B.2 Greenhouse effect

The solar constant at a planet with no atmosphere is S and the albedo of its surface is α.

What is the mean intensity of solar radiation absorbed by the planet, averaged over its whole surface?

The resonance model is used to explain why carbon dioxide is a greenhouse gas. Which statement best describes the mechanism in this model?

In a simplified annual-mean model, each square metre of Earth's surface has the following energy flows:
• it absorbs 160 W m⁻² of solar radiation;
• it absorbs 340 W m⁻² of infrared radiation from the atmosphere;
• it loses 100 W m⁻² by convection and evaporation.
The surface temperature is constant. What is the infrared intensity emitted by the surface?

The solar constant at the Earth (1.0 AU from the Sun) is 1.36 kW m⁻². An asteroid-mining probe is parked 2.0 AU from the Sun. What is the intensity of solar radiation at the probe?

An area of Arctic sea ice melts and is replaced by open ocean water. The solar intensity reaching the area is unchanged. Which describes the change in the albedo of the area and the change in the solar power absorbed by it?

In a resonance model of the greenhouse effect, the bonds in a molecule of carbon dioxide are treated as oscillators. One mode of vibration of the molecule has a natural angular frequency of 1.26 × 10¹⁴ rad s⁻¹.

Which row gives the wavelength of the electromagnetic radiation that drives this vibration at resonance, and the direction in which the molecule re-emits the absorbed energy?

Two planets orbit the same star. Planet Y is at distance d from the star and has an equilibrium surface temperature T. Planet X is at distance 2d and has the same albedo and the same emissivity as Y. Neither planet has an atmosphere.

What is the equilibrium surface temperature of X?

Two metal panels on the radiator of a spacecraft are at the same temperature. Panel X is coated with a paint of emissivity 0.30 and emits 150 W m⁻². Panel Y is coated with a paint of emissivity 0.90. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

Which row gives the power emitted per unit area by panel Y and the temperature of the panels?

At a high-latitude site, sunlight of intensity 950 W m⁻² (measured on a surface perpendicular to the rays) falls on level ground covered in snow. The rays make an angle of 65° with the vertical. The albedo of the snow is 0.80.

What is the power absorbed per unit area of the ground?

The graph shows the fraction of radiation absorbed by a layer of gas X in the atmosphere, for wavelengths from 0.20 μm to 12 μm.

The surface temperature of the Sun is 5800 K and the mean surface temperature of the Earth is 288 K. Treat both as black bodies. Which row is correct?

A moon with no atmosphere is in thermal equilibrium. When its only energy source is the solar radiation it absorbs, its mean surface temperature is T₀.

Tidal heating then generates thermal energy inside the moon at a rate equal to a fraction f of the solar power it absorbs. This energy also leaves as radiation from the surface. The albedo and the emissivity of the surface do not change.

What is the new mean surface temperature in equilibrium?

At noon on an equinox, the Sun is directly overhead at the equator. A weather station at latitude 60° N measures the solar intensity on horizontal ground. Ignore any difference in absorption by the atmosphere. What is the ratio \(\frac{\text{intensity on horizontal ground at }60^\circ}{\text{intensity on horizontal ground at the equator}}\)?

Nitrogen and oxygen together make up about 99% of dry air, yet neither is a significant greenhouse gas. What is the best explanation?

A simple model of a planet has a single atmospheric layer. The layer is completely transparent to incoming sunlight and absorbs all of the infrared radiation emitted by the surface. The layer radiates equally upwards and downwards. The surface and the layer both behave as black bodies. Without the layer, the surface temperature would be T₀. What is the surface temperature with the layer present?

A rocky planet with no atmosphere orbits a star in radiative equilibrium. Its equilibrium temperature does not depend on

Two spherical fuel tanks on a spacecraft are at the same temperature. Tank X has radius r and a surface emissivity of 0.90. Tank Y has radius 2r and a polished surface of emissivity 0.30. What is \(\frac{\text{power radiated by Y}}{\text{power radiated by X}}\)?

The front face of the heat shield on a solar probe faces the Sun. Which combination of surface properties for this face gives the lowest equilibrium temperature for the shield?

Carbon dioxide and water vapour absorb a large fraction of the radiation emitted by the Earth's surface but only a small fraction of the incoming solar radiation. What is the main reason for this?

A thermal camera calculates temperature by assuming that every surface has emissivity 1. It is pointed at a polished steel pan at 400 K whose emissivity is 0.20. Ignore any radiation reflected by the pan. What temperature does the camera display?

A star has a luminosity 16 times that of the Sun. An exoplanet orbits it at 2 AU and has the same albedo and emissivity as the Earth. The Earth's equilibrium temperature is T. What is the equilibrium temperature of the exoplanet?

The surface of a planet absorbs a mean intensity of 160 W m⁻² of solar radiation and 240 W m⁻² of infrared radiation emitted downwards by its atmosphere. The surface also loses 40 W m⁻² to the atmosphere by convection and evaporation. The emissivity of the surface is 0.90. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

What is the equilibrium temperature of the surface?

A thin, flat sunshade is in space at the same distance from the Sun as the Earth, with one face perpendicular to the Sun's rays. The intensity of sunlight there is 1.36 × 10³ W m⁻². The sunlit face has an albedo of 0.20. The sunshade conducts well, so both faces are at the same temperature, and both faces have an emissivity of 0.80. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

What is the equilibrium temperature of the sunshade?

During summer, a patch of Arctic sea ice of albedo 0.60 melts completely to leave open ocean of albedo 0.10. The incident solar intensity is unchanged. By what factor does the solar power absorbed by the patch increase?

A space agency models an exoplanet with the same albedo as Earth and no greenhouse effect. The exoplanet orbits a star whose luminosity is 4 times that of the Sun, at a distance 4 times the Earth–Sun distance. Take Earth's equilibrium temperature without a greenhouse effect to be 255 K. What is the best estimate of the exoplanet's equilibrium temperature?

Planets A and B receive the same solar intensity and have the same albedo. Both emit as black bodies. Planet A rotates rapidly and emits uniformly from its whole surface. Planet B is tidally locked: model it as re-emitting all the energy it absorbs from its sunlit hemisphere only, at a uniform temperature. What is \(\frac{T_B}{T_A}\)?

To estimate the solar constant, a student places a thin aluminium disc of diameter 4.00 cm and mass 12.0 g, blackened on its upper face, so that it faces the Sun directly at midday. The disc rests on polystyrene. Its temperature θ is read every 30 s with a digital thermometer of resolution 0.1 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. The table and the graph show the results with the line of best fit.

Three planets b, c and d orbit a faint star of mass 4.0 × 10²⁹ kg and luminosity 2.0 × 10²⁴ W in circular orbits. Their orbital periods are found from the regular dips in the star's brightness as each planet passes in front of it. Infrared observations of planet c suggest a mean surface temperature of 341 K. Astronomers want to know whether planet c could have an atmosphere. They model each planet as a black body with an albedo of 0.30 that radiates from its whole surface.

A satellite instrument measures the monthly mean albedo α of several regions of tropical ocean, together with the fraction f of each region covered by cloud. The graph shows α against f. Each albedo has an uncertainty of ±0.02, shown by the error bars. The solid line is the line of best fit; the dashed lines are the steepest and the shallowest lines that pass through all the error bars. The mean solar intensity arriving at the top of the atmosphere above these regions is 410 W m⁻².

A student measures the emissivity e of a matt paint. A metal plate painted on one face is heated electrically, and its temperature is measured with a thermometer embedded in the metal. A calibrated detector gives the net power radiated per unit area, I, from the painted face. The surroundings are at an unknown kelvin temperature T_r. The net intensity is modelled by

I = eσ(T⁴ − T_r⁴)

where T is the kelvin temperature of the plate. The graph shows I against T⁴ with the line of best fit, extended to lower values of T⁴.

In summer the mean intensity of sunlight reaching the surface of a region of the Arctic Ocean, averaged over day and night, is 300 W m⁻². Part of the region is covered by sea ice of albedo 0.60; the rest is open sea water of albedo 0.06.

Density of ice = 917 kg m⁻³; specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹.

A team planning a crewed Mars mission needs an estimate of the planet's equilibrium temperature. Mars orbits at 1.52 times the Earth–Sun distance and has a mean albedo of 0.25. The solar constant at Earth is 1360 W m⁻². Treat Mars as a black-body emitter with no greenhouse effect.

A cyclist at rest after a race has an exposed skin area of 1.8 m² at a temperature of 306 K. The emissivity of skin in the infrared is 0.97. The surroundings are at 293 K. Take σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

Mars orbits the Sun at a mean distance of 2.28 × 10¹¹ m. Luminosity of the Sun = 3.83 × 10²⁶ W; radius of the Earth's orbit = 1.50 × 10¹¹ m.

Carbon dioxide molecules absorb infrared radiation of wavelength 15.0 μm, which makes the molecules vibrate (bend). In a simple resonance model, this vibration is treated as an oxygen atom of mass 2.66 × 10⁻²⁶ kg oscillating on a spring of spring constant k.

A spacecraft studying the Sun carries a flat heat shield that always faces the Sun. At its closest approach the spacecraft is 6.9 × 10⁹ m from the centre of the Sun.

Luminosity of the Sun L_☉ = 3.83 × 10²⁶ W; solar constant at the Earth S = 1.36 × 10³ W m⁻²; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

A climate research group studies a region of the Arctic Ocean during one summer. The table shows typical albedos.
Surface | Albedo
Fresh snow on ice | 0.85
Bare sea ice | 0.60
Open ocean | 0.06
Over a 30-day period, the mean solar intensity reaching the surface is 200 W m⁻². During this period, an area of 2.0×10⁵ km² of bare sea ice becomes open ocean. The specific latent heat of fusion of ice is 3.3×10⁵ J kg⁻¹.

A building engineer is considering a reflective 'cool roof' coating for a flat roof of area 120 m². At midday, the solar intensity on the roof is 850 W m⁻². The existing roof has an albedo of 0.15. The coated roof has an albedo of 0.80. Both surfaces have an infrared emissivity of 0.90.

In hospitals, a capnograph measures the concentration of carbon dioxide in a patient's exhaled breath. It does this by measuring how much infrared radiation of wavelength 4.26 μm is absorbed as it passes through the breath sample. Exhaled breath also contains nitrogen, oxygen and water vapour.
Wien's constant b = 2.9×10⁻³ m K; c = 3.00×10⁸ m s⁻¹; mean surface temperature of the Earth = 288 K.

A space telescope surveys three rocky planets orbiting the same star. The table gives the stellar intensity at each planet's orbit, its albedo and its measured mean surface temperature.

Planet | Intensity / W m⁻² | Albedo | Surface temperature / K
b | 2600 | 0.10 | 370
c | 1100 | 0.35 | 300
d | 450 | 0.20 | 205

Assume each planet rotates rapidly. σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

A thin flat solar panel on a CubeSat in Earth orbit faces the Sun directly. The solar intensity is 1360 W m⁻². The panel reflects 15% of the incident sunlight. Both faces have emissivity 0.85, and the panel radiates from both faces at a uniform temperature. σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

A red dwarf star has a surface temperature of 3100 K and a radius of 1.46×10⁸ m. Astronomers look for a planet orbiting where the stellar intensity equals the Earth's solar constant, 1360 W m⁻². Treat the star as a black body. σ = 5.67×10⁻⁸ W m⁻² K⁻⁴; Wien constant = 2.9×10⁻³ m K.

A rocky planet with an atmosphere orbits a distant star. The intensity of the starlight at the distance of the planet is S = 2.00 × 10³ W m⁻².

The diagram shows the mean energy flows per unit area of the planet's surface, in W m⁻². Of the 520 W m⁻² emitted by the surface, 40 W m⁻² passes directly through the atmosphere to space and the rest is absorbed by the atmosphere. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

An oceanographer models noon absorption of sunlight by the sea at an equinox. Assume that 70% of the top-of-atmosphere intensity of 1360 W m⁻² reaches the surface as a direct beam at every latitude. The albedo of the ocean depends on the solar elevation angle (the angle of the Sun above the horizon):
Solar elevation | Ocean albedo
90° | 0.03
30° | 0.06
10° | 0.35
Sea ice has an albedo of 0.60 at all elevations. At noon on an equinox, the solar elevation is (90° − latitude).

A planetary model has a surface that absorbs a mean solar intensity of 240 W m⁻². The atmosphere is transparent to sunlight. The atmosphere is a single layer at temperature T_a that absorbs a fraction ε of the infrared emitted by the surface and has emissivity ε. The layer emits εσT_a⁴ both upwards and downwards. The surface, at temperature T_s, is a black body.

Satellite data show that 4.0×10¹² m² of Arctic sea ice of albedo 0.55 has been replaced by open ocean of albedo 0.07. The Earth's surface area is 5.1×10¹⁴ m², its mean albedo is 0.30 and the solar constant is 1360 W m⁻². Averaged over a year, the Arctic receives 0.55 times the global mean solar intensity per unit area. Ignore the effect of clouds.

A solar probe carries a flat heat shield whose front face points directly at the Sun. At closest approach the probe is 0.060 AU from the Sun. The solar constant at 1.0 AU is 1360 W m⁻². The front face has albedo 0.60 and emissivity 0.80. Assume the shield is thermally insulated at the back, so it radiates only from its front face. σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

A simple climate model treats the atmosphere as a single layer. The layer is transparent to sunlight and absorbs a fraction f of the infrared radiation emitted by the ground. The layer radiates equal power per unit area upwards and downwards. The ground behaves as a black body and absorbs a mean solar intensity of 240 W m⁻². σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

The Earth's equilibrium temperature of 255 K is given by \(T = \left(\frac{(1-\alpha)S}{4\sigma}\right)^{1/4}\), where S is the intensity of solar radiation arriving at the Earth (the solar constant, whose value varies slightly with solar activity) and α = 0.30 is the albedo.

A lunar rover must get rid of 300 W of heat produced by its electronics. It uses a flat radiator panel that emits from one face only. The panel's coating has emissivity 0.88 for infrared radiation. It reflects 80% of incident sunlight, so its albedo for sunlight is 0.80. In the design case the panel is in shadow at 290 K and absorbs no radiation. The solar intensity at the Moon is 1360 W m⁻². σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.