5.5 Black body radiation — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Sections 5.1 and 5.2.

Name the kind of radiation emitted by every warm object.
Work out $\dfrac{3.0 \times 10^8}{1.0 \times 10^{-5}}$ .
Which has the longer wavelength, infrared radiation or visible light?

Answers

Infrared radiation.
$3.0 \times 10^{13}\,\text{Hz}$ .
Infrared radiation has the longer wavelength.

If question 1 went wrong, look again at the order of the spectrum. If question 2 went wrong, take care with the powers of ten when you divide.

Practise with help

1. A metal block is heated until it glows. Explain how the radiation it emits changes as its temperature rises.

Hint

Two things change: how much radiation is emitted, and which wavelength is the most intense.

Next step

Say what happens to the total intensity. Then say what happens to the peak wavelength, and link this to the colour you see.

Full working

As the block gets hotter it emits more radiation at every wavelength, so the total intensity rises. The wavelength of the most intense radiation becomes shorter. The block first glows red, because red is the longest visible wavelength. As it gets hotter the peak moves to shorter wavelengths, so the colour changes through orange towards white.

2. A warm object emits most of its radiation at a wavelength of 10 μm. Calculate the frequency of this radiation.

Hint

Use the wave equation as $f = \dfrac{c}{\lambda}$ . Convert micrometres to metres first.

Next step

$10\,\mu\text{m} = 1.0 \times 10^{-5}\,\text{m}$ . Put this into $f = \dfrac{c}{\lambda}$ .

Full working

\begin{aligned} f &= \frac{c}{\lambda} \\ &= \frac{3.0 \times 10^8}{1.0 \times 10^{-5}} \\ &= 3.0 \times 10^{13}\,\text{Hz} \end{aligned}

3. A hotter object emits most of its radiation at a wavelength of 2.0 μm. Work out the frequency of the radiation from the hotter object.

Hint

This is the same calculation as problem 2, with a shorter wavelength. Check that your frequency comes out higher.

Next step

$2.0\,\mu\text{m} = 2.0 \times 10^{-6}\,\text{m}$ . Then divide the speed by this wavelength.

Full working

\begin{aligned} f &= \frac{c}{\lambda} \\ &= \frac{3.0 \times 10^8}{2.0 \times 10^{-6}} \\ &= 1.5 \times 10^{14}\,\text{Hz} \end{aligned}

The shorter wavelength gives the higher frequency.

4. A cup of hot tea cools to room temperature and then stays there. Explain why its temperature becomes constant.

Hint

The cup is always both absorbing and emitting radiation. Compare the two rates while the tea is hot, and again at room temperature.

Next step

First say why the tea is losing energy while it is hotter than the room. Then say what must be true about the two rates when the temperature stops changing.

Full working

While the tea is hotter than the room, it emits more infrared radiation than it absorbs, so its thermal store empties and it cools. Once the tea is at room temperature, it emits and absorbs radiation at the same rate. The net energy transfer is zero, so the temperature stays constant.

This answer is a model answer for you to compare with your own wording.

5. A black surface and a shiny surface are at the same temperature. Compare how much radiation each surface emits and absorbs.

Hint

A good absorber is also a good emitter. Think about what each surface does with infrared radiation.

Next step

Start with the black surface: decide whether it absorbs a lot or a little, then say what that means for how much it emits.

Full working

The black surface absorbs a lot of infrared radiation, and it also emits a lot. The shiny surface absorbs very little, because it reflects most of the radiation, and it also emits very little. The black surface is the better absorber and the better emitter.