krit.club logo

The Particulate Nature of Matter - Greenhouse Effect

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Greenhouse Effect is a process where certain gases in the Earth's atmosphere trap heat, preventing it from escaping into space and keeping the planet warmer than it would be otherwise.

•

Greenhouse gases (GHGs) include carbon dioxide (CO2CO_2), methane (CH4CH_4), water vapor (H2OH_2O), and nitrous oxide (N2ON_2O). These molecules have vibrational modes that resonate at frequencies corresponding to long-wavelength infrared radiation.

•

Solar radiation is primarily short-wavelength (visible and UV). The Earth absorbs this and re-emits energy as long-wavelength infrared (IR) radiation because the Earth's surface temperature is much lower than the Sun's.

•

The Albedo (aa) of a body is the ratio of the power of radiation reflected from the body to the total incident power. For Earth, the average albedo is approximately 0.300.30.

•

The Solar Constant (SS) is the intensity of solar radiation at the top of the Earth's atmosphere, approximately 1361 W m−21361 \text{ W m}^{-2}.

•

Blackbody Radiation: Earth and the Sun are modeled as blackbodies. The Stefan-Boltzmann Law states that the power emitted per unit area is proportional to the fourth power of the absolute temperature (T4T^4).

•

Wien's Displacement Law relates the peak wavelength (λmax\lambda_{max}) of emitted radiation to the temperature: λmax∝1T\lambda_{max} \propto \frac{1}{T}.

•

Energy Balance: For a planet with no atmosphere to be in thermal equilibrium, the power absorbed from the sun must equal the power radiated by the planet: S4(1−a)=σT4\frac{S}{4}(1-a) = \sigma T^4.

📐Formulae

λmaxT=2.90×10−3 m K\lambda_{max} T = 2.90 \times 10^{-3} \text{ m K}

P=eσAT4P = e \sigma A T^4

I=PAI = \frac{P}{A}

a=Reflected PowerIncident Powera = \frac{\text{Reflected Power}}{\text{Incident Power}}

Iin=S4(1−a)I_{in} = \frac{S}{4}(1 - a)

ΔE=hf\Delta E = hf

💡Examples

Problem 1:

Calculate the theoretical surface temperature of Earth if it had no atmosphere, given that the solar constant S=1360 W m−2S = 1360 \text{ W m}^{-2} and the average albedo a=0.30a = 0.30. (Use σ=5.67×10−8 W m−2 K−4\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4})

Solution:

The power absorbed per unit area is Iabs=S4(1−a)I_{abs} = \frac{S}{4}(1 - a). Setting this equal to the radiated power σT4\sigma T^4:

13604(1−0.30)=5.67×10−8×T4\frac{1360}{4}(1 - 0.30) = 5.67 \times 10^{-8} \times T^4

340×0.7=5.67×10−8×T4340 \times 0.7 = 5.67 \times 10^{-8} \times T^4

238=5.67×10−8×T4238 = 5.67 \times 10^{-8} \times T^4

T4=2385.67×10−8≈4.197×109T^4 = \frac{238}{5.67 \times 10^{-8}} \approx 4.197 \times 10^9

T=4.197×1094≈254.6 KT = \sqrt[4]{4.197 \times 10^9} \approx 254.6 \text{ K}

Explanation:

The factor of 14\frac{1}{4} accounts for the fact that the Earth intercepts solar radiation as a disk (πR2\pi R^2) but radiates it as a sphere (4πR24 \pi R^2). The resulting temperature of 255 K255 \text{ K} (−18∘C-18^\circ \text{C}) is much lower than the actual average temperature (288 K288 \text{ K}), demonstrating the warming effect of the atmosphere.

Problem 2:

Explain why CO2CO_2 is a greenhouse gas while N2N_2 is not, in terms of molecular resonance.

Solution:

N2N_2 is a homonuclear diatomic molecule with no permanent dipole moment, and its vibrations do not create a changing dipole moment. CO2CO_2, while linear and non-polar, has vibrational modes (bending and asymmetrical stretching) that create a transient dipole moment. These modes have natural frequencies in the infrared spectrum.

Explanation:

For a molecule to absorb infrared radiation, the photon's energy must match the difference between vibrational energy levels (ΔE=hf\Delta E = hf), and the vibration must cause a change in the dipole moment of the molecule.

Greenhouse Effect - Revision Notes & Key Formulas | IB Grade 12 Physics