Radiation

Question 1

What is the electromagnetic spectrum of an object

Answer 1

Characteristics distribution of electromagnetic radiation emitted or absorbed by that particular object

Question 2

What are the electromagnetic spectrums

Answer 2

-Radiowaves
-Microwaves
-Infrared
-Visible
-Ultraviolet
-X-ray
-Gamma Ray

Question 3

What is the trends of electromagnetic spectrum?

Answer 3

Wavelength gets shorter as you go down the spectrum
The frequency increases as you go down

Question 4

What 3 things happen when thermal radiation hits a surface

Answer 4

-Can be absorbed by the surface
-Can be reflected off the surface
-Can be transmitted through the surface

Question 5

How does temperature effect thermal radiation

Answer 5

Its intensity and spectrum

Question 6

What happens when temperature increases in relation to wavelengths and energy

Answer 6

Wavelengths decrease and energy increases

Question 7

What are the exchanges of radiation

Answer 7

-All bodies emit radiation to their surroundings
-All bodies receive radiation from their surrondings

Question 8

What is prevosts theory

Answer 8

Body emits radiation at a rate which depends only on its temperature and nature of surface

A body absorbs radiation at a rate which depends only on the temperature of its surroundings and on the nature of its surface

Question 9

“Consider a perfectly insulated body at a
temperature TB: it cannot lose or gain heat by
conduction or convection, only by radiation.
* If it is surrounded by some form of enclosure
which is maintained at a constant temperature TS”

Explain what happens in all situations of this?

Answer 9

If TB>TS the body emits radiation at a greater rate than it absorbs it so TB falls until it is equal to TS

If TS>TB the body absorbs radiation at a greater rate than it absorbs so TB rises until it is equal to TS

If TB=TS the rate of absorption is equal to the rate of emission and TB doesnt change

Question 10

What colours are the best absorbers and emitters?

Answer 10

-Black (Best)
-White
-Gray (worst)

Question 11

Difference between transmission and emission?

Answer 11

Transmission: Part of the radiation passes through the body
Emission: The amount of thermal radiation released or given off by a body

Question 12

What is the Black Body?

Answer 12

An object that absorbs all the radiation which strikes it

Question 13

What does emissivity describe?

Answer 13

Describes the surface radiation properties of a body

Question 14

What is emissivity?

Answer 14

The ratio of energy radiated/absorbed by a particular material to energy radiated/absorbed by a black body at the same temperature

Question 15

What do the emissivity range mean?

Answer 15

0-1
0:A body that does not emit or absorb at all
1:A body that emit at the maximum possible rate an absorbs all the energy striking it

Question 16

A mirror has an emissivity of  = 0.1. A laser beam with a
power of 5 mW is incident on the mirror. What is the
power of the reflected beam

Answer 16

Since the emissivity is 0.1, so is the absorption factor.
10% of the power is absorbed, and 90% is reflected.
So the reflected power is 50.9 = 4.5 mW.
The rest of the power has been absorbed into the material of the mirror
and converted to heat

Question 17

Explain how punching a small hole in the lid of an empty meal tin is a close approximation to a black body?

Answer 17

Light enters the hole and is reflected many times by the wall which will get a % absorbed each time before it reaches the hole again.

Question 18

Is the distribution of energy even?

Answer 18

No, the distributions of energy are not even and it depends on the source temperature
(for ex: heating metal red will appear first as it the longest wavelength then yellow then orange then blue and the combination appears white)

Question 19

What is Weins Law

Answer 19

Product of the wavelength of the peak of the spectrum and the temperature of the black body is a constant

Question 20

Describe the radiated power over wavelength graph

Answer 20

At lower wavelengths there is little photons so less radiated power and as wavelength increases there is a peak radiated power around the middle as there is a high number of photons at high energy

Question 21

What is weins formula?

Answer 21

Wavelengthp=0.0029/T (In kelvin)

Question 22

If the temperature of a blackbody is 500K what will the peak be?

Answer 22

p = 0.0029/500 m
= 5.8x10^-6 = 5.8 micrometer

Question 23

What is Stefans Law?

Answer 23

Gives the total intensity of the radiation over all the over all wavelengths

Question 24

Formula for Stefans Law

Answer 24

Q=oAT^4 (o=5.7x10^-8 Stefans constant, A is the area radiating and T is absolute temperature)

Question 26

What is Weins law?

Answer 26

the product of the wavelength of the peak of the
spectrum and the temperature of the blackbody is a constant

Question 27

What is the formula for weins law?

Answer 27

Wavelength= Weins constant/Temperature in Kelvin (0.0029/T)

Question 28

Describe the Radiated Power vs Wavelength graph

Answer 28

Due to weins law curves with a higher radiated power peaks are at lowers wavelengths because as temperature increases wavelength max decreases

Lower temperature objects emit mainly in the infrared region

Question 29

What is Stefans Law

Answer 29

gives the total intensity (power) of
the radiation over all wavelengths

Question 30

How will Stefans Law be modified if the object isnt a perfect black body

Answer 30

Q=e (emissivity) x o (stefans consstant) x Area x Temperature

Question 31

What is the value of Stefans constant

Answer 31

5.7x10^-8Wm^-2K^-4

Question 32

A 100 W light bulb has a filament which is 0.60 m long with a
diameter of 8.010−5 m. The filament may be regarded as a black
body. Estimate its working temperature. If the filament is not a
perfect blackbody, would the working temperature be higher or
lower? Why?

Answer 32

Surface area of filament (excluding ends) = pixdxL
= pi x 8.0x10^−5x0.60 = 1.51x10−4 m2
For a blackbody, Stefan’s law:
The power of the light bulb is 100 W,
So 100 = 5.710−81.5110−4T4
Hence T4 = 1.161013 or T = 1846 K
The working temperature of the filament is 1846 K
If the filament is not a perfect black body, i.e. <1, since the electrical power
supplied stays the same, the radiated power stays the same, the working
temperature will be higher. For example, if
 = 0.9, T will rise to 1895 K.
However, the filament will be a less efficient radiator.

Question 33

What will happen if a black body at T1 is placed in an enclosure T2

Answer 33

-Heat is radiated at: Q=oxSxT1^4
Heat is absorbed at: oAT2^4
Net energy flow from the enclosure to the blackbody:oxA(T2^4-T1^4) (Heat absorbed-Heat radiated)

Question 34

A potato at a temperature of 150ºC is placed in a refrigerator whose
walls are at 2ºC. Assuming the potato is a perfect black body, and is
spherical with a diameter of 7 cm, and all heat transfer is by
radiation, calculate the rate at which it loses energy.

Answer 34

Surface area of potato =pi x d^2 = 1.54x10^−2 m2.
Power emitted is 5.7x10^−8x1.54x10^−2 x (273+150)^4 = 28.1 W
Power absorbed is
5.7x10^−8x1.54x10^−2x(273+2)^4 = 5.0 W
Thus, the potato therefore initially loses heat at 23.1 W.
This will fall with time, as it cools.
– If instead the potato was at 2ºC, and was in an oven at 150ºC, the calculation
would be the other way round: the potato would gain energy at 23.1 W.
– Microwave ovens supply much more power by restricting the wavelengths
radiated to a very narrow band which excites water molecules – they are not
black bodies

Question 35

Formula for energy of a photon

Answer 35

E=hf
E= Energy of photon
h= Plancks Constant (6.626x10^34)
f= Frequency

Question 36

Speed equation

Answer 36

c=f x wavelength
c= speed
f=frequency
wavelength

Question 37

Relationship between Energy, Frequency, and plancks constant and wavelength

Answer 37

E=hc/wavelength
E=energy
h= Wavelength (6.626x10^34)
c=speed of light (3x10^8)

Question 38

Formula for intensity

Answer 38

I=P/A
I=Intensity
P=Power
A=Area

Question 39

Surface area for a sphere?

Answer 39

A=4xpixr^2

Question 40

Surface area of a filament

Answer 40

A = 2 π r L

Question 41

What is the area of a cylinder? and what is it without the ends?

Answer 41

2nr^2+2nrh
n=pi
r=radius
h=height

Without ends:
n x d x L