# Chapter 7.2. The Black Body Law All objects that have temperature radiate energy The amount of energy per second radiated is P = eσΑΤ 4 A = Surface Area.

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Chapter 7.2

The Black Body Law All objects that have temperature radiate energy The amount of energy per second radiated is P = eσΑΤ 4 A = Surface Area (m 2 ), σ = Stefan-Boltzman Constant and is equal to 5.67 x 10 -8, T is temperature (K) e is emissivity of the surface and ranges from 0 to 1 The better the emitter the closer e is to 1. Good emitters are good absorbers of energy The perfect emitter is known as a black body and e = 1

Practice with Stefan Boltzman Law If a hot piece of iron (400 K) is in the open air at a temperature of (293 K), then it will both receive and radiate energy. Assume that e for iron is.8 and surface area is.1 m 2.This can be combined as follows: P net = P out – P in = eσA (T 4 out – T 4 in ) P net = P out – P in = (.8)(5.67 x 10 -8) (.1)(400 4 – 293 4 )= P net = 82.7 Watts

Black Body Spectra Wein’s law: λ max T = 2.9 x 10 -3 m K Different objects emit different wavelength radiation based off of their temperature. High temperature emits high frequency radiation

Solar Radiation The sun emits 3.9 x 10 26 W. This power is spread spherically to space. Our planet receives this power based off of this equation: Intensity = P/4πd 2 The average distance from the Earth to the Sun is 1.5 x 10 11 m. The intensity is 1400 W/m 2 Power delivered to Earth is given by P=IA

Solar Radiation Albedo, α, – is the ratio of reflected power/total incident power. It is a unitless number. Examples… Snow has an albedo of.85 whereas charcoal has an albedo of.04. On average the earth’s albedo is.3 Power delivered to a surface area A is governed by this equation P = (1 – α)IA Power reflected off a surface is governed by this equation P = αIA

Solar Radiation Let’s calculate how much energy on average the Earth really receives with everything included: Intensity = P/ 4 π d 2 = 3.9 x 10 26 / 4 π (1.5 x 10 11 ) 2 = 1400 W/m 2 However this radiation is going to only part of the earth like so Sun is only heating ¼ of the earth’s total area Disc of area πr 2 Area of Earth is 4 πr 2

Solar Radiation So the 1400 W/m 2 is really ¼ the value = 1400/4 = 350 W/m 2 Also the albedo reflection is 30% (remember α =.3) So only 70% of the 350 is absorbed. The net absorption of the 1400 W/m 2 is 245 W/m 2.

Energy Balance The Earth has a constant average temperature and therefore has as much energy coming in as is coming out. This is complicated by the Earth’s atmosphere which creates an additional green house effect. The greenhouse effect makes the Earth 32 K hotter than it would be otherwise. Overall, there is still energy balance coming in and leaving Earth.

Incoming Intensity 100 Reflected from surface 5 Reflected from atmosphere 25 Radiation from Clouds and Atmosphere 25 + 40= 65 Radiation from surface with no atmospheric absorption 5 Absorbed Infrared Re radiated to Earth 96 Surface Infrared Radiation 106 Transmitted to Surface 50 Convection and Evaporation 30 Energy Flow Diagram for the Earth – Atmosphere System Atmosphere Earth Surface

Mechanism for Photon Absorption There are 4 major green house gasses: H2O, CO2, CH4 (methane), N2O (Nitrous Oxide) These gasses will absorb specific wavelengths of energy due to their electron bonding levels. Think in terms of resonance and that molecules have natural frequencies that they want to receive waves in. Look at the trasmittance curves on pgs 443 of the book.

The remaining content In order to cover the remaining content please read pgs 444 – 450 and write a summary sentence for each paragraph.

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