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All objects emit and absorb radiation. The net rate of heat transfer by radiation (absorption minus emission) is related to both the temperature of the object and the temperature of its surroundings. Assuming that an object with a temperature T 1 size 12{T rSub { size 8{1} } } {} is surrounded by an environment with uniform temperature T 2 size 12{T rSub { size 8{2} } } {} , the net rate of heat transfer by radiation    is

Q net t = σ e A T 2 4 T 1 4 , size 12{ { {Q rSub { size 8{"net"} } } over {t} } =σ`e`A` left (T rSub { size 8{2} } rSup { size 8{4} } - T rSub { size 8{1} } rSup { size 8{4} } right )} {}

where e size 12{e} {} is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray, or black; the balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When T 2 > T 1 size 12{T rSub { size 8{2} }>T rSub { size 8{1} } } {} , the quantity Q net / t size 12{Q rSub { size 8{"net"} } /t} {} is positive; that is, the net heat transfer is from hot to cold.

Take-home experiment: temperature in the sun

Place a thermometer out in the sunshine and shield it from direct sunlight using an aluminum foil. What is the reading? Now remove the shield, and note what the thermometer reads. Take a handkerchief soaked in nail polish remover, wrap it around the thermometer and place it in the sunshine. What does the thermometer read?

Calculate the net heat transfer of a person: heat transfer by radiation

What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22 . C size 12{"22" "." 0°C} {} . The person has a normal skin temperature of 33 . C and a surface area of 1 . 50 m 2 size 12{1 "." "50"`m rSup { size 8{2} } } {} . The emissivity of skin is 0.97 in the infrared, where the radiation takes place.

Strategy

We can solve this by using the equation for the rate of radiative heat transfer.

Solution

Insert the temperatures values T 2 = 295 K and T 1 = 306 K , so that

Q t = σ e A T 2 4 T 1 4 size 12{ { {Q} over {t} } σ`e`A` left (T rSub { size 8{2} } rSup { size 8{4} } - T rSub { size 8{1} } rSup { size 8{4} } right )} {}
= 5 . 67 × 10 8  J/s  m 2  K  4 0 . 97 1 . 50  m 2 295  K 4 306 K 4 size 12{ {}= left (5 "." "67" times "10" rSup { size 8{ - 8} } `"J/s" cdot m cdot K right ) left (0 "." "97" right ) left (1 "." "50"`m rSup { size 8{2} } right ) left [ left ("295"`K right ) rSup { size 8{4} } - left ("306"`K right ) rSup { size 8{4} } right ]} {}
= 99  J/s = 99 W. size 12{ {}= - "100"`"J/s"= - "100"`W "." } {}

Discussion

This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at rest may produce energy at the rate of 125 W and that conduction and convection will also be transferring energy to the environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer to the environment by many methods, because clothing slows down both conduction and convection, and has a lower emissivity (especially if it is white) than skin.

The Earth receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Because the Sun is hotter than the Earth, the net energy flux is from the Sun to the Earth. However, the rate of energy transfer is less than the equation for the radiative heat transfer would predict because the Sun does not fill the sky. The average emissivity ( e size 12{e} {} ) of the Earth is about 0.65, but the calculation of this value is complicated by the fact that the highly reflective cloud coverage varies greatly from day to day. There is a negative feedback (one in which a change produces an effect that opposes that change) between clouds and heat transfer; greater temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing the temperature. The often mentioned greenhouse effect    is directly related to the variation of the Earth’s emissivity with radiation type (see the figure given below). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for life on Earth. The Earth’s relatively constant temperature is a result of the energy balance between the incoming solar radiation and the energy radiated from the Earth. Most of the infrared radiation emitted from the Earth is absorbed by carbon dioxide ( CO 2 size 12{"CO" rSub { size 8{2} } } {} ) and water ( H 2 O size 12{H rSub { size 8{2} } O} {} ) in the atmosphere and then re-radiated back to the Earth or into outer space. Re-radiation back to the Earth maintains its surface temperature about 40º C size 12{"40"°C} {} higher than it would be if there was no atmosphere, similar to the way glass increases temperatures in a greenhouse.

Practice Key Terms 5

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Source:  OpenStax, Physics 105: adventures in physics. OpenStax CNX. Dec 02, 2015 Download for free at http://legacy.cnx.org/content/col11916/1.1
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