Radiation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a white plate with a black spot on it, heat it until it is very hot, and carry it into a dark room. Which part glows more brightly: the white plate, or the black spot? The white part, surely, right? It looks brighter in daylight. Hmm, hold that thought. Today: radiation. By the end, you will know the answer, and exactly why. This comes up often in JEE, so let's begin.

0:32Radiation: heat that needs no medium

So, what is radiation? Every body above zero kelvin, the capital K on screen stands for kelvin, emits electromagnetic radiation. Okay? Now think. Take a block of ice, freezing cold. Does it radiate too? Yes, it does! Ice is far above zero kelvin. How much a body radiates depends on its temperature. And no medium is needed, which is how the Sun's heat crosses nearly empty space to reach us. Right? Now, radiation falling on a body is partly absorbed, partly reflected, and partly passed through. These fractions are the absorptivity, small a, the reflectivity, small r, and the transmissivity, small t. So a plus r plus t equals one. Clear? And the perfect absorber, with a equal to one, absorbs everything that falls on it. We call it a black body. And it is also the best emitter. Got it?

1:28Kirchhoff's law: good absorbers are good emitters

Now, Kirchhoff's law. Fix the temperature, capital T, and the wavelength, lambda, the Greek letter on screen. Let capital E be how strongly a body emits, its emissive power. Then capital E over a is the same for all bodies, equal to capital E of a black body. With me? So think. Capital E over a is fixed. If a body has a large absorptivity, what happens to its emissive power? It is large too! Yes, exactly! Good absorbers are good emitters. We call the emitting ability the emissivity, small e, and Kirchhoff's law tells us that e equals a. Okay? Now, back to our puzzle. Plate and spot are equally hot. The white part reflects most light, while the black spot absorbs it. So in the dark room, which part glows more brightly? The black spot! Correct! It is the better absorber, so, by Kirchhoff's law, it is also the better emitter. Puzzle solved! Right?

2:34Stefan-Boltzmann law: power grows as T to the power 4

Now the big one: the Stefan-Boltzmann law. The power radiated, capital P, equals e sigma A T to the power four. Sigma is the Greek letter on screen, a constant we will meet in a moment. Capital A is the surface area, and capital T is the absolute temperature, in kelvin, never in Celsius. The emissivity e lies between zero and one; the signs mean less than or equal to. Clear? Sigma is the Stefan-Boltzmann constant: five point six seven times ten to the power minus eight watt per square metre per kelvin to the power four. For a black body, e equals one. Okay? Now tell me. Double the absolute temperature of a black body. By what factor does its radiated power change? Did you say two times? Oh no, careful! Power goes as T to the power four, and two to the power four is sixteen. So the power becomes sixteen times. With me? Every body also absorbs radiation from its surroundings, at temperature T naught, the T with a small zero below. So the net loss is e sigma A, times T to the power four minus T naught to the power four. Got it?

3:47Wien's displacement law

Now look at this graph: black-body spectra at three temperatures. As the temperature rises, does the peak move to shorter wavelengths, or longer? Shorter! Yes, exactly! That is Wien's displacement law: lambda m times T equals b. Lambda with a small m below is the wavelength of peak emission, and b is Wien's constant, two point nine times ten to the power minus three metre kelvin. Okay? Lambda m times T is constant, so double T, and lambda m is halved. Clear? And from the graph: the area under each curve is sigma T to the power four, the power per unit area of a black body. So the hotter curve has a much bigger area. Right?

4:31Topper's trick: compare with ratios

Now a topper's trick. When a question compares two bodies, skip sigma completely. For spheres of the same material, P is proportional to r squared T to the power four. The sign like a stretched alpha means proportional to. Okay? Try this one. Two such spheres have radii in the ratio one is to two, and absolute temperatures in the ratio two is to one. Compare their powers. Radius squared gives one and four; temperature to the power four gives sixteen and one. So P one is to P two, the small one and two below are just labels, is one times sixteen, is to four times one: four is to one. Got it?

5:11Example 1: power radiated by a black sphere at 500 K

Now, Example one. Find the power radiated by a black sphere of radius ten centimetres, at five hundred kelvin. Pause the video, and try it first. Done? Okay, let's check. Black sphere means black body, so e is one. And the radius, ten centimetres, is zero point one metre. Now, which area goes into the formula? Pi r squared, or four pi r squared? Four pi r squared! Correct! The sphere radiates from its whole surface, not from a flat disc. So capital A is four pi times zero point one squared, about zero point one two five seven square metre. The wavy equals sign means approximately. With me? Now, five hundred to the power four is six point two five times ten to the power ten. Times sigma, that makes sigma T to the power four about three thousand five hundred and forty four watt per square metre. Okay? Finally, P equals three thousand five hundred and forty four, times zero point one two five seven. About four hundred and forty five watts. Clear?

6:19Example 2: net radiative heat loss of a person in a room

Now, Example two, a real-life one. Here is a person of surface area one point five square metres and emissivity zero point nine, at three hundred and ten kelvin, in a room at three hundred kelvin. Find the net radiative loss. Pause and try. Done? Okay, let's check. P net, with small net below, equals e sigma A, times T to the power four minus T naught to the power four, with T three hundred and ten and T naught three hundred kelvin. The difference is only ten kelvin, so the bracket is just ten to the power four, right? No, no, careful! This is the trap. The difference of fourth powers is not the fourth power of the difference. Three hundred and ten to the power four is about nine point two three five times ten to the power nine, and three hundred to the power four is eight point one times ten to the power nine. Difference: one point one three five times ten to the power nine, kelvin to the power four. Got it? Next, e sigma A: zero point nine, times five point six seven times ten to the power minus eight, times one point five. About seven point six five times ten to the power minus eight, watt per kelvin to the power four. Right? Multiply: seven point six five times one point one three five is about eight point seven, and the powers of ten give ten. So the net loss is about eighty seven watts, just by radiation! Clear?

7:49Quick revision: Kirchhoff, Stefan-Boltzmann and Wien laws

So, quick revision. Kirchhoff: good absorbers are good emitters, e equals a. Stefan-Boltzmann: P equals e sigma A T to the power four, and the net loss is e sigma A, times T to the power four minus T naught to the power four, with T in kelvin. Wien: lambda m times T equals b, so a hotter body peaks at a shorter wavelength. You have done really well today. Now go and try the practice problems on this page. All the best!