
Thermal radiation comprises electromagnetic waves, including sunlight and ultraviolet and infrared light, that are emitted as a result of thermal motion of atoms and molecules. Thermal radiation is emitted by all objects with temperatures above absolute zero and thermal radiation is also absorbed by all objects. The power in thermal radiation over all wavelengths increases as the absolute temperature to the fourth power. A perfect blackbody is a surface that absorbs all incident radiation and emits only thermal radiation. The blackbody spectrum describes the relationship between power and wavelength for thermal radiation. An object in thermal equilibrium balances heat gained with heat lost. In space, conduction and convection cannot occur so radiation dominates heat transfer. Radiative heat balance determines the average temperature of Earth, the planets, and all other Solar System objects except the Sun itself.

emissivity, blackbody


$$P=\epsilon \sigma A{T}^{4}$$

 $$P=\epsilon \sigma A({T}_{\infty}^{4}{T}^{4})$$



Review problems and questions 

 The Stefan–Boltzmann equation describes how much energy every object in the universe radiates away. But in the universe, some objects are warming up, while others are cooling down. How can this be?

Every object is absorbing thermal energy from its surroundings and also radiating thermal energy to its surroundings. The difference between the two determines whether an object is warming or cooling.

 If you heat steel from room temperature (20°C) to its melting point at 1,510°C, by what increased factor will it radiate thermal energy?

Answer: The amount of thermal energy radiated will increase by a factor of 1370. To answer such questions, you must always start by converting temperatures to the Kelvin scale! The initial temperature is 293 K, while the final temperature is 1783 K. The ratio of these temperatures is (1,783/293) = 6.09. In the Stefan–Boltzmann equation, the power radiated is proportional to the temperature to the fourth power, so the radiated thermal energy will increase by a factor of (6.09)^{4} = 1370! This high radiated power is why it isn’t even safe to get close to a large quantity of very hot metal.

 Telescope mirrors are coated with a thin layer of a shiny metal so that light striking them will mostly reflect, rather than be absorbed. For infrared astronomy, however, there is another important reason: As the mirror becomes less reflective, it radiates more energy at infrared wavelengths—which can overwhelm the images from distant stars or galaxies. Why does a lessreflective mirror radiate more infrared radiation?

As a mirror becomes less reflective, whether from a lower reflectivity metal or increased buildup of dirt, its emissivity increases. The temperature of the mirror might be 280 K, so it radiates like a blackbody at that temperature with that increased emissivity. The peak emission from a 280 K blackbody is at infrared wavelengths. Astronomers prevent this radiation from being emitted by keeping their mirrors free of dirt and dust!

 The dwarf planet Eris has an average orbital distance from the Sun of 68 AU. Estimate the average temperature of Eris based on a radiation balance between the thermal energy it receives from the Sun and the energy Eris radiates as a blackbody. Express your answer in degrees Celsius.

Answer: −239°C By using the equation on page 1421, the temperature at Eris can be related to its orbital radius by$$T=\frac{280}{\sqrt{R/\text{AU}}}\text{K}=\frac{280}{\sqrt{68}}=34\text{K}$$This corresponds to −239°C!

 While the Earth has a wide range of temperatures across its surface, its average temperature is approximately 15°C or so. How much power is radiated by the Earth at that temperature, assuming the Earth is 100% emissive? (The radius of the Earth is 6,370 km.) Compare this power to the Hoover Dam, which can generate up to 2,080 MW.

Answer: The power is 1.99×10^{17} W, or 95.6 million times more than the generating capacity of the Hoover Dam.
The assumption is that the Earth has an emissivity of ε = 1.00 (or 100%). The power radiated by the Earth is then$$P=\epsilon \sigma A{T}^{4}=(1.00)(5.67\times {10}^{8}{\text{Wm}}^{2}{\text{K}}^{4})\mathrm{4\pi}{\left(6.37\times {10}^{6}\text{m}\right)}^{2}{\left(288\text{K}\right)}^{4}=1.99\times {10}^{17}\text{W}$$In comparison, the Hoover Dam generates up to 2,080 MW (or 2,080×10^{6} W = 2.08×10^{9} W), so$$\frac{1.99\times {10}^{17}\text{W}}{2.08\times {10}^{9}\text{W}}=9.56\times {10}^{7}$$The Earth radiates power equivalent to the generating power of 95.6 million Hoover Dams!

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