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Temperature and Radiation
Mike Luciuk
The three main methods of heat transfer resulting in change of temperature are conduction,
convection and radiation. In conduction, energy is transferred by physical contact, like when one
burns a finger while touching a hot pot. In convection, energy transfer occurs by fluid motion like
water boiling in a kettle. In radiation, energy is transferred by the absorption or emission of
electromagnetic radiation like the warmth from the Sun. This paper will illustrate some examples
of how radiation determines temperature of objects in space and on Earth.
Electromagnetic Radiation
All bodies with a temperature greater than absolute zero radiate energy. Absolute zero is the
temperature at which there is no molecular or atomic random motion. It’s denoted by 0 Kelvin
degrees, which is equivalent to -273.15° C or -459.67° F. Late in the nineteenth century, Stefan
experimentally and Boltzmann theoretically developed a relationship between the temperature
of a body and the amount of power it radiates.
To determine outgoing radiation power , we utilize the Stefan-Boltzmann Law: 4
P = A εσ T (1)
Where P (watts) is the radiated power from a body of area A (m^2 ) at temperature T (K). ε is emissivity, a dimensionless number between 0 and 1 that determines the efficiency of a body to radiate and absorb energy. A perfect radiator and absorber has an emissivity of 1. Soil, ice, rock, asphalt and human skin have emissivities slightly less than 1. σ is the Stefan-Boltzmann constant, 5.67 x 10 -8^ Wm-2T- T is the body temperature in Kelvin.
So if absolute temperature (in Kelvin degrees) doubles, radiated power increases by a factor of sixteen. Also, changes in temperature alter radiation peak wavelengths. Temperature increases move peak radiation to smaller wavelengths and vice-versa.
Equilibrium and Solid Angles
Assume we have several bodies of different temperatures and we want to determine temperature of a specific body from their radiation. At steady state or equilibrium, radiation received must equal radiation emitted. The amount of radiation received depends on the emitting body’s temperature, its size and its distance to the receiving body. Size and distance are quantified by calculating the solid angle. Solid angles are defined as the area of the emitting body divided by distance squared from the receiving body, with units called steradians:
Solid angle in steradians, (^2)
area
distance
For example, the solid angle of a body in space absorbing cosmic microwave background (CMB) radiation which comes from all directions is: (area of a sphere perceived from its
center)/(radius^2 ) =
2 4 π r (^) r 2 = 4 πsteradians. The solid angle of a body receiving radiation from the
Sun at 1 astronomical unit (AU) distance is: (solar disk area)/(distance^2 ) or
Ω =
5 2 5 2 8 2
sun sun
area x x distance x
π (^) − = = steradians, where the Sun’s radius is ̴700,000 km and
an astronomical unit is ̴150 million kilometers.
Let’s assume we have a body in space receiving radiation from other sources. It will in turn radiate power based on its temperature. At equilibrium, radiation absorbed will equal radiation emitted: 4 4 4 Ω (^) body Tbody = Ω cmb (^) Tcmb + Ω (^) sun Tsun + others (3)
We can solve for Tbody by inserting the (^) Ω and T factors in the equation. For example, Tcmb = ̴2.
K, Tsun = ̴5800 K, Ω (^) body = 4 π for effective temperature, Ω (^) cmb= 4 π , and Ω (^) sun= 6.84x10- steradians at one astronomical unit (AU).
Temperatures Due to the Sun
In space, the major factor for temperatures of solar system planets and asteroids is the contribution from the Sun. However, other sources might include radiation from the cosmic microwave background, other nearby bodies, tidal effects, or internal sources of heat from elements with long term radioactivity. We’ll assume ideal conditions: emissivity = 1 for all bodies and no radiation attenuation through space. For example, assume a body orbits the Sun with a semi-major axis of 1 AU (150 million km). The Sun’s “surface” temperature is about 5800 K. We’ll assume the body’s albedo is zero and it has no greenhouse atmosphere. Recall that albedo is the fraction of radiation reflected back to space. CMB radiation will also be ignored. We can calculate the body’s effective temperature as follows: 4 5 4 4 π Teff 6.84 10 x (5800) −
5 4 14 (6.84 10 )(5800) 280 4
eff
x T K π
− = (^) =
If the body has an albedo A, and has a greenhouse factor G, we can calculate its effective temperature as follows: 4 5 4 4 π (1 G T ) (^) avg 6.84 10 x (1 A )(5800) − − = − (^) (6)
For example, Earth’s albedo, has a value of about 0.3 due to clouds. Its greenhouse factor, which reduces outgoing radiation is about 0.4, mainly due to water vapor in its atmosphere. Earth’s predicted average temperature is: 1 5 4 14 1 1 1 4 (6.84 10 )(5800) 1 4 0.7 4 (280 ) 291 avg (^) 1 4 1 eff 0.
A x A T T K K
G π G
= ^ ^ = ^ ^ = ^ =
Temperature from Radiation on Earth
In this section, we’ll explore maximum temperature changes when hot or cold objects are encountered when surrounded by ambient temperature conditions. Accuracy is compromised by ignoring radiation losses that could be caused by the atmosphere.
Nuclear Fireball Assume an observer is 100 km from a nuclear explosion, a one megaton device whose fireball’s temperature is 10,000 K and diameter is one kilometer. Ambient temperature is 300 K (80° F). What is the temperature facing the device at the observer’s location? The solid angle from the observer’s perspective to the nuclear device is π steradians, a situation analogous to subsolar temperature planetary calculations above. The device’s solid
angle is
(^2 ) 2
π (^) = x − steradians. The observer’s temperature is affected by ambient
and device temperatures: 4 4 5 4 π Tobserver π (300) (7.85 10 x )(10000) − = + (^) (12)
This results in a temperature of 713 K, or 824° F, a change of 413° C or 744° F, which would cause fires and severe burns for anyone facing the blast. However, anything in the shade of the nuclear device would only be exposed to the ambient 80° F temperature.
Encountering an Iceberg A ship passes 0.5 km from a 1 km long, 100m high iceberg. The iceberg’s temperature is 273 K (32° F) and the ambient temperature is 300 K. What is the temperature drop on the ship’s side facing the iceberg as it passes by?
The iceberg’s perceived solid angle is (1 0.1)^2 0.
x (^) = steradians.
This is a significant percentage of the ambient π solid angle. Therefore, a reduction of 0. steradians will be required. 4 4 4 π Tship = (π − 0.4)(300) + 0.4(273) (13)
As it moves by the iceberg, the ship’s temperature would gradually drop to 297 K (75° F). The drop of 5° F would likely be noticed by passengers on the ship’s side facing the iceberg. The other side of the ship would remain at 80° F.
References
http://en.wikipedia.org/wiki/Electromagnetic_spectrum http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law http://en.wikipedia.org/wiki/Steradian http://en.wikipedia.org/wiki/Sun http://ocw.nd.edu/physics/nuclear-warfare/images-1/comparative-nuclear-fireball-size.jpg/view http://www.fas.org/nuke/intro/nuke/thermal.htm
See related tutorial on this website titled, “The Dewing or Frosting of Telescope Optics”.
