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Lecture 6

PHYS 1041 Lecture Notes - Lecture 6: Boltzmann Constant, Thermal Conductivity, Calorie

3 Pages
112 Views
Winter 2014

Department
Physics
Course Code
PHYS 1041
Professor
Dr.David Fleming
Lecture
6

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Physics 1041
Notes
March 12th
Electromagnetic Radiation
light can act as a wave or a particle, and can have properties of both of these
aspects at different times
there are countless possible wavelengths of light
visible light is a narrow section towards the middle of the light spectrum
the temperature range of sun corresponds with the visible light spectrum
When:
v= velocity (m/s)
Then:
λ
fv
=
When:
c = speed of light (3.00 x 108m/s)
f=frequency (Hz)
λ= wavelength (m)
Then:
λ
fc
=
Blackbody Radiation
any object that is a perfect absorber of electromagnetic radiation
if you know the temperature of blackbody, you can predict its radiation spectrum
Wien's Law:
The relationship between the temperature (in Kelvins) and the peak wavelength of a light
wave is:
mKT
peak
3
10898.2
×=×
λ
Blackbody Absorber
When:
P= power of light radiation (W)
σ= the Stefan Boltzmann constant, equal to 5.67 x 10-5W/(m2K4)
A= surface area of absorber (m2)
T= temperature in Kelvins
Then:
4
ATP
σ
=
If radiator is not perfect, an ε sign is used to indicate its percentage efficiency in decimals
as follows:
4
ATP
εσ
=

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Description
Physics 1041 Notes March 12th Electromagnetic Radiation • light can act as a wave or a particle, and can have properties of both of these aspects at different times • there are countless possible wavelengths of light • visible light is a narrow section towards the middle of the light spectrum • the temperature range of sun corresponds with the visible light spectrum When: • v= velocity (m/s) Then: • v = fλ When: 8 • c = speed of light (3.00 x 10 m/s) • f=frequency (Hz) • λ= wavelength (m) Then: • c = fλ Blackbody Radiation • any object that is a perfect absorber of electromagnetic radiation • if you know the temperature of blackbody, you can predict its radiation spectrum Wien's Law: The relationship between the temperature (in Kelvins) and the peak wavelength of a light wave is: −3 λ peak× T = 2.898 ×10 mK Blackbody Absorber When: • P= power of light radiation (W) • σ= the Stefan Boltzmann constant, equal to 5.67 x 10 W/(m K ) 4 2 • A= surface area of absorber (m ) • T= temperature in Kelvins Then: 4 • P = σAT If radiator is not perfect, an ε sign is used to indicate its percentage efficiency in decimals as follows: 4 • P = εσAT March 14th • the peak wavelength of a light wave increases with temperature • higher temperatures mean a short
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