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Topic 2.2Physics SL45 flashcards

Greenhouse effect

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Card 1 of 452.2.1
2.2.1
Question

Define intensity.

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2.2.111 cards

Card 1definition
Question

Define intensity.

Answer

The radiation **power received per unit area** (perpendicular to the rays). Unit: **W m⁻²**.

Card 2formula
Question

Formula for intensity?

Answer

$I = \dfrac{P}{A}$ — power ÷ area. **Given** in the data booklet.

Card 3definition
Question

What is the unit of intensity?

Answer

**W m⁻²** (watts per square metre).

Card 4formula
Question

Intensity a distance d from a source radiating equally in all directions?

Answer

$I = \dfrac{P}{4\pi d^{2}}$ — the power spread over a sphere of radius d (so I ∝ 1/d²).

Card 5definition
Question

State what is meant by the solar constant.

Answer

The **intensity of the Sun's radiation arriving at Earth's distance** (just above the atmosphere): **S = 1.36 × 10³ W m⁻²**.

Card 6definition
Question

Value of the solar constant?

Answer

**1.36 × 10³ W m⁻²** — given in the data booklet.

Card 7concept
Question

Why does intensity fall with distance?

Answer

A fixed power spreads over an ever-larger **sphere** (A = 4πd²); same power ÷ bigger area = smaller intensity.

Card 8concept
Question

Double the distance from a source — what happens to the intensity?

Answer

It drops to a **quarter** (× 1/4), because I ∝ 1/d² (inverse-square law).

Card 9formula
Question

How do you find a source's total power from the intensity at distance d?

Answer

Multiply by the whole sphere area: **P = I × 4πd²**.

Card 10formula
Question

Useful power output of a solar panel?

Answer

Incident **intensity × panel area × efficiency** (efficiency as a decimal).

Card 11concept
Question

Whose power is the solar constant — the Sun's total, or per m²?

Answer

**Per m²** — it is an intensity (W m⁻²) at Earth's distance, not the Sun's total power (W).

2.2.212 cards

Card 12definition
Question

What is a black body?

Answer

A perfect **absorber and emitter** of radiation — it absorbs every wavelength that hits it and, when hot, radiates over all wavelengths. Stars are a good model.

Card 13formula
Question

State the Stefan-Boltzmann law.

Answer

The total power (luminosity) radiated by a black body is $L = \sigma A T^4$ — surface area × temperature to the fourth power × the Stefan-Boltzmann constant. **Given** in the data booklet.

Card 14definition
Question

In L = σAT⁴, what is σ and its value?

Answer

The **Stefan-Boltzmann constant**, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given).

Card 15concept
Question

How does radiated power depend on temperature?

Answer

As **T⁴** — doubling the kelvin temperature multiplies the power by 2⁴ = **16**.

Card 16formula
Question

State Wien's displacement law.

Answer

The peak wavelength and absolute temperature multiply to a constant: $\lambda_{max} T = 2.9 \times 10^{-3}$ m K. **Given** in the data booklet.

Card 17concept
Question

In Wien's law, how do λ_max and T relate?

Answer

They are **inversely** related — a **hotter** body has a **shorter** peak wavelength (bluer light).

Card 18concept
Question

What unit must temperature be in for these laws?

Answer

**Kelvin (K)** — never °C. Convert with K = °C + 273.

Card 19concept
Question

What happens to the black-body curve when T rises?

Answer

It gets **taller** (more total power, Stefan-Boltzmann) and its **peak shifts to a shorter wavelength** (Wien).

Card 20example
Question

Find the peak wavelength of a 5800 K star.

Answer

$\lambda_{max} = \dfrac{2.9 \times 10^{-3}}{5800} = 5.0 \times 10^{-7}$ m (500 nm).

Card 21concept
Question

How do you compare the power of two black bodies?

Answer

Write $L = \sigma A T^4$ for each and **divide** one by the other — σ cancels, leaving a ratio of areas and T⁴.

Card 22example
Question

Why does an iron bar glow red then white as it heats?

Answer

Rising T shifts the spectrum's peak to shorter wavelengths (Wien) and adds power across all wavelengths (Stefan-Boltzmann), so the visible colour shifts red → orange → white.

Card 23formula
Question

For a black-body sphere, what is the area A in L = σAT⁴?

Answer

The sphere's surface area, $A = 4\pi r^2$, so $L \propto r^2 T^4$.

2.2.311 cards

Card 24definition
Question

Define albedo.

Answer

The **fraction of incident sunlight that a surface reflects** (scatters back). A number between 0 and 1, with no unit.

Card 25formula
Question

Formula for albedo?

Answer

$\text{albedo} = \dfrac{\text{total scattered power}}{\text{total incident power}}$ — the reflected fraction. **Given** in the data booklet.

Card 26concept
Question

If the albedo is 0.30, what fraction is absorbed?

Answer

**0.70** — the absorbed fraction is 1 − albedo.

Card 27definition
Question

Roughly what is Earth's average albedo?

Answer

About **0.30** — roughly 30% of sunlight is reflected back to space.

Card 28comparison
Question

Which surfaces have a high albedo? A low albedo?

Answer

**High:** fresh snow/ice (~0.8), thick cloud (~0.7). **Low:** dark ocean (~0.06), forest/asphalt (~0.1–0.2).

Card 29concept
Question

Why is the average incoming intensity S ÷ 4?

Answer

Sunlight lands on the **disc** Earth shows the Sun (πr²) but is shared over the whole **sphere** (4πr²): πr² ÷ 4πr² = 1/4.

Card 30example
Question

What is the average absorbed intensity for Earth?

Answer

About **240 W m⁻²**: (1 − 0.30) × (S ÷ 4) = 0.70 × 340 ≈ 240 W m⁻².

Card 31concept
Question

What does 'energy balance' mean for a planet?

Answer

At a steady temperature the power **absorbed** from the Sun equals the power **radiated** away. Energy in = energy out.

Card 32concept
Question

What does Earth's albedo depend on?

Answer

The **surface** (ice/cloud high, ocean/forest low), **cloud cover**, and the Sun's angle — so **latitude** and **time of day**.

Card 33concept
Question

Common albedo mistake to avoid?

Answer

Treating albedo as the **absorbed** fraction. Albedo is the **reflected** fraction; absorbed = 1 − albedo.

Card 34concept
Question

How does emissivity enter the balance?

Answer

A real surface radiates **emissivity ×** the black-body value. Use it when the surface is not a perfect black body (emissivity < 1).

2.2.411 cards

Card 35definition
Question

What is the greenhouse effect?

Answer

Greenhouse gases let **sunlight in** but absorb the **infrared** the warm surface radiates out, sending some **back down** — so the surface stays **warmer**.

Card 36definition
Question

Name the four main greenhouse gases.

Answer

**Carbon dioxide (CO₂)**, **methane (CH₄)**, **water vapour (H₂O)** and **nitrous oxide (N₂O)**.

Card 37concept
Question

Which radiation do greenhouse gases trap — incoming or outgoing?

Answer

**Outgoing infrared** from the warm surface. Incoming sunlight (mostly visible) passes straight through.

Card 38process
Question

Outline the mechanism (2-mark answer).

Answer

Greenhouse gases **absorb** the **infrared** the surface emits, then **re-emit** it in all directions, so some returns **back down** to the surface, keeping it warmer.

Card 39concept
Question

Why do CO₂ and CH₄ absorb infrared but N₂ and O₂ don't?

Answer

CO₂/CH₄ bonds **resonate** (vibrate) at infrared frequencies, so they absorb infrared; the simple N₂/O₂ bonds do not.

Card 40definition
Question

What does 'resonate' mean here?

Answer

The infrared radiation's frequency **matches** the natural vibration frequency of the gas molecule's bonds, so the bond absorbs the energy.

Card 41comparison
Question

Natural vs enhanced greenhouse effect?

Answer

**Natural** = warming from gases always present (Earth ~33 °C warmer, needed for life). **Enhanced** = **extra** warming from human-added gases.

Card 42example
Question

Main human cause of the enhanced greenhouse effect?

Answer

**Burning fossil fuels** (coal, oil, gas), which releases extra **CO₂**.

Card 43example
Question

Roughly how much warmer is Earth because of the greenhouse effect?

Answer

About **33 °C** warmer than it would be with no atmosphere — without it, Earth would be far too cold for life.

Card 44concept
Question

Common greenhouse-effect mistake to avoid?

Answer

Saying the gases block **incoming sunlight**. They don't — sunlight passes in; the gases trap the **outgoing infrared**.

Card 45example
Question

Where does the extra methane (CH₄) mostly come from?

Answer

**Farming** (cattle), **rice fields**, **landfill** and **gas leaks** — a strong infrared absorber per molecule.

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