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IZhO 2022, Теория — Задача 2. Problem 2. Greenhouse effect

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XVIII International Zhautykov Olympiad/Theoretical Competition · 16 февруари 2022 г. · 10 т.

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The duration of the theoretical competition is 4 hours. There are three problems.
You can use your own calculator for numerical calculations.
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Условие

Introduction

Any heated body, whose temperature $T$ is above absolute zero, radiates electromagnetic waves. The spectrum of this radiation, called thermal, depends on the optical properties of the body surface and its temperature. Despite the fact that the radiation of each body is its individual characteristic, general laws are well known that describe the thermal radiation.

Kirchhoff's law. In a state of thermodynamic equilibrium, the ratio of the emissivity of a body $r(\lambda,T)$ to its absorptivity $k(\lambda,T)$ is a universal function $r_0(\lambda,T)$ that does not depend on individual characteristics

\frac{r(\lambda,T)}{k(\lambda,T)} = r_0(\lambda,T).

The value $r(\lambda,T)\Delta\lambda$ has the meaning of the energy emitted per unit area per unit time in a narrow wavelength range from $\lambda$ to $\lambda + \Delta\lambda$ . The absorptivity of the body $k(\lambda,T)$ is a dimensionless absorption coefficient equal to the ratio of the radiation energy absorbed by the body to the total radiation energy incident on the body surface, if the wavelengths of the incident radiation lie in a narrow wavelength range from $\lambda$ to $\lambda + \Delta\lambda$ . If the body completely absorbs all the incident electromagnetic radiation $k(\lambda,T) = 1$ , then such a body is called an absolute black body.

Wien's displacement law. The wavelength $\lambda_{max},$ , at which the function $rr_0(\lambda,T)$ has a maximum, is related to the absolute temperature by the relation

\lambda_{max}T = b,

where $b = 2.898\cdot10^{-3}\ \mathrm{m/K}$ is called the Wien constant.

Stefan-Boltzmann law. The total emissivity of a black body over all wavelengths is described by the formula

R(T) = \int_0^{\infty} r_0(\lambda,T)d\lambda = \sigma T^4,

where $\sigma = 5.670\cdot10^{-8}\ W/(m^2\cdot K^4)$ stands for the Stefan-Boltzmann constant.

Using the Stefan-Boltzmann law, the formula for $r_0(\lambda,T)$ can be represented as

r_0(\lambda,T) = \sigma T^4\varphi(\lambda,T).

Here, the Planck function $\varphi(\lambda,T)$ describes the energy distribution in the black body radiation spectrum; the value $\varphi(\lambda,T)\Delta\lambda$ is equal to the fraction of the thermal radiation energy per narrow spectral interval from $\lambda$ to $\lambda + \Delta\lambda$ . The total area under the graph of the function $\varphi(\lambda,T)$ is equal to unity. In this problem, it is recommended to use the graphs of this function, shown in the figure below and plotted at temperatures $t = 0^{\circ}\mathrm{C}$ and $t = 50^{\circ}\mathrm{C}$ .

Graph of fi(lambda) versus wavelength in micrometer from 0 to 20; two bell-shaped curves plotted at 0°C and 50°C, with maxima around 9–10 μm and values up to about 0.074; the vertical axis runs 0,00 to 0,08. — The Planck function

Model of the Earth and its atmosphere

Climate change associated with an increase in the average temperature of the atmosphere is now an established scientific fact. It is believed that the main cause of the global warming is the greenhouse effect. Solar radiation, whose main part lies in the visible region of the spectrum, passes almost completely through the atmosphere, and then is absorbed by the earth's surface. On the contrary, the thermal radiation of the earth's surface, which lies mainly in the infrared region of the spectrum, is significantly absorbed by certain atmospheric gases, mainly water vapor and carbon dioxide. In this problem, the simplest model of the greenhouse effect is considered and some numerical estimates are made of its influence on the atmospheric temperature.

The surface of the Earth is assumed to be an absolutely black body of a spherical shape, covered with a layer of the atmosphere, whose thickness is much less than the Earth radius. Conventionally, the atmosphere is divided into two parts: 1) the lower layer, directly adjacent to the earth's surface and having the same temperature as the earth's surface; 2) the upper (greenhouse) layer, capable of absorbing thermal radiation coming from the earth's surface. It is assumed that the transfer of energy between the Sun, the earth's surface and the atmosphere is carried out only through radiation, and the temperatures of the earth's surface and atmospheric layers are the same at all their points and do not depend on time, for example, on the time of day or year season. In the following, the solar constant $W = 1.40\cdot10^3\ \mathrm{W/m^2}$ is considered known, which is the power of solar radiation incident on the unit area of the Earth, oriented perpendicular to the incident light.

Atmosphere without the greenhouse effect

2.1 Evaluate the wavelength $\lambda_{max\ S}$ , which corresponds to the maximum in the thermal radiation of the Sun, if the surface temperature of the Sun is approximately equal to $T_{\mathrm{S}} = 6500\ \mathrm{K}$ .

2.2 Neglecting the absorption of the atmosphere and considering the earth's surface as an absolutely black body, evaluate the steady-state temperature of the Earth's surface $T_0$ , and also determine this temperature $t_0$ in the Celsius scale. This temperature is called below as the "black earth" temperature.

2.3 Calculate the wavelength $\lambda_{max\ E}$ , which corresponds to the maximum in the radiation of the "black earth".

2.4 Calculate the power of solar radiation $w$ per unit area of the Earth's surface.

Various atmosphere models

In reality, the upper layer of the atmosphere effectively absorbs electromagnetic radiation of certain wavelengths in the infrared region of the spectrum. For example, the figure on the right shows the real dependence of the atmospheric transmission as a function of the wavelength. As can be seen, this dependence is rather complicated, therefore, in the problem below, several simplified models are considered.

Atmospheric transmittance (percent) versus wavelength (microns) from 0 to 15, with shaded absorption bands; below the plot, arrows indicate absorbing molecules H2O, CO2, O2, HDO, N2O, CH4, O3.

Let $K$ be the total absorption coefficient of the upper layer of the atmosphere for the thermal radiation of the Earth, i.e. the ratio of the energy of the thermal radiation absorbed by the upper layer of the atmosphere to the total energy incident on the upper layer of atmosphere from the earth's surface.

2.5 Obtain the formula for the steady-state temperature of the Earth's surface $T_1$ and express it in terms of the "black earth" temperature $T_0$ and the absorption coefficient $K$ .

Maximum greenhouse effect

Let the upper layer of the atmosphere completely transmit the solar radiation and completely absorb the thermal radiation of the Earth.

2.6 Calculate how much the temperature of the Earth's surface $\Delta t_1 = T_1 - T_0$ increases compared to the temperature of the "black earth" due to the maximum greenhouse effect.

Water greenhouse effect

Let the spectral absorption coefficient $k(\lambda)$ of the upper layer of the atmosphere be a known function of the wavelength $\lambda$ of the incident radiation and be independent of its temperature.

2.7 Express the total absorption coefficient of the upper atmosphere $K$ in terms of $k(\lambda)$ and the Planck distribution function $\varphi(\lambda,T_1),$ , where $T_1$ denotes the temperature of the Earth's surface.

Assume that the absorption in the upper layer of the atmosphere is completely due to water vapor. Approximately, it can be considered that water vapor completely absorbs radiation whose wavelengths lie in the range from 5.00 to 8.00 μm, whereas the rest is completely transmitted.

2.8 Using the plots of the Planck distribution function given in the introduction section of this problem, calculate the numerical values of the total absorption coefficient $K$ of the upper atmosphere for two values of the Earth's surface temperatures $t_1 = 0^{\circ}\mathrm{C}$ and $t_1 = 50^{\circ}\mathrm{C}$ .

In the above specified temperature range from $t_1 = 0^{\circ}\mathrm{C}$ to $t_1 = 50^{\circ}\mathrm{C}$ , the dependence of the total absorption coefficient on the ground temperature $t_1$ is approximately described by a linear function of the temperature itself: $K(t_1) = K_0(1 + \alpha t_1),,$ , where $K_0, \alpha$ are some constants.

2.9 Calculate the numerical values of the parameters $K_0$ and $\alpha$ .

In the following, assume that the temperature changes under question are small, so formulas of approximate calculus can be used.

2.10 Neglecting the dependence of the absorption coefficient of the atmosphere on the temperature and assuming it to be equal to the absorption coefficient at the temperature of the "black Earth" $T_0$ , calculate the change in the temperature of the Earth's surface $\Delta t_1 = T_1 - T_0$ .

2.11 Calculate the change in the temperature of the Earth's surface $\Delta t_1 = T_1 - T_0 \ll T_0$ if the dependence of the atmospheric absorption coefficient on the temperature is described by the linear function of the earth's temperature as defined above.

Amplification of the greenhouse effect by carbon dioxide

Let us take into account the effect of the absorption by carbon dioxide present in the atmosphere. At the current concentration of carbon dioxide in the atmosphere (approximately 0.05%), it can be considered that carbon dioxide completely absorbs the radiation of the Earth in the wavelength ranges from 2.50 to 3.00 μm and from 6.50 to 7.00 μm, and in the range from 16.0 to 18.0 μm the absorption coefficient equals 0.500. For other wavelengths, the absorption of radiation by carbon dioxide can be neglected.

2.12 Estimate how much, as compared to the water greenhouse effect model, the temperature of the Earth's surface increases due to the absorption of radiation by carbon dioxide.

2.13 Estimate how much, as compared to 2.12, the temperature of the Earth's surface increases if the concentration of carbon dioxide in the atmosphere increases by $\eta = 2.00$ times as compared to its current concentration.

Отговори

Решение

Problem 2. Greenhouse effect (10.0 points)

Atmosphere without greenhouse effect

2.1 Direct calculation by Wien's formula gives the following result

\lambda_{\max S} = \frac{b}{T_S} = 0.446\ \mu\mathrm{m}.

(1)

Earth illuminated from one side by parallel solar rays and radiating in all directions.

2.2 In the steady state, the power of solar radiation incident on the Earth is equal to the power of the thermal radiation of the Earth. When writing the energy balance equation, it must be taken into account that the Sun illuminates the Earth from one side, and the Earth radiates in all directions, i.e.

W\cdot\pi R^2 = \sigma T_0^4\cdot 4\pi R^2.

(2)

It follows from this relation that

T_0 = \sqrt[4]{\frac{W}{4\sigma}} = 280.3\ \mathrm{K},

(3)

and the same temperature in degrees Celsius is equal to

t_0 = 7.15\ {}^{\circ}\mathrm{C}.

(4)

2.3 According to the Wien's formula, we find that at the given temperature, the maximum radiation corresponds to the wavelength

\lambda_{\max E} = \frac{b}{T_E} = 10.3\ \mu\mathrm{m}.

(5)

2.4 The same geometric relationships that lead to equation (2) allow one to conclude that the power of solar radiation per unit area of the Earth's surface is found as

w = \frac{W\cdot\pi R^2}{4\pi R^2} = \frac{W}{4} = 350\ \mathrm{W/m^2}.

(6)

Two-layer atmosphere model: solar flux w, atmospheric layer with coefficient k at temperature t2, fluxes R1 and R2, Earth surface at t1.

Various atmosphere models

2.5 We introduce the following notation:

$t_1$ (or $T_1$ in the Kelvin scale) – the emperature of the Earth's surface and the lower layer of the atmosphere immediately adjacent to it; $t_2$ (or $T_2$ ) – the temperature of the upper layer of the atmosphere; $w$ – the flux density of solar radiation, i.e. the energy incident on a unit area of the Earth's surface per unit time (or irradiated); $R_1$ – the thermal radiation power per unit area of the Earth; $R_2$ – the thermal radiation power per unit area of the atmospheric layer; the radiation fluxes of this layer towards the Earth and into outer space are equal.

The energy balance equation for a unit area of the Earth's surface has the following form

w + R_2 = R_1.

(7)

A similar equation for the upper layer of the atmosphere gives rise to

KR_1 = 2R_2.

(8)

Using the laws of thermal radiation, energy fluxes can be expressed in terms of the temperatures of the radiating surfaces as follows

R_1 = \sigma T_1^4,

(9)

R_2 = K\sigma T_2^4.

(10)

Therefore, taking into account formulas (2) and (3), we obtain from expressions (7)-(10) the temperature of the Earth's surface in the form

T_1 = \frac{T_0}{\sqrt[4]{1-\frac{K}{2}}}.

(11)

Maximum greenhouse effect

2.6 For the maximum greenhouse effect $K = 1$ , therefore, it is obtained for this model

T_1 = T_0\sqrt[4]{2} = 333.3\ \mathrm{K} = 60.2\ {}^{\circ}\mathrm{C}.

(12)

Thus, the maximum increase in temperature due to the greenhouse effect on the "black earth" is equal to

\Delta t_1 = 53.0\ {}^{\circ}\mathrm{C}.

(13)

Water greenhouse effect

2.7 The Earth as a black body irradiates the energy

W_0 = \int\limits_0^{\infty} r_0(\lambda, T_1)\, d\lambda,

(14)

The absorbed energy can be expressed in terms of the spectral absorption coefficient and the spectral density of the Earth's radiation as follows

W_A = \int\limits_0^{\infty} k(\lambda) r_0(\lambda, T_1)\, d\lambda,

(15)

then the total absorption coefficient of terrestrial radiation by theupper layer of the atmosphere is calculated by the formula

K = \frac{W_A}{W_0} = \frac{\int\limits_0^{\infty} k(\lambda) r_0(\lambda, T_1)\, d\lambda}{\int\limits_0^{\infty} r_0(\lambda, T_1)\, d\lambda} = \frac{\sigma T_1^4\int\limits_0^{\infty} k(\lambda)\varphi(\lambda, T_1)\, d\lambda}{\sigma T_1^4\int\limits_0^{\infty}\varphi(\lambda, T_1)\, d\lambda} = \int\limits_0^{\infty} k(\lambda)\varphi(\lambda, T_1)\, d\lambda.

(16)

2.8 Since in the indicated wavelength range from 5.0 to 8.0 μm the water vapor absorbs all incident radiation, the total absorption coefficient is equal to the fraction of radiation energy falling into this interval. This fraction of energy is evalulated as the areas under the graphs given in the problem introduction.

Two graphs of spectral energy fraction φ versus wavelength (4–10 μm), for t = 0°C and t = 50°C, with shaded area between 5 and 8 μm; y-axis 0–0.08.

The calculations carried out for 4 points gives the following values for the absorption coefficients

t_1 = 0\ {}^{\circ}\mathrm{C}:\quad K_0 = 0.092,

(17)

t_1 = 50\ {}^{\circ}\mathrm{C}: \quad K_{50} = 0.158. \qquad (18)

2.9 It follows from the proposed relationship $K\left(t_1\right) = K_0\left(1 + \alpha t_1\right)$ that

K_0 = 0.092, \qquad (19)

\alpha = \frac{1}{t_{50}}\left(\frac{K_{50}}{K_0} - 1\right) = 0.014\ \mathrm{K}^{-1}. \qquad (20)

2.10 At the temperature of $t_1 = 5{,}4\,{}^\circ C$ , the absorption coefficient of the upper layer of the atmosphere is found as

K\left(t_0\right) = K_0\left(1 + \alpha t_0\right) = 0.101. \qquad (21)

Since the absorption coefficient is rather small, formula (12) for the steady temperature can be simplified to

T_1 = \frac{T_0}{\sqrt[4]{1-\frac{K}{2}}} \approx T_0\left(1 + \frac{K}{8}\right), \qquad (22)

and the rise in temperature is obrained as

\Delta t_1 = T_0\,\frac{K\left(t_0\right)}{8} = 3.55\ {}^{\circ}\mathrm{C}. \qquad (23)

2.11 To accurately answer the question, it is necessary to solve the nonlinear equation

T_1 = \frac{T_0}{\sqrt[4]{1-\frac{K\left(T_1\right)}{2}}}. \qquad (24)

However, the relative change in the absolute temperature is small, so we represent the sought temperature in the form

T_1 = T_0 + \Delta t, \qquad (25)

from which we find the value of the temperature change in view of the condition $\Delta t \ll T_0$

\Delta t = \frac{T_0\,\dfrac{K_0\left(1+\alpha t_0\right)}{8}}{1 - T_0\,\dfrac{\alpha K_0}{8}} = \frac{\Delta t_1}{1 - T_0\,\dfrac{\alpha K_0}{8}} \approx 3.73\ {}^{\circ}\mathrm{C}. \qquad (26)

Amplification of the greenhouse effect by carbon dioxide

2.12 Let us calculate the absorption coefficient due to carbon dioxide. To make estimates, we can assume that the air temperature differs slightly from $0\,{}^\circ C$ . To do this, we take into account that: 1) in the range from 2.5 to 3.0 μm, the energy of the thermal radiation of the Earth is negligible; 2) in the range from 6.5 μm to 7.0 μm all radiation is absorbed by water vapor; 3) in the range from 16 μm to 18 μm, the fraction of radiation energy is equal to $\Phi = 0.08$ (calculated according to the graph for $t = 0\,{}^\circ C$ ). Therefore, the additional absorption coefficient due to the presence of carbon dioxide is found as

K_2 = 0.04. \qquad (27)

Since the absorption of carbon dioxide and water vapor lie in different spectral ranges, the total absorption coefficient is equal to the sum of the absorption coefficients of water and carbon dioxide. Then the change in the steady-state surface temperature (taking into account absorption by carbon dioxide) increases by the value

\Delta t_1 = T_0\,\frac{K_2}{8} \approx 1.4\ {}^{\circ}\mathrm{C}. \qquad (28)

2.13 To calculate the absorption coefficient with increased concentration, we use the obvious reasoning: in the presence of several absorbing layers, the total transmission is equal to the product of the transmission coefficients of individual layers, therefore

1 - k_1 = \left(1 - k_0\right)^2. \qquad (29)

Hence it follows that if the concentration is doubled, the spectral absorption coefficient is expected to increase from 0.50 to

k_1 = 2k_0 - k_0^2 = 0.75. \qquad (30)

Therefore, the total absorption coefficient becomes equal to

K_2 = k\Phi = 0.06. \qquad (31)

i.e. increases by $\Delta K_2 = 0.02$ . Therefore, the additional rise in temperature is finally obtained as

\Delta t_1' = T_0\,\frac{\Delta K_2}{8} \approx 0.7\ {}^{\circ}\mathrm{C}. \qquad (32)

	Content	Points
2.1	Formula (1): $\lambda_{\max S} = \dfrac{b}{T_S}$	0.1	0.2
	Numerical value in formula (1): $\lambda_{\max S} = 0.446\ \mu m$	0.1
2.2	Formula (2): $W\cdot\pi R^2 = \sigma T_0^4\cdot 4\pi R^2$	0.4	1.0
	Formula (3): $T_0 = \sqrt[4]{\dfrac{W}{4\sigma}}$	0.2
	Numerical value in formula (3): $T_0 = 280.3\ \mathrm{K}$	0.2
	Numerical value in formula (4): $t_0 = 7.15\ {}^\circ \mathrm{C}$	0.2
2.3	Formula (5): $\lambda_{\max E} = \dfrac{b}{T_E}$	0.1	0.2
	Numerical value in formula (5): $\lambda_{\max E} = 10{,}3\ \mu\mathrm{m}$	0.1
2.4	Formula (6): $w = \dfrac{W}{4}$	0.1	0.2
	Numerical value in formula (6): $w = 350\ \mathrm{W/m^2}$	0.1
2.5	Formula (7): $w + R_2 = R_1$	0.2	1.2
	Formula (8): $KR_1 = 2R_2$	0.2
	Formula (9): $R_1 = \sigma T_1^4$	0.2
	Formula (10): $R_2 = K\sigma T_2^4$	0.2
	Formula (11): $T_1 = \dfrac{T_0}{\sqrt[4]{1-\dfrac{K}{2}}}$	0.4
2.6	Direct use of $K=1$	0.1	0.5
	Formula (12): $T_1 = T_0\sqrt[4]{2}$	0.2
	Numerical value in formula (13): $\Delta t_1 = 53.0\ {}^\circ \mathrm{C}$	0.2
2.7	Formula (14): $W_0 = \int\limits_0^{\infty} r_0\left(\lambda, T_1\right)d\lambda$	0.2	0.8
	Formula (15): $W_A = \int\limits_0^{\infty} k(\lambda) r_0\left(\lambda, T_1\right)d\lambda$	0.2
	Formula (16): $K = \int\limits_0^{\infty} k(\lambda)\varphi\left(\lambda, T_1\right)d\lambda$	0.4
2.8	Numerical value in (17): $t_1 = 0\ {}^\circ \mathrm{C}$ : $K_0 = 0.092$	0.6	1.2
	Numerical value in (18): $t_1 = 50\ {}^\circ \mathrm{C}$ : $K_{50} = 0.158$	0.6
2.9	Numerical value in (19): $K_0 = 0.092$	0.2	0.4
	Numerical value in (20): $\alpha = 0.031\ \mathrm{K}^{-1}$	0.2
2.10	Numerical value in (21): $K\left(t_0\right) = 0.0757$	0.4	0.8
	Numerical value in (23): $\Delta t_1 = 2.65\ {}^\circ \mathrm{C}$	0.4
2.11	Formula (24): $T_1 = \dfrac{T_0}{\sqrt[4]{1-\dfrac{K\left(T_1\right)}{2}}}$	0.2	1.0
	Formula (25): $T_1 = T_0 + \Delta t$ at $\Delta t \ll T_0$	0.2
	Formula (26): $\Delta t = \dfrac{T_0\,\dfrac{K_0\left(1+\alpha t_0\right)}{8}}{1 - T_0\,\dfrac{\alpha K_0}{8}} = \dfrac{\Delta t_1}{1 - T_0\,\dfrac{\alpha K_0}{8}}$	0.4
	Numerical value in formula (26): $\Delta t \approx 2.84\ {}^\circ \mathrm{C}$	0.2
2.12	Numerical value in (27): $K_2 = 0.04$	0.5	1.0
	Numerical value in (28): $\Delta t_1 \approx 1.4\ {}^\circ \mathrm{C}$	0.5
2.13	Formula (29): $1-k_1 = \left(1-k_0\right)^2$	0.5	1.5
	Formula (30): $k_1 = 2k_0 - k_0^2$	0.2
	Numerical value in formula (31): $K_2 = k\Phi = 0.06$	0.4
	Numerical value in (32): $\Delta t_1' \approx 0.7\ {}^\circ \mathrm{C}$	0.4
Total			10.0

Mathematical hints for the theoretical problems

The following formulas may be useful:

\int x^n dx = \frac{x^{n+1}}{n+1} + C,

where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant

\int \frac{dx}{x} = \ln \ln |x| + C,

where $C$ stands for an arbitrary constant

(1 + x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2,

for $|x| \ll 1$ and any value of $\gamma$

\ln \ln (1 + x) \approx x,

for $|x| \ll 1$

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