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IPhO 2024, Теория — Задача 1. The Greenhouse Effect

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The Greenhouse Effect · 21 юли 2024 г. · 30 т.

Условие

In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.

All objects, at different temperatures, emit thermal radiation. The quantity $u(\lambda, T)d\lambda$ indicates the thermal radiative power per unit area of an object at temperature $T$ between the wavelengths $\lambda$ and $\lambda + d\lambda$ . According to Planck's theory of blackbody radiation, we have

u(\lambda, T) = \frac{2\pi h c^2}{\lambda^5}\, \frac{1}{\exp(\tfrac{hc}{\lambda k_{\mathrm{B}} T}) - 1}, \qquad (1)

in which $hc = 1.24 \times 10^3\ \mathrm{eV \cdot nm}$ and $k_{\mathrm{B}} = 8.62 \times 10^{-5}\ \mathrm{eV/K}$ . The wavelength corresponding to the maximum of $u(\lambda, T)$ comes from the relation $\lambda_{\max} T = b$ (Wien's displacement law). Indeed, using equation (1), it can be shown that $b = \tfrac{hc}{x_{\mathrm{m}} k_{\mathrm{B}}}$ , where the dimensionless quantity $x_{\mathrm{m}}$ is the non-trivial root of an equation of the form $f(x) = 0$ ; you are asked to find the function $f(x)$ in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as $U(T) = \sigma T^4$ where $\sigma = 5.67 \times 10^{-8}\ \mathrm{W/m^2 K^4}$ . Moreover, according to Kirchhoff's law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.

Throughout this problem assume that the Sun is a blackbody at its average surface temperature of $T_{\mathrm{S}} = 5.77 \times 10^3\ \mathrm{K}$ . The Sun's radius is $R_{\mathrm{S}} = 6.96 \times 10^8\ \mathrm{m}$ and the average distance between the Earth and the Sun is $d = 1.50 \times 10^{11}\ \mathrm{m}$ . We denote by $\tilde{u}_{\mathrm{S}}(\lambda)$ , the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. $S_0 = \int \tilde{u}_{\mathrm{S}}(\lambda) d\lambda$ , is called the solar constant.

In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.

A. Earth as a Blackbody

In this part, consider the Earth's surface as a blackbody and neglect the Earth's atmosphere.

In figure 1 the functions $\gamma \tilde{u}_{\mathrm{S}}(\lambda)$ and $u(\lambda, T_{\mathrm{E}})$ are plotted versus $\lambda$ , where $\gamma$ is a dimensionless coefficient to rescale $\tilde{u}_{\mathrm{S}}(\lambda)$ such that the values of the two peaks coincide.

Graph of $u$ versus $\lambda$ showing a narrow, tall blue peak at short wavelength ($\gamma \tilde{u}_{\mathrm{S}}(\lambda)$, the solar spectrum) and a broad red peak at long wavelength ($u(\lambda, T_{\mathrm{E}})$, the Earth's thermal spectrum); the two peaks reach the same maximum value marked by a horizontal line. — Figure 1 - The plot of $u(\lambda, T_{\mathrm{E}})$ (red) and $\gamma \tilde{u}_{\mathrm{S}}(\lambda)$ (blue) versus $\lambda$

B. The Greenhouse Effect

In this part, we introduce a simple model in which the Earth's atmosphere is modeled as a thin layer at a small distance above the Earth's surface so that the difference between the area of the atmosphere's layer and the area of the Earth's surface can be neglected (see figure 2). In what follows assume that the major part of the thermal radiation from the Earth and the Sun are emitted at wavelengths near the $\lambda_{\max}$ for each one. Also assume that the "atmosphere layer" reflects a fraction $r_{\mathrm{A}} = 0.255$ of the visible-ultraviolet radiation incident from above or below, and completely transmits the rest. Assume that the atmosphere does not reflect any part of the infrared radiation, however, it absorbs a fraction $\varepsilon$ of the infrared radiation and transmits the rest. This behavior, known as the greenhouse effect, changes the average temperature of the Earth. The Earth's surface, on the other hand, reflects a fraction $r_{\mathrm{E}}$ of the visible-ultraviolet radiation and absorbs the rest of this radiation and all the infrared radiation.

Schematic drawing of incoming yellow solar radiation rays, some reflected; a thin blue horizontal band labelled ATMOSPHERE above a green Earth's surface; red wavy arrows labelled Infrared heat radiation going up from the surface and both up and down from the atmosphere; a large pink arrow labelled Non-radiative thermal flow pointing upward from the surface to the atmosphere. — Figure 2 - Thermal flows between the Earth and the atmosphere

Now assume that $r_{\mathrm{E}} \neq 0$ . In this case, the combined system of "Earth + atmosphere" reflects a different fraction of the solar radiation, called "albedo" and denoted by $\alpha$ .

Assume $T_{\mathrm{A}} = 245\ \mathrm{K}$ and $T_{\mathrm{E}} = 288\ \mathrm{K}$ . These values come from real data and may differ from the results which you have obtained in the previous tasks. Now suppose that a non-radiative (e. g. convective) thermal flow $J_{\mathrm{NR}} = k(T_{\mathrm{E}} - T_{\mathrm{A}})$ is maintained from the Earth to the atmosphere, where $k$ is a constant. The quantity, $J_{\mathrm{NR}}$ , is the transmitted power per unit area.

A-1 Find the solar constant, $S_0$ . [0,6 т.]

A-2 Find the Earth's temperature, $T_{\mathrm{E}}$ . [0,6 т.]

A-3 Find the function $f(x)$ . [0,4 т.]

A-4 Calculate the numerical value of $x_{\mathrm{m}}$ , and from this value $x_{\mathrm{m}}$ , find the value of $b$ . [0,4 т.]

A-5 Find $\lambda_{\max}$ for the Sun and the Earth. [0,2 т.]

A-6 Determine $\gamma$ . [0,8 т.]

B-1 Assume that $\varepsilon = 1$ and $r_{\mathrm{E}} = 0$ , and calculate the Earth's temperature $T_{\mathrm{E}}$ and the atmosphere's temperature $T_{\mathrm{A}}$ . [1 т.]

B-2 Determine the albedo, $\alpha$ , in terms of $r_{\mathrm{E}}$ and $r_{\mathrm{A}}$ . Then calculate its numerical value assuming $r_{\mathrm{E}} = 0.102$ (and $r_{\mathrm{A}} = 0.255$ ). [1,6 т.]

B-3 a) Express the Earth's temperature in terms of $\sigma$ , $\alpha$ , $S_0$ , and $\varepsilon$ .

b) Using the given data and the calculated albedo, find the numerical value of $\varepsilon$ which leads to the current average temperature of $T_{\mathrm{E}} = 288\ \mathrm{K}$ for the Earth. [1 т.]

B-4 Find $\dfrac{dT_{\mathrm{E}}}{d\varepsilon}$ and determine by how much the Earth's temperature increases if $\varepsilon$ increases by one percent. [0,8 т.]

B-5 Calculate $\varepsilon$ and $k$ in terms of $T_{\mathrm{E}}$ , $T_{\mathrm{A}}$ , $\sigma$ , $\alpha$ , and $S_0$ . [1,6 т.]

B-6 a) Differentiating the equations obtained in part B-5 with respect to $\varepsilon$ , find the two algebraic equations satisfied by $\dfrac{dT_{\mathrm{A}}}{d\varepsilon}$ and $\dfrac{dT_{\mathrm{E}}}{d\varepsilon}$ .

b) Use these equations to find the numerical value of change in the Earth's temperature as a result of a one percent increase in the value of $\varepsilon$ . [1 т.]

Решение

A. Earth as a blackbody

A-1. All the energy emitted from the surface of the Sun, will reach a sphere of radius $d$ , therefore:

\sigma T_{\mathrm{S}}^4.\,(4\pi R_{\mathrm{S}}^2) = (4\pi d^2).\,S_0

S_0 = \sigma T_{\mathrm{S}}^4.\left(\frac{R_{\mathrm{S}}}{d}\right)^2 = 5.67\times 10^{-8}\ \frac{\mathrm{W}}{\mathrm{m^2 K^4}}\times (5.77\times 10^{3}\ \mathrm{K})^4\times\left(\frac{6.96\times 10^{8}\ \mathrm{m}}{1.5\times 10^{11}\ \mathrm{m}}\right)^2 = 1.35\times 10^{3}\ \frac{\mathrm{W}}{\mathrm{m^2}}

A-1 (0.6 pt)

$S_0 = \sigma T_{\mathrm{S}}^4.\left(\dfrac{R_{\mathrm{S}}}{d}\right)^2$ , Numerical value of $S_0 = 1.35\times 10^{3}\ \mathrm{W/m^2}$

A-2. It is assumed that the Earth is in thermal equilibrium. Therefore, the energy it receives per unit time should be equal to the energy it radiates per unit time. The Earth’s cross-section intercepting the solar radiation at this distance has an area of $\pi R_{\mathrm{E}}^2$ , but the Earth radiates heat from all points on its surface with an area of $4\pi R_{\mathrm{E}}^2$ , so:

\pi R_{\mathrm{E}}^2.\,S_0 = 4\pi R_{\mathrm{E}}^2\sigma T_{\mathrm{E}}^4\ \to\ T_{\mathrm{E}} = \left(\frac{S_0}{4\sigma}\right)^{\frac{1}{4}} = 278\ \mathrm{K}

A-2 (0.6 pt)

$T_{\mathrm{E}} = \left(\dfrac{S_0}{4\sigma}\right)^{\frac{1}{4}} = \sqrt{\dfrac{R_{\mathrm{S}}}{2d}}\,T_{\mathrm{S}}$ , Numerical value of $T_{\mathrm{E}} = 278\ \mathrm{K}$

A-3. The radiation is maximum at the wavelength for which the derivative of $u$ with respect to $\lambda$ is zero:

\frac{du}{d\lambda} = \frac{2\pi h c^2}{\lambda^6}.\frac{1}{exp(\frac{hc}{\lambda k_{\mathrm{B}}T}) - 1}.\left[-5 + \frac{hc}{\lambda k_{\mathrm{B}}T}\,\frac{exp(\frac{hc}{\lambda k_{\mathrm{B}}T})}{exp(\frac{hc}{\lambda k_{\mathrm{B}}T}) - 1}\right]

\frac{du}{d\lambda}\Big|_{\lambda = \lambda_{\mathrm{m}}} = 0 \quad\Rightarrow\quad \frac{hc}{\lambda_{\mathrm{m}} k_B T}\,\frac{exp\left(\frac{hc}{\lambda_{\mathrm{m}} k_{\mathrm{B}}T}\right)}{exp\left(\frac{hc}{\lambda_{\mathrm{m}} k_{\mathrm{B}}T}\right) - 1} = 5

Defining $x_{\mathrm{m}} \equiv \frac{hc}{\lambda_{\mathrm{m}} k_{\mathrm{B}}T}$ we obtain the following transcendental equation:

5(1 - e^{-x_{\mathrm{m}}}) - x_{\mathrm{m}} = 0

A-3 (0.4 pt)

$f(x) = 5(1 - e^{-x}) - x$

A-4. The first guess is $x_{\mathrm{m}}^{(1)} = 5$ . Substituting repeatedly for $x_{\mathrm{m}}$ we can continue as follows:

x_{\mathrm{m}}^{(2)} = 5(1 - e^{-5}) = 4.97

x_{\mathrm{m}}^{(3)} = 5(1 - e^{-4.97}) = 4.97

Further iterations do not change the value of $x_{\mathrm{m}}$ to three significant figures, so:

\lambda_{\mathrm{m}} T = \frac{hc}{x_{\mathrm{m}} k_{\mathrm{B}}} = b = 1240\ \mathrm{eV\cdot nm}\times\frac{1}{4.97\times 8.62\times 10^{-5}\ \mathrm{eVK^{-1}}} = 2.89\times 10^{6}\ \mathrm{nm\cdot K}

A-4 (0.4 pt)

$x_{\mathrm{m}} = \{4.96, 4.97\}$ , Numerical value of $b = [2.89, 2.90]\times 10^{6}\ \mathrm{nm\cdot K}$

A-5. Using Wien’s displacement law and the constant $b$ obtained in the previous part, we can calculate the wavelength at which the radiation from the Sun and the Earth reaches its maximum:

\lambda_{\max}^{\mathrm{Sun}} = \frac{b}{T_{\mathrm{S}}} = \frac{2.89\times 10^{6}\ \mathrm{nm\cdot K}}{5.77\times 10^{3}\ \mathrm{K}} = [5.01, 5.02]\times 10^{2}\ \mathrm{nm}

\lambda_{\max}^{\mathrm{Earth}} = \frac{b}{T_{\mathrm{E}}} = \frac{2.89\times 10^{6}\ \mathrm{nm\cdot K}}{278\ \mathrm{K}} = 1.04\times 10^{4}\ \mathrm{nm}

A-5 (0.2 pt)

$\lambda_{\max}^{\mathrm{Sun}} = [5.01, 5.02]\times 10^{2}\ \mathrm{nm}$ , $\lambda_{\max}^{\mathrm{Earth}} = 1.04\times 10^{4}\ \mathrm{nm}$

A-6. From the diagram, it can clearly be seen that $\gamma \tilde{u}_{\mathrm{S}}(\lambda_{\max}^{\mathrm{S}}) = u(\lambda_{\max}^{\mathrm{Earth}}, T_{\mathrm{E}})$ , so we have:

\tilde{u}_{\mathrm{S}}(\lambda_{\max}^{\mathrm{Sun}}) = \left(\frac{R_{\mathrm{S}}}{d}\right)^2 \frac{2\pi h c^2}{\left(\lambda_{\max}^{\mathrm{Sun}}\right)^5}\,\frac{1}{exp\left(\frac{hc}{\lambda_{\max}^{\mathrm{Sun}} k_{\mathrm{B}}T_{\mathrm{S}}}\right) - 1} = \left(\frac{R_{\mathrm{S}}}{d}\right)^2 \frac{2\pi h c^2}{\left(\lambda_{\max}^{\mathrm{Sun}}\right)^5}\,\frac{1}{exp\left(\frac{hc}{k_{\mathrm{B}}b}\right) - 1}

u(\lambda_{\max}^{\mathrm{Earth}}, T_{\mathrm{E}}) = \frac{2\pi h c^2}{\left(\lambda_{\max}^{\mathrm{Earth}}\right)^5}\,\frac{1}{exp\left(\frac{hc}{\lambda_{\max}^{\mathrm{Earth}} k_{\mathrm{B}}T_{\mathrm{E}}}\right) - 1} = \frac{2\pi h c^2}{\left(\lambda_{\max}^{\mathrm{Earth}}\right)^5}\,\frac{1}{exp\left(\frac{hc}{k_{\mathrm{B}}b}\right) - 1}

Dividing these two quantities we’ll find:

\gamma = \left(\frac{d}{R_{\mathrm{S}}}\right)^2 \times \left(\frac{T_{\mathrm{E}}}{T_{\mathrm{S}}}\right)^5 = [1.20, 1.21]\times 10^{-2}

A-6 (0.8 pt)

$\gamma = \left(\dfrac{d}{R_{\mathrm{S}}}\right)^2 \times \left(\dfrac{T_{\mathrm{E}}}{T_{\mathrm{S}}}\right)^5 = \left(\dfrac{d}{R_{\mathrm{S}}}\right)^2 \times \left(\dfrac{\lambda_{\max}^{\mathrm{Sun}}}{\lambda_{\max}^{\mathrm{Earth}}}\right)^5$ , Numerical value of $\gamma = [1.20, 1.21]\times 10^{-2}$

B. The Greenhouse Effect

B-1. Both the Earth and its atmosphere are in thermal equilibrium, so one can write an equation that balances the input and output powers. For the Earth we have:

(\pi R_{\mathrm{E}}^2)(1 - r_{\mathrm{A}})S_0 + (4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{A}}^4 = (4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{E}}^4,

and for the atmosphere:

(4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{E}}^4 = 2(4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{A}}^4.

Note that the coefficient 2 on the right-hand side of the equation is due to the atmosphere radiating heat on both sides (above and below). Eliminating $T_E$ from the two relations we obtain:

T_{\mathrm{A}} = \left(\frac{(1 - r_{\mathrm{A}})\frac{S_0}{4}}{\sigma}\right)^{\frac{1}{4}} = 2.58\times 10^{2}\ \mathrm{K} \quad\Rightarrow\quad T_{\mathrm{E}} = (2T_{\mathrm{A}}^4)^{\frac{1}{4}} = 3.07\times 10^{2}\ \mathrm{K}

B-1 (1.0 pt)

$T_A = \left(\dfrac{(1-r_{\mathrm{A}})\frac{S_0}{4}}{\sigma}\right)^{\frac{1}{4}}$ , Numerical value of $T_A = 2.58\times 10^{2}\ \mathrm{K}$

$T_{\mathrm{E}} = \left(\dfrac{(1-r_{\mathrm{A}})\frac{S_0}{2}}{\sigma}\right)^{\frac{1}{4}}$ , Numerical value of $T_{\mathrm{E}} = 3.07\times 10^{2}\ \mathrm{K}$

B-2. As can be seen in the figure, a fraction $(1 - r_{\mathrm{A}})$ of the solar radiation reaches the Earth’s surface after traversing the atmosphere. A fraction $r_{\mathrm{E}}$ of this light is reflected back and reaches the atmosphere, where a fraction $r_{\mathrm{A}}$ is reflected and returns to the Earth’s surface. This process repeats ad infinitum and the sum of the powers transmitted at all these instances, determines the albedo. Denoting the power returned to space after $n$ reflections by $\tilde{S}_n$ , we’ll have $\tilde{S}_0 = r_{\mathrm{A}}S_0$ and the remaining power i.e. $(1 - r_{\mathrm{A}})S_0$ , reaches the Earth’s surface. From this power, $(1 - r_{\mathrm{A}})r_{\mathrm{E}}S_0$ is reflected, and a fraction $1 - r_{\mathrm{A}}$ of it is transmitted through the atmosphere to the space, hence:

\tilde{S}_1 = (1 - r_{\mathrm{A}})^2 r_{\mathrm{E}}S_0 = \frac{(1 - r_{\mathrm{A}})^2}{r_{\mathrm{A}}}\,r_{\mathrm{E}}\tilde{S}_0

The power that is reflected back to the Earth by the atmosphere after $(n-1)$ reflections is $\tilde{S}_{n-1}\left(\frac{r_{\mathrm{A}}}{1-r_{\mathrm{A}}}\right)$ , of which a fraction $r_{\mathrm{E}}$ is again sent back towards the atmosphere on the $n$ ’th reflection, and the atmosphere allows a fraction $1 - r_{\mathrm{A}}$ of this reflected power to escape into the space, thus:

\tilde{S}_n = \frac{\tilde{S}_{n-1}}{1 - r_{\mathrm{A}}}\,r_{\mathrm{A}}r_{\mathrm{E}}\times(1 - r_{\mathrm{A}}) = r_{\mathrm{A}}r_{\mathrm{E}}\tilde{S}_{n-1} = (r_{\mathrm{A}}r_{\mathrm{E}})^{n-1}\tilde{S}_1

By adding all these terms, one obtains the power returned per unit area from the Earth-atmosphere system:

\tilde{S} = \sum_{n=0}^{\infty}\tilde{S}_n = \tilde{S}_0 + \tilde{S}_1\sum_{n=1}^{\infty}(r_{\mathrm{A}}r_{\mathrm{E}})^{n-1} = r_{\mathrm{A}}S_0 + (1 - r_{\mathrm{A}})^2 r_{\mathrm{E}}S_0\times\frac{1}{1 - r_{\mathrm{A}}r_{\mathrm{E}}} = \left[r_{\mathrm{A}} + \frac{(1 - r_{\mathrm{A}})^2 r_{\mathrm{E}}}{1 - r_{\mathrm{A}}r_{\mathrm{E}}}\right]\times S_0

Dividing by the solar constant we get the value for albedo:

\alpha = \frac{\tilde{S}}{S_0} = r_{\mathrm{A}} + \frac{(1 - r_{\mathrm{A}})^2 r_{\mathrm{E}}}{1 - r_{\mathrm{A}}r_{\mathrm{E}}} = 3.13\times 10^{-1}

B-2 (1.6 pt)

$\alpha = r_{\mathrm{A}} + \dfrac{(1-r_{\mathrm{A}})^2 r_{\mathrm{E}}}{1 - r_{\mathrm{A}}r_{\mathrm{E}}}$ , Numerical value of $\alpha = 3.13\times 10^{-1}$

B-3. Again, thermal equilibrium requires the input and output powers to be equal both for the Earth and for the atmosphere, the only difference being that the Earth absorbs now a fraction $1-\alpha$ of the Sun’s radiation. Thus, for Earth we have:

(4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{A}}^4 + (\pi R_{\mathrm{E}}^2)(1 - \alpha)S_0 = (4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{E}}^4,

and for the atmosphere:

(4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{E}}^4 = 2(4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{A}}^4

T_{\mathrm{E}} = \left[\frac{(1 - \alpha)}{2\sigma(2 - \epsilon)}S_0\right]^{\frac{1}{4}}\ ,\qquad T_{\mathrm{A}} = \left(\frac{T_{\mathrm{E}}^4}{2}\right)^{\frac{1}{4}}

\epsilon = \frac{\left[\sigma T_{\mathrm{E}}^4 - \frac{(1-\alpha)}{4}S_0\right]}{\sigma T_{\mathrm{A}}^4} = 2\,\frac{\left[\sigma T_{\mathrm{E}}^4 - \frac{(1-\alpha)}{4}S_0\right]}{\sigma T_{\mathrm{E}}^4} = [8.07, 8.11]\times 10^{-1}

B-3 (1.0 pt)

$T_{\mathrm{E}} = \left[\frac{(1-\alpha)}{2\sigma(2-\epsilon)}S_0\right]^{\frac{1}{4}}$ , Numerical value of $\epsilon = [8.07, 8.11]\times 10^{-1}$

B-4.

\frac{dT_{\mathrm{E}}}{d\epsilon} = \frac{1}{4}\left[\frac{(1 - \alpha)S_0}{2\sigma(2 - \epsilon)}\right]^{\frac{1}{4}}\frac{1}{(2 - \epsilon)}

dT_{\mathrm{E}} = \frac{dT_{\mathrm{E}}}{d\epsilon}\,\epsilon\,\frac{d\epsilon}{\epsilon} = \left[\frac{4\sigma T_{\mathrm{E}}^4}{(1 - \alpha)S_0} - 1\right]\frac{T_{\mathrm{E}}}{4}\times 0.01 = [4.87, 4.92]\times 10^{-1}

B-4 (0.8 pt)

$\dfrac{dT_{\mathrm{E}}}{d\epsilon} = \dfrac{1}{4}\left[\dfrac{(1-\alpha)S_0}{2\sigma(2-\epsilon)}\right]^{\frac{1}{4}}\dfrac{1}{(2-\epsilon)}$ , Numerical value of $\delta T_{\mathrm{E}} = [4.87, 4.92]\times 10^{-1}\ \mathrm{K}$

B-5. The equations for thermal equilibrium are similar to those for Part B.3, only a non-radiative thermal current needs to be added. For the Earth:

(\pi R_{\mathrm{E}}^2)(1 - \alpha)S_0 + (4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{A}}^4 = (4\pi R_{\mathrm{E}}^2)\sigma T_{\mathrm{E}}^4 + (4\pi R_{\mathrm{E}}^2)k(T_{\mathrm{E}} - T_{\mathrm{A}}),

and for the atmosphere:

(4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{E}}^4 + (4\pi R_{\mathrm{E}}^2)k(T_{\mathrm{E}} - T_{\mathrm{A}}) = 2(4\pi R_{\mathrm{E}}^2)\epsilon\sigma T_{\mathrm{A}}^4.

After completing the calculations, we will have:

\epsilon = \frac{\sigma T_{\mathrm{E}}^4 - (1 - \alpha)\,\dfrac{S_0}{4}}{\sigma\,(T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4)} = [8.47, 8.52]\times 10^{-1}

k = \frac{\epsilon\sigma(2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4)}{T_{\mathrm{E}} - T_{\mathrm{A}}} = \frac{(2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4)\times\left[\sigma T_{\mathrm{E}}^4 - (1 - \alpha)\dfrac{S_0}{4}\right]}{(T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4)\times(T_{\mathrm{E}} - T_{\mathrm{A}})} = [3.57, 3.66]\times 10^{-1}\ \mathrm{W/m^2 K}

B-5 (1.6 pt)

$\epsilon = \dfrac{\sigma T_{\mathrm{E}}^4 - (1-\alpha)\frac{S_0}{4}}{\sigma(T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4)}$ , Numerical value of $\epsilon = [8.47, 8.52]\times 10^{-1}$

$k = \dfrac{(2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4)\times\left[\sigma T_{\mathrm{E}}^4 - (1-\alpha)\frac{S_0}{4}\right]}{(T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4)\times(T_{\mathrm{E}} - T_{\mathrm{A}})}$ , Numerical value of $k = [3.57, 3.66]\times 10^{-1}\ \mathrm{W/m^2K}$

B-6. In order to find the change in the temperatures of the Earth and the atmosphere in terms of $\epsilon$ and $k$ , we take the logarithm of both sides of the relations before taking the derivative:

ln\,\epsilon = ln\left[\sigma T_{\mathrm{E}}^4 - (1 - \alpha)\frac{S_0}{4}\right] - ln\,\sigma - ln\,(T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4)

ln\,k = ln\,\epsilon + ln\,\sigma + ln(2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4) - ln(T_{\mathrm{E}} - T_{\mathrm{A}})

\frac{1}{\epsilon} = \frac{4\sigma T_{\mathrm{E}}^3\,\dfrac{dT_{\mathrm{E}}}{d\epsilon}}{\sigma T_{\mathrm{E}}^4 - (1 - \alpha)\dfrac{S_0}{4}} - \frac{4T_{\mathrm{E}}^3\,\dfrac{dT_{\mathrm{E}}}{d\epsilon} - 4T_{\mathrm{A}}^3\,\dfrac{dT_{\mathrm{A}}}{d\epsilon}}{T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4}

0 = \frac{1}{\epsilon} + \frac{8T_{\mathrm{A}}^3\,\dfrac{dT_{\mathrm{A}}}{d\epsilon} - 4T_{\mathrm{E}}^3\,\dfrac{dT_{\mathrm{E}}}{d\epsilon}}{2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4} - \frac{\dfrac{dT_{\mathrm{E}}}{d\epsilon} - \dfrac{dT_{\mathrm{A}}}{d\epsilon}}{T_{\mathrm{E}} - T_{\mathrm{A}}}

\epsilon\left[\frac{1}{T_{\mathrm{E}} - T_{\mathrm{A}}} + \frac{4T_{\mathrm{E}}^3}{2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4}\right]\frac{dT_{\mathrm{E}}}{d\epsilon} = 1 + \epsilon\left[\frac{8T_{\mathrm{A}}^3}{2T_{\mathrm{A}}^4 - T_{\mathrm{E}}^4} + \frac{1}{T_{\mathrm{E}} - T_{\mathrm{A}}}\right]\frac{dT_{\mathrm{A}}}{d\epsilon}

1 + \epsilon\left[\frac{4T_{\mathrm{E}}^3}{T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4} - \frac{4\sigma T_{\mathrm{E}}^3}{\sigma T_{\mathrm{E}}^4 - (1 - \alpha)\dfrac{S_0}{4}}\right]\frac{dT_{\mathrm{E}}}{d\epsilon} = \frac{4T_{\mathrm{A}}^3}{T_{\mathrm{E}}^4 - T_{\mathrm{A}}^4}\,\epsilon\,\frac{dT_{\mathrm{A}}}{d\epsilon}

Solving this set of linear equations and substituting $\epsilon$ in B-5, we find:

\frac{dT_\mathrm{E}}{d\epsilon} = \frac{\left[\dfrac{\sigma\left(T_\mathrm{E}^4 - T_\mathrm{A}^4\right)}{\sigma T_\mathrm{E}^4 - (1-\alpha)\dfrac{S_0}{4}}\right]\left[1 + \left(\dfrac{T_\mathrm{E}^4 - T_\mathrm{A}^4}{4T_\mathrm{A}^3}\right)\left[\dfrac{8T_\mathrm{A}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4} + \dfrac{1}{T_\mathrm{E} - T_\mathrm{A}}\right]\right]}{\left[\dfrac{1}{T_\mathrm{E} - T_\mathrm{A}} + \dfrac{4T_\mathrm{E}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4}\right] - \left(\dfrac{\sigma T_\mathrm{A}^4 - (1-\alpha)\dfrac{S_0}{4}}{\sigma T_\mathrm{E}^4 - (1-\alpha)\dfrac{S_0}{4}}\right)\left(\dfrac{T_\mathrm{E}}{T_\mathrm{A}}\right)^3\left[\dfrac{8T_\mathrm{A}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4} + \dfrac{1}{T_\mathrm{E} - T_\mathrm{A}}\right]}

\epsilon\frac{dT_\mathrm{E}}{d\epsilon} = \frac{1 + \left(\dfrac{T_\mathrm{E}^4 - T_\mathrm{A}^4}{4T_\mathrm{A}^3}\right)\left[\dfrac{8T_\mathrm{A}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4} + \dfrac{1}{T_\mathrm{E} - T_\mathrm{A}}\right]}{\left[\dfrac{1}{T_\mathrm{E} - T_\mathrm{A}} + \dfrac{4T_\mathrm{E}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4}\right] - \left(\dfrac{\sigma T_\mathrm{A}^4 - (1-\alpha)\dfrac{S_0}{4}}{\sigma T_\mathrm{E}^4 - (1-\alpha)\dfrac{S_0}{4}}\right)\left(\dfrac{T_\mathrm{E}}{T_\mathrm{A}}\right)^3\left[\dfrac{8T_\mathrm{A}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4} + \dfrac{1}{T_\mathrm{E} - T_\mathrm{A}}\right]}

dT_\mathrm{E} = \epsilon\,\frac{dT_\mathrm{E}}{d\epsilon}\,\frac{d\epsilon}{\epsilon} = [5.21\,, 5.28] \times 10^{-1}\ \mathrm{K}

B-6 (1.0pt)

(a)

\begin{cases} \epsilon\left[\dfrac{1}{T_\mathrm{E} - T_\mathrm{A}} + \dfrac{4T_\mathrm{E}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4}\right]\dfrac{dT_\mathrm{E}}{d\epsilon} = 1 + \epsilon\left[\dfrac{8T_\mathrm{A}^3}{2T_\mathrm{A}^4 - T_\mathrm{E}^4} + \dfrac{1}{T_\mathrm{E} - T_\mathrm{A}}\right]\dfrac{dT_\mathrm{A}}{d\epsilon} \\[6pt] 1 + \epsilon\left[\dfrac{4T_\mathrm{E}^3}{T_\mathrm{E}^4 - T_\mathrm{A}^4} - \dfrac{4\sigma T_\mathrm{E}^3}{\sigma T_\mathrm{E}^4 - (1-\alpha)\dfrac{S_0}{4}}\right]\dfrac{dT_\mathrm{E}}{d\epsilon} = \dfrac{4T_\mathrm{A}^3}{T_\mathrm{E}^4 - T_\mathrm{A}^4}\,\epsilon\,\dfrac{dT_\mathrm{A}}{d\epsilon} \end{cases}

(b) $\delta T_\mathrm{E} = [5.21\,, 5.28] \times 10^{-1}\ \mathrm{K}$

Оригинал в Архива: IPhO_2024_Q1.pdf · официални решения: IPhO_2024_S1.pdf