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Energy Production & Thermal Energy Transfer Flashcards
Energy Production: Thermal Energy Transfer
Black Body Radiation
Black-body radiation: Radiation emitted from a theoretical perfect emitter at a given temperature.
It depends on temperature and size but not the surface's nature.
All objects above absolute zero emit black-body radiation across all wavelengths of the electromagnetic spectrum.
A perfect emitter is also a perfect absorber.
The theoretical perfect emitter is called black body radiation because a black object absorbs all light energy falling on it.
The graph of wavelengths and intensity demonstrates:-
Varied intensity among wavelengths.
Intensity between two wavelengths represents intensity emitted in that range.
Total area under the graph indicates total power radiated.
Approximation:-
A small hole in a matte black painted enclosed container is a very good approximation.
The hole appears very black inside.
Radiation color depends on temperature when heated (red to yellow to white with increasing temperature).
Emission depends on temperature and size, irrespective of cavity material.
Black-body Radiation and Wien’s Displacement Law
Wien’s Displacement Law:-
Relates the wavelength at maximum intensity (\lambda[max) to the Kelvin temperature (T) of the black body.
Formula: \lambda[max \times T = constant = 2.9 \times 10^{-3} m⋅K
Where:
\lambda[max is the wavelength at maximum intensity (in meters).
T is the Kelvin temperature (in Kelvin).
Implication: Wavelength at which emitted radiation is maximal is inversely proportional to its Kelvin temperature.
Lower temperature corresponds to longer wavelength.
Applications:-
Analyzing star light to determine its surface temperature.
Hot stars emit all visible light frequencies, appearing white.
Cooler stars emit higher wavelengths (lower frequencies), appearing red.
Temperature Increase Effects:-
Overall intensity at each wavelength increases (curve is higher).
Total power emitted per square meter increases (total area is greater).
The peak shifts to shorter wavelengths (higher frequencies).
Examples
Example 1:-
Using the black-body radiation spectrum of the sun to calculate its effective surface temperature.
Formula: T = \frac{2.9 \times 10^{-3}}{\lambda[max} = \frac{2.9 \times 10^{-3}}{500 \times 10^{-9}} = 5800 K
Where:
T is the calculated surface temperature (in Kelvin).
\lambda[max is the wavelength at maximum intensity (in meters).
Example 2:-
Explanation of black-body radiation.
Example 3:-
Cosmic microwave background radiation spectrum.
Using the graph to estimate the temperature of the black body.-
T = \frac{2.9 \times 10^{-3}}{\lambda[max} = \frac{2.9 \times 10^{-3}}{1.1 \times 10^{-3}} \approx 2.7 K
Where:
T is the estimated temperature (in Kelvin).
\lambda[max is the wavelength at maximum intensity (in meters).
Example 4:-
Cosmic microwave background radiation (CMB) and density of the universe.
Temperature estimation:-
T = \frac{2.9 \times 10^{-3}}{\lambda[max} = \frac{2.9 \times 10^{-3}}{1 \times 10^{-3}} \approx 3 K
Where:
T is the estimated temperature (in Kelvin).
\lambda[max is the wavelength at maximum intensity (in meters).
Example 5:-
Peak wavelength of CMB when the universe's temperature is 4100K (\lambda[max = 700nm).-
\lambda[max = \frac{2.9 \times 10^{-3}}{T} = \frac{2.9 \times 10^{-3}}{4100} = 700 nm
Where:
\lambda[max is the peak wavelength (in nanometers).
T is the temperature (in Kelvin).
Intensity of Radiation
Intensity of radiation (I):-
Power per unit area received by an object (W/m²).
The energy of a wave at a distance (d) from the source spreads over a sphere's area: \I = \frac{P}{A} = \frac{P}{4\pi d^2}
Where:
\I is the intensity of radiation (in W/m²).
P is the power (in Watts).
A is the area (in m²).
d is the distance from the source (in meters).
For stars:-
P: Luminosity (L) in Watts (W).
I: Apparent brightness (b) in W/m².
Formula: b = \frac{L}{4\pi d^2}
Where:
b is the apparent brightness (in W/m²).
L is the luminosity (in Watts).
d is the distance from the star (in meters).
Luminosity (L):-
Total energy emitted by a star per second (total power radiated), measured in Watts (W).
Apparent brightness (b):-
Incident power per unit area received by an observer on Earth (W/m²).
Ratio of brightness:-
\frac{b1}{b2} = \frac{L1}{L2} \times \frac{d2^2}{d1^2}
Where:
b1 and b2 are the apparent brightnesses of two stars.
L1 and L2 are the luminosities of two stars.
d1 and d2 are the distances to two stars.
At the same distance, greater luminosity means greater brightness.
Brightness is inversely proportional to the square of the distance.
Stars can have the same apparent brightness with different luminosities, depending on their distances.-
Example: A star's luminosity is 6 \times 10^{31} W, and its apparent brightness is 1.9 \times 10^{-9} W/m^2. Calculate the distance:
d = \sqrt{\frac{L}{4\pi b}} = \sqrt{\frac{6 \times 10^{31}}{4 \pi \times 1.9 \times 10^{-9}}} \approx 5 \times 10^{19} m
Stefan-Boltzmann Law
Stefan-Boltzmann Law:-
Relates the total power radiated by a black body per unit area to its Kelvin temperature.
Energy emitted per unit time (power) per unit area is proportional to (Kelvin temperature in K)^4 : \I \propto T^4
Formula: \I = \sigma T^4
Where:
\I is the intensity of radiation (in W/m²).
\sigma: Stefan-Boltzmann constant (5.67 \times 10^{-8} Wm^{-2}K^{-4}).
T is the Kelvin temperature (in Kelvin).
Total power radiated (L) by a black-body:-
L = \sigma A T^4
Where:
L is the total power radiated (in Watts).
\sigma is the Stefan-Boltzmann constant (5.67 \times 10^{-8} Wm^{-2}K^{-4}).
A is the surface area (in m²).
T is the Kelvin temperature (in Kelvin).
Assuming a spherical star (radius r):-
L = 4 \pi r^2 \sigma T^4
Where:
L is the luminosity (in Watts).
r is the radius of the star (in meters).
\sigma is the Stefan-Boltzmann constant (5.67 \times 10^{-8} Wm^{-2}K^{-4}).
T is the Kelvin temperature (in Kelvin).
Luminosity depends on temperature and size.
Ratio:-
\frac{L1}{L2} = \frac{r1^2}{r2^2} \times \frac{T1^4}{T2^4}
Where:
L1 and L2 are the luminosities of two stars.
r1 and r2 are the radii of two stars.
T1 and T2 are the temperatures of two stars.
Example: Sun's radius is 6.96 \times 10^8 m and surface temperature is 5800 K. Calculate luminosity:-
L = 4 \pi r^2 \sigma T^4 = 4 \pi (6.96 \times 10^8)^2 (5.67 \times 10^{-8}) (5800)^4 \approx 3.9 \times 10^{26} W
Grey Bodies and Emissivity
Grey bodies:-
Objects close to black body behavior but not perfectly (100%).
Emit less energy per second than a perfect black body of the same dimensions at the same temperature.
Emissivity (e):-
Ratio of power emitted by a radiating object to power emitted by a black body of same dimensions at same temperature.
Formula: e = \frac{Power \ emitted \ by \ radiating \ object}{Power \ emitted \ by \ a \ black \ body \ of \ the \ same \ dimensions \ at \ the \ same \ temperature}
Note:-
Emissivity is dimensionless.
Perfect black body: e = 1.
Perfect reflector: e = 0.
Real objects: 0 < e < 1.
Sun and Solar Constant
Solar constant:-
Incident solar power per unit perpendicular area at Earth's average distance from the sun (top of the atmosphere).
Average ground level value: ≈ 1400 W/m².
Calculation:-
Sun's temperature: 5800 K, radius: 7 \times 10^8 m, Earth-Sun distance: 1.5 \times 10^{11} m.
Energy spreads over an imaginary sphere (radius = Earth-sun distance).-
Area: A = 4 \pi r^2 = 4 \pi (1.5 \times 10^{11})^2 = 2.8 \times 10^{23} m^2
Where:
A is the area of the imaginary sphere (in m²).
r is the Earth-Sun distance (in meters).
Total power radiated from the sun:-
P = e \sigma A[Sun} T^4 = 1 \times 5.7 \times 10^{-8} \times (4 \pi (7 \times 10^8)^2) \times (5800)^4 \approx 4 \times 10^{26} W
Where:
P is the total power radiated (in Watts).
e is the emissivity of the Sun (assumed to be 1).
\sigma is the Stefan-Boltzmann constant (5.67 \times 10^{-8} Wm^{-2}K^{-4}).
A[Sun} is the surface area of the Sun (in m²).
T is the temperature of the Sun (in Kelvin).
Solar constant:-
I = \frac{P{of \ radiation \ emitted \ by \ sun}}{A{of \ the \ imaginary \ sphere \ over \ which \ energy \ is \ spread}} = \frac{4 \times 10^{26}}{2.8 \times 10^{23}} \approx 1400 W/m^2
Where:
I is the solar constant (in W/m²).
P is the total power radiated by the Sun (in Watts).
A is the area of the imaginary sphere (in m²).
Average power received per unit area of Earth:-
(\frac{1}{4}) \times (Intensity \ of \ Sun's \ radiation \ at \ Earth's \ position) = \frac{1400}{4} = 350 W/m^2
Sources of errors in measuring solar constant:
Earth's albedo reflects radiation.
Earth's surface is not always normal to incident radiation.
Atmospheric scattering and absorption reduce energy.
Weather conditions affect radiation arrival.
Periodic variations of the solar constant:-
The sun's output varies by ≈ 0.1% during its 11-year sunspot cycle.
Earth's elliptical orbit causes ≈ 7% difference (closer in January).-
This difference is not the reason for summer in the southern Hemisphere- the seasons occur because the axis of rotation of the earth is not perpendicular to the plane of its orbit around the sun.
Energy Balance in the Earth Surface-Atmosphere System
Thermal equilibrium state of a planet:-
Constant temperature implies absorbed power equals radiated energy.
If it absorbs more energy than radiates, then the temperature must go up and if the rate of loss of energy is greater than its rate of absorption then its temperature must go down.
Simplified model: average equilibrium temperature of the Earth’s surface is TE and that of the atmosphere is TA.
\I{Absorbed \ by \ earth} = \I{Radiated \ by \ earth}
\I{Transmitted \ through \ atmosphere} + \I{Radiated \ by \ atmosphere} = \I[Radiated \ by \ earth}
\I{Transmitted \ through \ atmosphere} + e{atmosphere} \times \sigma \times T{a}^4 = \sigma \times T{E}^4
Where:
\I*{Transmitted \ through \ atmosphere}
is the intensity of radiation transmitted through the atmosphere.e*{atmosphere} is the emissivity of the atmosphere.
\sigma is the Stefan-Boltzmann constant (5.67 \times 10^{-8} Wm^{-2}K^{-4}).
T*{a} is the temperature of the atmosphere (in Kelvin).
T*{E} is the temperature of the Earth’s surface (in Kelvin).
Radiation losses:
Absorption.
Scattering by atmosphere.
Factors affecting absorption and scattering:-
The degree to which this absorption and scattering occurs depends on the position of the sun in the sky at a particular place..
Albedo:-
The extent to which a particular surface can reflect energy is known as its Albedo.
Albedo
Albedo (\alpha):-
Ratio of total scattered power to total incident power.
Formula: \alpha = \frac{Total \ scattered \ power}{Total \ incident \ power}
Where:
\alpha is the albedo.
\alpha = \frac{P{scattered}}{P{incident}} = \frac{E{reflected}}{E{incident}}
Where:
P*{scattered} is the total scattered power.
P*{incident} is the total incident power.
E*{reflected} is the energy reflected.
E*{incident} is the energy incident.
Note:-
Albedo is dimensionless.
Fraction of radiation reflected back into space.
Varies daily and depends on season (cloud formation) and altitude.
A material with high albedo reflects more amount from the incident visible radiation. Oceans have a low value of Albedo, so the fraction of reflection will be smaller than the absorbed one as they are good absorber. On the other hand, snow (ice) has a high value of Albedo so the fraction of reflection will be much greater than the absorbed one as snow is a good reflector.
Global annual average albedo on Earth: 30%.
Examples:-
Oceans vs. Polar Ice Cap:-
Oceans have low albedo (good absorbers).
Ice caps have high albedo (good reflectors).
Greenhouse Effect
Green House Effect: Natural process by which short wavelength radiation received from the sun causes the surface of the Earth to warm up, and thus it emits longer wavelength in the form of infra-red radiation because the Earth is cooler than the Sun. Some of this infra- red radiation will be absorbed by green house gases in the atmosphere and re-radiated in all directions. The part of the radiation that is re-radiated back to Earth will cause the surface temperature of the earth to rise and so re-radiated at a different frequency. As a result, the upper atmosphere and the surface of the earth are warmed (Without green house effect, the temperature of the earth would be much lower). The temperature of the Earth’s surface will be constant if the rate at which it radiates energy equals the rate at which it absorbs energy.
Green House Gases: gases in the atmosphere that transmit ultraviolet radiation but absorb infra-red radiation and then re-radiate it in all directions. Such as: Carbon dioxide (CO2), Methane (CH4), Water (H2O), Nitrous oxide (N2O), Ozone (O3) and Chlorofluorocarbons (CFCs).
Source and Sinks of Green House Gases
Carbon dioxide (CO2):-
Source: Emissions from volcanoes/ Burning of fossil fuels in power plants/cars / breathing.
Sink: Oceans / rivers / lakes / seas / plants (trees).
Methane (CH4), Nitrous oxide (N2O) are : Plants and livestock , Livestock and industries respectively as thier souce.
Greenhouse mechanism explaination
Energy level differences corresponding to infra-red energies and so the bonds within the molecules of the gas will oscillate/vibrate with natural frequencies in the infrared region; therefore the infra-red photons radiated by the surface of the earth will be absorbed as a result of resonance.
Global Warming Effect
Enhanced greenhouse effect:-
Increased concentration of infra-red radiation from atmosphere to Earth.
Due to human activities like burning fossil fuels, incineration of waste.
Causes:
Changes in greenhouse gas composition (natural or human).
Changes in intensity of solar radiation (increased solar activity).
Cyclical changes in Earth's orbit.
Volcanic activity (CO2 emission).
Increased carbon dioxide leads to:-
Decreased albedo.
\*\\*Ways for reducing global warming:
Replace the use of coal and oil with natural gas / nuclear power station.
Isotropic analysis allows the temperature to be estimated and air bubbles trapped in the ice cores can be used to measure the atmospheric concentrations of greenhouse gases. Measurements show that variations of temperature and carbon dioxide are very closely correlated.
Evidence for global warming:
Antarctic ice core data.
Tree ring analysis.
Analysis of water levels.
Melting Ice sheets: Oceanic ice floats on sea water and displaces a certain amount of sea water, and therefore when it melts the sea water level does not change.
Mechanisms that may increase the rate of global warming:
Increased solar (flare) activity.
Cyclical changes in the Earth’s orbit.
Volcanic activity increases global warming through CO2 emission.
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NoteStudied by 172 peopleNoteStudied by 7 peopleNoteStudied by 4 peopleNoteStudied by 19 peopleNoteStudied by 25 peopleNoteStudied by 66 people