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The Particulate Nature of Matter - Greenhouse Effect

Grade 11IBPhysics

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

🔑Concepts

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The Greenhouse Effect is the process by which certain gases in the atmosphere absorb and re-emit infrared radiation, thereby warming the Earth's surface.

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Solar radiation is primarily short-wavelength (visible, UV, and near-IR), corresponding to the Sun's high surface temperature (T≈5800 KT \approx 5800 \, K).

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Earth's surface absorbs this energy and re-radiates it as long-wavelength infrared (IR) radiation, corresponding to its lower temperature (T≈288 KT \approx 288 \, K).

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Greenhouse gases such as CO2CO_2, H2OH_2O, CH4CH_4, and N2ON_2O possess molecular energy levels that allow them to absorb IR photons through resonance. This occurs when the frequency of the radiation matches the natural frequency of the molecular vibrations.

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The Greenhouse Effect relies on the molecules' ability to change their dipole moment during vibration, which is why diatomic gases like N2N_2 and O2O_2 are not greenhouse gases.

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Albedo (α\alpha) is the ratio of the power of radiation reflected from a surface to the total incident power. For Earth, the average albedo is approximately 0.300.30.

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Surface Heat Capacity (CsC_s) is the energy required to raise the temperature of a unit area of a planet's surface by 1 K1 \, K, measured in J m−2 K−1J \, m^{-2} \, K^{-1}.

📐Formulae

λmax=2.90×10−3T\lambda_{max} = \frac{2.90 \times 10^{-3}}{T}

P=eσAT4P = e \sigma A T^4

α=Reflected PowerIncident Power\alpha = \frac{\text{Reflected Power}}{\text{Incident Power}}

I=P4πr2I = \frac{P}{4 \pi r^2}

Iabsorbed=(1−α)S4I_{absorbed} = \frac{(1 - \alpha) S}{4}

ΔT=(Iin−Iout)ΔtCs\Delta T = \frac{(I_{in} - I_{out}) \Delta t}{C_s}

💡Examples

Problem 1:

Given the solar constant S=1360 W m−2S = 1360 \, W \, m^{-2} and an average planetary albedo of α=0.30\alpha = 0.30, calculate the average intensity of solar radiation absorbed by the Earth's surface.

Solution:

Iavg=(1−α)S4I_{avg} = \frac{(1 - \alpha) S}{4} Iavg=(1−0.30)×13604I_{avg} = \frac{(1 - 0.30) \times 1360}{4} Iavg=238 W m−2I_{avg} = 238 \, W \, m^{-2}

Explanation:

The factor of 44 arises because the Earth intercepts solar radiation as a disk (area πR2\pi R^2) but distributes that energy over its entire spherical surface area (4πR24 \pi R^2).

Problem 2:

A planet is modeled as a blackbody (e=1e = 1) with no atmosphere. If the absorbed solar intensity is 238 W m−2238 \, W \, m^{-2}, determine the equilibrium surface temperature TT.

Solution:

I=σT4I = \sigma T^4 238=5.67×10−8×T4238 = 5.67 \times 10^{-8} \times T^4 T4=2385.67×10−8T^4 = \frac{238}{5.67 \times 10^{-8}} T=4.197×1094≈255 KT = \sqrt[4]{4.197 \times 10^9} \approx 255 \, K

Explanation:

Using the Stefan-Boltzmann Law, we equate the absorbed power per unit area to the emitted power per unit area. The resulting temperature (255 K255 \, K or −18∘C-18^\circ C) is much lower than Earth's actual average temperature (288 K288 \, K), illustrating the significant warming role of the greenhouse effect.

Greenhouse Effect Grade 11 Notes & Examples | IB Physics