.

Well statement A, each colour of paper emitted similar intensities of infrared.

Well that could, that is a possible explanation of why there's no difference in these results.

Maybe each surface really did emit the same amount of infrared.

Statement B, any differences in radiation were too small to be detected against other factors that cause cooling like evaporation or conduction for example.

That's a possibility two.

And statement C, the paper was too far separated from the hot water by glass, which is the thermal insulator to significantly affect the cooling of the water.

I think that is a possibility two.

So we could really do with a better method for investigating how the colour or nature of the surface affects the amount of infrared emitted.

Well, a Leslie cube is a better way of investigating how the nature of the surface affects the infrared emitted.

And John Leslie, who is named after, was a Scottish physicist in the 1800s.

So a Leslie cube is a hollow metal cube that can be filled with hot water and each side is painted with a different colour or finish normally four from that list below.

An infrared detector can detect the different intensities of infrared emitted from each side of Leslie cube.

Instead of an infrared detector, you could use an infrared thermometer, which would measure a higher temperature when more infrared is emitted from a surface, even when every surface on a Lesley cube is the same temperature, the infrared thermometer will measure them as different temperatures if they're giving out different amounts of infrared because that's how the infrared thermometer works.

Let's do a check from what I just said by infrared thermometers.

A Leslie cube is filled with hot water.

An alcohol thermometer is inserted into the hot water as you can see it in the picture.

After five minutes the alcohol thermometer of read 71 degrees, The following readings were then taken using an infrared thermometer at the same distance from each side.

So you can look at the results from each side in the table.

Which of the following is the best estimate of the temperature of the shiny metal side? Choose which you think from A, B, C, or D.

Five seconds to decide.

Right, I'll tell you the right answer.

It is D.

The best estimate of the shiny metal side is 70 degrees C because that is the temperature of the water.

The alcohol thermometer read, 71 degrees C.

So 70 is the best choice.

And that metal side has been in contact with that hot water for five minutes.

So it's gonna be the same temperature.

The shiny metal side is giving out less infrared radiation than the other side of the cube when it's at 70 degrees C.

That's why the infrared thermometer reads it as having a lower temperature around 40 degrees C, when in fact it must be have a temperature of around 70 degrees C.

Well done if you've got that.

Now instead of a infrared detector or an infrared thermometer, you could use a thermal camera to detect the different intensities of infrared from each side of a Leslie cube.

And that would then be shown as an image with different colours for different intensities of infrared emitted from each side.

Now all faces of the cube in the images on the right are the same temperature, but the shiny metal side appears to be a much lower temperature in the thermal image because it's emitting much less infrared.

And what you can also see in this picture, which is quite nice, is you can see that the shiny metal side of the Leslie cube is reflecting infrared that's emitted from the hand.

So you can see a reflection of the hand in that shiny metal side in the thermal image.

So in the infrared.

So infrared thermometers and infrared detectors and thermal cameras can't provide accurate temperature measurements of surfaces that reflect infrared because they might really be measuring or showing the infrared emitted from a different object and reflected from that object.

So how do we use the Leslie cube to do this experiment? The field Leslie cube, obviously it should be filled with hot water, should then be sealed with a bung to reduce the rate of cooling by evaporation.

A heatproof mat will also reduce conduction to the tabletop below.

If you use a bung with a fitted alcohol thermometer, that means you can check what temperature the water is when the measurements are taken.

That means you can repeat measurements at the same temperature of water if needed.

You can also check that the temperature stays constant.

That should be a control variable.

Of course, as you measure each side of the Leslie cube.

You should wait three to five minutes after adding the hot water before taking the measurements.

That should be enough time to allow all the sides to come to the same temperature, the temperature of the water.

And the infrared thermometer must be at the same distance from the cube and at the same angle when it takes each measurement 'cause those factors would affect the infrared intensity measurements that it's making.

Let's do a quick check on what we've just said.

A Leslie cube and infrared thermometer are used to investigate the emission of infrared by different surfaces.

Which of the following are control variables? Things that must be the same every test so they don't affect the results.

Pause video and choose from this list, which are control variables? Okay, I'll give you some feedback now.

The actual temperature of each side of the cube, that should be the same every time 'cause if one was one side of the cube was hotter than the others, then it would emit more infrared.

But that's 'cause it was hotter, not 'cause it's a better emitter.

D should be a control variable and E, because they would affect the measurements of infrared by the infrared thermometer if that angle or the distance was different.

C is the independent variable.

That's what you're changing in each test.

B is the dependent variable.

That's what you're seeing if it's different each time.

And F is another possible independent variable, something you can change to see if it has an effect on your measurements of the emission of infrared.

Makes sense to mention at this point that using a Leslie cube and an infrared thermometer is clearly gonna be a better method for investigating the emission of infrared from different surfaces than covering glass test tubes in different colours of paper and measuring the temperature drop in a set time.

And there's at least three reasons why.

Firstly, in the Leslie Cube, the different surfaces are all at the same temperature at the same time.

Secondly, the sides of Leslie cube are metal, which is a good thermal conductor.

So the surfaces tested are in better thermal contact with the hot water in Leslie cube than in the experiment with glass test tubes.

And thirdly, using an infrared thermometer directly measures the infrared emitted, whereas a temperature drop is affected by other factors too, like evaporation and conduction, which are difficult to reduce to make differences in infrared emission noticeable.

Let's do a last check on ideas using about using a Leslie cube.

Andeep used a Leslie cube and an infrared thermometer to investigate the emission of infrared from different surface.

Here are his results.

What type of graph should Andeep draw and why? Fill in the gaps to answer that question.

Pauses video now have a good go at that.

Let's see how he got on.

Andeep should draw for this experiment, a bar chart.

What's the reason why? It's because his independent variable, what he changed every time, which is the nature and colour and of the surface, the independent variable in the experiment is categoric.

It comes in categories.

So you need a different bar for each category.

The other kind of variable is a continuous variable, which comes as a scale of of numbers like temperature.

So the y axis will be a scale of numbers, but the x axis, you need a bar for each category 'cause the independent variable is categoric you need to draw a bar chart.

So have a go at doing that now.

Here's Andeep's data again.

Draw a bar chart of his results.

Pause video now have a good go at doing that.

You should do it on graph paper or square paper if you don't have graph paper.

So it's a proper to to scale graph.

Here's some feedback on that task.

So here is an example graph and the first thing is got that makes it a good graph is axis titles and units.

That's a surface and apparent temperature in degree C.

It's got an x-axis label, which is the label for each bar.

It's got a good y-axis scale for different temperatures going up in even amounts.

It's got accurate neat bars and it's got a meaningful graph title.

Well done if yours has got all those things as well.

This takes us to the final section of the lesson where we'll draw some conclusions.

We'll look at theoretical objects called black body radiators and we'll look at how emitting and absorbing infrared radiation can affect an object's temperature.

Our first conclusion from this lesson should be the good absorbers of infrared.

Like dull dark surfaces are always also good emitters of infrared and vice versa.

Good emitters are good absorbers and good absorbers are good emitters of infrared radiation.

And that's because if substances have the right structure to strongly absorb infrared, they also have the right structure to strongly emit it.

So that was Sofia's idea from earlier in the lesson.

Well done if you agreed with Sofia, 'cause that's right.

Dark and dull surfaces are good emitters and absorbers of infrared, whereas shiny metal surfaces are poor emitters, absorbers of infrared.

And apart from those two observations, actually the colour of a surface alone is not always a good guide to how well it emits and absorbs infrared.

White paint and water, which is transparent are actually good emitters and absorbers of infrared.

So the rules for absorbing infrared are different to the rules for absorbing visible light.

White paint would reflect a lot visible light, but it actually absorbs just as much infrared as black paint does.

So material may be a good absorber and emitter of some wavelengths of electromagnetic wave and a poor absorber and emitter of other wavelengths of electromagnetic wave.

Now it's possible to imagine an object that would absorb 100% of any wavelength of electromagnetic radiation that falls upon it.

So that means for this object that we're imagining in our minds, no radiation at all that lands on it would be reflected from it or would be transmitted through it.

100% of all radiation, any wavelength would be absorbed.

Such an object that we can imagine would be called a perfect black body.

And black signifies absorbing all wavelengths of light and reflecting none body is just an old fashioned term for object.

So it's important to emphasise that a perfect black body object would be a theoretical idea.

We do not think perfect black body objects actually exist.

However, treating some objects as if they are perfect black bodies that can help to simplify scientific models, for example of stars and planets.

And it can still produce accurate predictions because some objects are actually pretty close to being a perfect black body for a good range of wavelengths.

So it's a good model sometimes.

A perfect black body object would also be the best possible emitter of radiation because good absorbers are also good emitters.

And we're saying a perfect black body object would be a perfect absorber of all radiation.

So it would also be the best possible limiter of radiation.

Let's do a check on what we've just said.

Match each kind of surface to how well it emits and absorbs radiation.

Pause the video and have a go at this.

Let's see how you got on.

So dark and dull surfaces.

They're usually good emitters and absorbers of infrared white surface is sometimes good and sometimes poor emitters and absorbers of infrared shiny metallic surfaces are usually poor emitters and absorbers of infrared.

And a perfect black body surface would be a theoretical surface that absorbs 100% of any radiation that falls upon it.

Well done if you got those right.

We're going to look now at how emitting and absorbing infrared radiation can affect the temperature of an object.

Because all objects are not just constantly emitting infrared radiation, they're also constantly absorbing infrared radiation from their surroundings at the same time as emitting their own radiation.

So an object's temperature is gonna be affected by the power of the radiation being absorbed, which would warm it up compared to the power of radiation being emitted, which would cool it down.

So if the power absorbed is greater than the power emitted, then the object is gonna get hotter.

And an example of that is a rock heating up in the sun in the desert, the power absorbed from the sun's radiation is greater than the power emitted by its own infrared radiation.

If the power absorbed is less than the power emitted by radiation, then an object is gonna get cooler.

For example, a hot baking tray that's come outta the oven is gonna be emitting more infrared radiation into the surroundings than it gains infrared radiation from the surroundings.

So it's gonna cool down.

And if the power absorbed is equal to the power emitted, then an object's temperature is going to be constant.

You could also think about that the other way around.

If an object's temperature is constant, then the power absorbed by infrared radiation is likely to be equal to the power emitted by infrared radiation.

For example, a tomato in a fridge is gonna be temperature, it's gonna be emitting just as much radiation into the surroundings of the fridge than it absorbs from the surroundings of the fridge.

Because it is constant temperature, it can't be gaining more than it's losing and vice versa.

And these ideas apply to the Earth as a whole in space as well.

Now for the past 2000 years, Earth's average global temperature was stable and that's because the radiation absorbed by Earth was balanced by the levels of radiation emitted by Earth.

So Earth temperature generally didn't rise or cool.

However, in the present day, although the incident radiation from the sun is the same, Earth's average global temperature is rising.

And you probably already know why? It is because less radiation is being emitted by Earth into space.

So we've got the radiation absorbed and the radiation emitted with the same, but now less radiation is being emitted by Earth into space.

We've got imbalance between the radiation absorbed and the radiation emitted.

And why is that? Well, it's due to human activity having caused increased levels of greenhouse gases in the atmosphere.

Greenhouse gases like carbon dioxide and methane.

What those gases do in the atmosphere, one of the things they do in the atmosphere is they absorb some of the radiation, which is emitted by Earth and reradiate it back down to Earth surface, preventing its escaping.

So increased levels of greenhouse gases in the atmosphere are going to reduce the radiation emitted from Earth into space.

So Earth is absorbing more radiation than it's emitting and warming up.

And there's another reason as well because less instant radiation is actually being reflected by Earth because the ice caps have begun to melt and shrink because of their increased temperature.

So more incident radiation from the sun is actually being absorbed as well.

So not only have we got less radiation being emitted than is absorbed, the amount of radiation being absorbed has actually increased as well.

Making even bigger imbalance between the power of radiation absorbed and the power of radiation emitted.

And obviously what I'm describing here is climate change and global warming, which lots of scientists and politicians are debating about the best way forward to prevent the most harmful effects that climate change could potentially cause in the future.

Let's do a check on some of the ideas we just talked about.

Reptiles cannot maintain their body temperature without help from the environment they bask in the sun to stay warm by absorbing radiation, a particular lizards body emits radiation with a power of 120 watts.

Assume the lizard has no other source of heating except for the radiation it absorbs.

What power of radiation would the lizard need to absorb in order to keep its body temperature constant? Pause video and come up with an idea.

The answer is 120 watts.

And that's because to maintain a constant temperature, the power of radiation absorbed must be equal to the power of radiation emitted.

Time for a task to complete this lesson.

Sort the following list of statements into the three categories listed in the table.

Pause video now and give that task a good go.

Okay, I'll give you some feedback now.

Here are the correct groups to sort each statement into statements.

A, D, G, and J are all objects which were warming up because the radiation absorbed at the power of the radiation absorbed must be greater than the power of the radiation emitted statements B, E, F, and H were all objects whose temperature would be decreasing because the power of the radiation absorbed must be less than the power of the radiation emitted.

And statement C and I described objects which would be maintaining a steady or constant temperature because the power of the radiation absorbed would be equal to the power of radiation emitted.

Well done if you got those right.

Here's a summary of the lesson.

All bodies or objects of all temperatures emit and absorb infrared radiation all the time.

Good absorbers of infrared are always also good emitters of infrared.

The range of wavelengths emitted and their intensities depends on an object's temperature.

Hotter bodies emit more infrared in the same time.

The emission of infrared by different surfaces can be investigated using a Leslie cube and an infrared thermometer.

Dark and dull surfaces are good emitters and absorbers unlike shiny metal surfaces.

A black body is a theoretical object that would absorb all incident radiation that falls on it.

And the temperature of a body depends on the rates, the powers of which radiation is absorbed and emitted.