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Условие Fourier's law of thermal conductivity Electrothermal analogy The first estimate of the Earth age Отговори Решение

IZhO 2025, Теория — Задача 2. First estimate of the age of the Earth

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XXI International Zhautykov Olympiad/Theoretical Competition · 14 януари 2025 г. · 10 т.

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The time available for the theoretical competition is 4 hours. There are three questions.
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Условие

Fourier's law of thermal conductivity

Thermal conductivity characterizes the ability of a body to transfer thermal energy from one point to another if there is a temperature difference between them. Let us consider a flat homogeneous plate with a cross-sectional area $S$ and a thickness $dx$ , whose surfaces have the temperature difference $dT$ . The phenomenon of thermal conductivity in a one-dimensional stationary case is described by Fourier's law, which has the following form:

dQ = -\kappa\frac{dT}{dx}Sdt,

where $dQ$ is the amount of heat transferred during time $dt$ through the plate surface in the direction of the normal coinciding with the direction of decreasing temperature. The constant $\kappa$ is called the thermal conductivity coefficient and depends on the plate material, and the value $j_Q = \frac{1}{S}\frac{dQ}{dt}$ refers to the modulus of the heat flux, which is a vector directed towards the heat transfer.

2.1 The left end of the rod, enclosed in a heat-insulating shell, is maintained at a temperature of $t_1 = 25°\mathrm{C}$ , whereas its right end is kept at a temperature of $t_2 = 100°\mathrm{C}$ . The rod itself consists of two parts of the same length, but different thermal conductivities. The thermal conductivity coefficient of the left part is twice as high as the thermal conductivity coefficient of the right part. Calculate the temperature $t$ of the contact surface of the rod parts. [1 т.]

2.2 A rod of length $l$ with a heat-insulated side surface consists of a material whose thermal conductivity coefficient varies along according to the law $\kappa = a/T^2$ , where $a$ is a certain constant. The ends of the rod are maintained at temperatures $T_1$ and $T_2$ . Determine the dependence of temperature $T(x)$ on the coordinate $x$ , which is the distance from the rod end with temperature $T_1$ . [1 т.]

2.3 Estimate the time it takes for an ice crust of thickness $h = 4.00$ cm to form on the surface of a lake at an ambient air temperature of $t = -20.0°\mathrm{C}$ . The thermal conductivity coefficient of ice is $\kappa = 2.20\ \mathrm{W/(m\cdot K)}$ , its density is $\rho = 900\ \mathrm{kg/m^3}$ , and the specific melting heat is $\lambda = 3.33\cdot 10^5\ \mathrm{J/kg}$ . [1 т.]

Electrothermal analogy

Electrothermal analogy is a method for calculating thermal systems based on replacing thermal quantities with their electrical analogues. The method is based on the identity of the mathematical apparatus of thermal physics and electrical engineering: the propagation of heat and electric current are described by the same differential equations, while measuring the electrical characteristics of real objects is much simpler.

Wire-frame cube; vertex A at the front bottom left, C at the front bottom right (adjacent to A), B at the back top right, diagonally opposite to A. — Figure 2.1

Circle (sphere) with centre O; point A at the top and point B at the right on the sphere, joined to O by dashed radii; point C lies on a dashed line from O towards the viewer, perpendicular to OA and OB. — Figure 2.2

2.4 Find Below a table of correspondence between thermal and electrical quantities. Thermal quantities are written out explicitly, and you need to transfer the table to your solution and write the corresponding electrical quantities in front of them: electric current density $j_E$ , specific conductivity $\sigma$ , electric charge $q$ , electric field potential $\varphi$ .

Thermal quantity	Electrical quantity
temperature $T$
amount of heat $Q$
heat flux $j_Q$
thermal conductivity coefficient $\kappa$

[0,8 т.]

2.5 A cube consists of identical thin rods with thermally insulated side surfaces. Two vertices of the cube $A$ and $B$ are maintained at constant temperatures $t_A = 20°\mathrm{C}$ and $t_B = 120°\mathrm{C}$ respectively. Determine the temperature $t_C$ at vertex $C$ , see Figure 2.1 below. [1 т.]

2.6 A thin heat-conducting sphere of radius $R$ is made of a material with the wall thickness $h \ll R$ . Thermal power $P$ is supplied to point $A$ of the sphere, and the same thermal power is removed at point $B$ of the sphere. Neglecting heat losses, determine the modulus of the heat flux at point $C$ of the sphere, see Figure 2.2 below ( $O$ is the sphere center, $AO \perp OB, OC \perp OA, OC \perp OB$ ). [1 т.]

The first estimate of the Earth age

In the 18th century, there lived a famous naturalist, Georges-Louis Leclerc, Count de Buffon (1707-1788), who was a member of the Paris Academy of Sciences, the French Academy, a member of the Royal Society of London, and also a foreign honorary member of the St. Petersburg Academy of Sciences. He published his works in a multi-volume work, Natural History, in which he attempted to systematize the knowledge gained to that time in a variety of fields of science, and also expressed a number of advanced ideas. His interests included geology, and he was the first to formulate the problem of determining the age of the Earth. In particular, Leclerc developed a hypothesis about the origin of the planets through the collision of celestial bodies, which should have led to their heating and subsequent cooling. It was through the process of temperature change over time that the first estimate of the Earth age in history was obtained.

Of course, Leclerc did not know the laws of physics in their modern form and did not have a sufficient command of the mathematical apparatus, but he had a keen physical sense.

2.7 Let a teapot have a spherical shape with a radius $r_0 = 15.0$ cm and after heating water in it to the boiling point, it cools down to the room temperature in a time of $\tau_0 = 6.00$ h. Let us imagine a hypothetical situation: the teapot is proportionally increased to the size of the Earth, the radius of which is $R_E = 6400$ km. Estimate the cooling time τ of such a hypothetical "teapot" under the same external conditions. [1 т.]

Unlike Count de Buffon, we are familiar with the modern physics laws, so let us consider the simplest model of the Earth, whose radius is equal to $R_E = 6400$ km. Inside the Earth there is a core with a radius of $R_0 = 3500$ km, whose substance perfectly conducts heat ( $\kappa = \infty$ ) and has a specific heat capacity of $c = 500$ J / (kg · K) at a density of $\rho = 5.46\ \mathrm{g/cm^3}$ . Outside the core, right up to the Earth's surface, is the mantle, whose substance has zero specific heat capacity and a thermal conductivity coefficient of $\kappa = 50.3\ \mathrm{W/(m\cdot K)}$ . Neglect the influence of the Earth's crust, and assume that the Sun has no thermal effect on the Earth, which can be considered an absolutely black body being in a vacuum.

2.8 At the present moment in time, the Earth's surface temperature is known to be $T = 300$ K. Calculate the temperature $T_0$ of the Earth's core within the framework of the presented model. [1 т.]

2.9 Calculate the temperature of the mantle $T_H$ at a depth of $H = 2.00$ km from the Earth's surface. [1 т.]

2.10 Calculate how much the temperature $\Delta T$ of the Earth's surface differs from what it was $\tau = 1.00\cdot 10^6$ years ago. [1,2 т.]

Отговори

Решение

Fourier's Law of Thermal Conductivity

2.1 According to Fourier's law, the heat flux in the left rod is equal to

j_{Q1} = -2\kappa\frac{T - T_1}{l}, \qquad (1)

and the heat flux in the right rod is found as

j_{Q2} = -\kappa\frac{T_2 - T}{l}. \qquad (2)

In steady state the heat flux should be the same in both rods

j_{Q1} = j_{Q2}, \qquad (3)

from which we find

T = \frac{2T_1 + T_2}{3} = 50°\mathrm{C}. \qquad (4)

2.2 The heat flux along the rod must remain constant, which yields the differential equation

\frac{1}{T^2}\frac{dT}{dx} = b = const, \qquad (5)

whose solution has the form

-\frac{1}{T} = bx + C, \qquad (6)

where $C$ is a constant of integration.

Taking into account the boundary conditions

T(0) = T_1, \qquad (7)

T(l) = T_2, \qquad (8)

we obtain the following final expression for the temperature dependence

T(x) = \frac{T_1}{1 - \left(1 - \frac{T_1}{T_2}\right)\frac{x}{l}}. \qquad (9)

2.3 When ice forms near its contact with the lake water, the temperature is $0°\mathrm{C}$ , which causes a temperature difference of $\Delta t = |t|$ . Let us consider a portion of ice surface of the area $S$ , assuming its thickness to be $h$ . Then the amount of heat released into the environment at the point of contact during time $d\tau$ is equal to

\delta Q_1 = \kappa\frac{\Delta t}{h}Sd\tau. \qquad (10)

During the process of water crystallization at the point of contact with ice, the amount of heat released is found as

\delta Q_2 = \lambda\rho Sdh. \qquad (11)

By equating these amounts of heat $\delta Q_1 = \delta Q_2$ , we obtain the formula

d\tau = \frac{\lambda\rho}{\kappa|t|}hdh, \qquad (12)

which after integration gives the final answer

\tau = \frac{\lambda\rho h^2}{2\kappa|t|} = 5.45\cdot 10^3\ s. \qquad (13)

Electrothermal analogy

2.4 Ohm's law for the electric current density has the form $j_E = \sigma E = -\sigma\frac{d\varphi}{dx}$ , and the electric current density itself is associated with the flow of electric charge, so the table of correspondence between thermal and electrical quantities takes the form shown below.

Thermal value	Electrical quantity
temperature $T$	electric field potential $\varphi$
amount of heat $Q$	electric charge $q$
heat flux $j_Q$	electric current density $j_E$
thermal conductivity coefficient $\kappa$	specific conductivity $\sigma$

2.5 It follows from the electrothermal analogy that the cube can be represented as an equivalent electrical circuit, shown in the figure below.

Equivalent electrical circuit of the cube: node T_A, three equal resistors in parallel to node T_C, then six equal resistors in parallel, then three equal resistors in parallel to node T_B.

Since the temperature in the electrothermal analogy is a potential, and the resistances of all ribs are the same, we obtain the relation

\frac{T_B - T_A}{\frac{5}{6}} = \frac{T_C - T_A}{\frac{1}{3}}, \qquad (14)

whence

T_C = \frac{1}{5}(3T_A + 2T_B) = 60°\mathrm{C}. \qquad (15)

2.6 Let us add another heat source to the system at point $A'$ , which is located diametrically opposite to point $B$ , and also another heat sink at point $B'$ , diametrically opposite to point $A$ , see the figure below, which shows the side view.

Side view of the sphere: A at the top, B′ at the bottom, A′ on the left, B on the right, joined by dashed diameters through the centre C; at C two perpendicular arrows j_Q0 (towards B and towards B′) and their resultant j_Q along the diagonal.

Obviously, each of the pairs $AB'$ and $A'B$ makes the same contribution to the heat flux at point $C$ , equal in magnitude

j_{Q0} = \frac{P}{2\pi Rh}. \qquad (16)

On the other hand, the total heat flux at point $C$ is equal to the vector sum of two mutually perpendicular fluxes $j_{Q0}$ , and it, as follows from the symmetry, is exactly twice the desired one, therefore

j_Q = \frac{\sqrt{2}}{2}j_{Q0} = \frac{P}{2\sqrt{2}\pi Rh}. \qquad (17)

The first estimate of the Earth age

2.7 The amount of heat $Q$ stored in a kettle is proportional to the cube of its size $R$ , that is,

Q \sim R^3. \qquad (18)

and the power $P$ lost by it due to heat transfer is proportional to its surface area

P \sim R^2. \qquad (19)

Therefore, the required time is proportional to the size of the teapot and is

\tau = \frac{R_E}{r_0}\tau_0 = 2.56\cdot 10^8\ h = 2.92\cdot 10^4\ years. \qquad (20)

2.8 Since the thermal conductivity of the core is infinite, its temperature $T_0$ is the same everywhere, including at the boundary with the mantle. According to the electrothermal analogy, the temperature difference $T_0 - T$ is the potential difference (voltage), and the radiation power from the Earth's surface is determined by the Stefan-Boltzmann law

P = 4\pi R_E^2\sigma T^4 \qquad (21)

and is similar to electric current, so there is a linear relationship between them

T_0 - T = PR_T. \qquad (22)

where the quantity $R_T$ represents the so-called thermal resistance.

Using the same electrothermal analogy, thermal resistance is written as

R_T = \frac{1}{\kappa}\int_{R_0}^{R_E}\frac{dR}{4\pi R^2} = \frac{1}{4\pi\kappa}\left(\frac{1}{R_0} - \frac{1}{R_E}\right), \qquad (23)

from which we finally obtain the temperature of the Earth's core

T_0 = T + \frac{\sigma T^4 R_E(R_E - R_0)}{\kappa R_0} = 4.84\cdot 10^7\mathrm{K}. \qquad (24)

2.9 Since the transferred heat (current strength) is the same, we obtain the proportion for the temperature

\frac{T_0 - T}{R_T} = \frac{T_H - T}{R_H}, \qquad (25)

where the thermal resistance of a spherical layer of thickness $H$ is given by the expression

R_H = \frac{H}{4\pi\kappa R_E^2}. \qquad (26)

Thus, we obtain the temperature at depth $H$ as

T_H = T + (T_0 - T)\frac{HR_0}{R_E(R_E - R_0)} = 1.86\cdot 10^4\mathrm{K}. \qquad (27)

The actual core temperature is, of course, much lower, since the thermal balance with the Sun must be taken into account.

2.10 The change over time in the amount of heat stored in the core is equal to

\frac{dQ}{dt} = cm\frac{dT_0}{dt}, \qquad (28)

where the mass of the core is obtained as

m = \frac{4}{3}\pi\rho R_0^3. \qquad (29)

According to the law of energy conservation we have

\frac{dQ}{dt} = -P, \qquad (28)

where $P$ is given by expression (21).

The core temperature $T_0$ is given by formula (24), in which the first term can be neglected ( $T_0 \gg T$ ), so we finally obtain

\frac{dT}{T} = -\frac{\kappa R_E}{c\rho R_0^2(R_E - R_0)}d\tau. \qquad (29)

By integrating and expanding the exponential, we find

\Delta T = \frac{\kappa R_E T\tau}{c\rho R_0^2(R_E - R_0)} = 3.14\cdot 10^{-2}\mathrm{K}. \qquad (30)

	Content	Points
2.1	Formula (1): $j_{Q1} = -2\kappa\frac{T - T_1}{l}$	0.2	1.0
	Formula (2): $j_{Q2} = -\kappa\frac{T_2 - T}{l}$	0.2
	Formula (3): $j_{Q1} = j_{Q2}$	0.2
	Formula (4): $T = \frac{2T_1 + T_2}{3}$	0.2
	Numerical value in formula (4): $T = 50°\mathrm{C}$	0.2
2.2	Formula (5): $\frac{1}{T^2}\frac{dT}{dx} = b = const$	0.2	1.0
	Formula (6): $-\frac{1}{T} = bx + C$	0.2
	Formula (7): $T(0) = T_1$	0.2
	Formula (8): $T(l) = T_2$	0.2
	Formula (9): $T(x) = \frac{T_1}{1 - \left(1 - \frac{T_1}{T_2}\right)\frac{x}{l}}$	0.2
2.3	Formula (10): $\delta Q_1 = \kappa\frac{\Delta t}{h}Sd\tau$	0.2	1.0
	Formula (11): $\delta Q_2 = \lambda\rho Sdh$	0.2
	Formula (12): $d\tau = \frac{\lambda\rho}{\kappa\lvert t\rvert}hdh$	0.2
	Formula (13): $\tau = \frac{\lambda\rho h^2}{2\kappa\lvert t\rvert}$	0.2
	Numerical value in formula (13): $\tau = 5.45\cdot 10^3\ s$	0.2
2.4	0.2 for each correct value in the table	4x0.2	0.8
2.5	Correct equivalent circuit	0.4	1.0
	Formula (14): $\frac{T_B - T_A}{\frac{5}{6}} = \frac{T_C - T_A}{\frac{1}{3}}$	0.2
	Formula (15): $T_C = \frac{1}{5}(3T_A + 2T_B)$	0.2
	Numerical value in formula (15): $T_C = 60°\mathrm{C}$	0.2
2.6	Formula (16): $j_{Q0} = \frac{P}{2\pi Rh}$	0.5	1.0
	Formula (17): $j_Q = \frac{P}{2\sqrt{2}\pi Rh}$	0.5
2.7	Formula (18): $Q \sim R^3$	0.2	1.0
	Formula (19): $P \sim R^2$	0.2
	Formula (20 $\tau = \frac{R_E}{r_0}\tau_0$	0.3
	Numerical value in formula (20): $\tau = 2.56\cdot 10^8\ h = 2.92\cdot 10^4\ years$	0.3
2.8	Formula (21): $P = 4\pi R_E^2\sigma T^4$	0.2	1.0
	Formula (22): $T_0 - T = PR_T$	0.2
	Formula (23): $R_T = \frac{1}{4\pi\kappa}\left(\frac{1}{R_0} - \frac{1}{R_E}\right)$	0.2
	Formula (24): $T_0 = T + \frac{\sigma T^4 R_E(R_E - R_0)}{\kappa R_0}$	0.2
	Numerical value in formula (24): $T_0 = 4.84\cdot 10^7\mathrm{K}$	0.2
2.9	Formula (25): $\frac{T_0 - T}{R_T} = \frac{T_H - T}{R_H}$	0.3	1.0
	Formula (26): $R_H = \frac{H}{4\pi\kappa R_E^2}$	0.2
	Formula (27): $T_H = T + (T_0 - T)\frac{HR_0}{R_E(R_E - R_0)}$	0.3
	Numerical value in formula (27): $T_H = 1.86\cdot 10^4\mathrm{K}$	0.2
2.10	Formula (28): $\frac{dQ}{dt} = cm\frac{dT_0}{dt}$	0.2	1.2
	Formula (29): $m = \frac{4}{3}\pi\rho R_0^3$	0.2
	Formula (30): $\frac{dQ}{dt} = -P$	0.2
	Formula (31): $\frac{dT}{T} = -\frac{\kappa R_E}{c\rho R_0^2(R_E - R_0)}d\tau$	0.2
	Formula (32): $\Delta T = \frac{\kappa R_E T\tau}{c\rho R_0^2(R_E - R_0)}$	0.2
	Numerical value in formula (32): $\Delta T = 3.14\cdot 10^{-2}\mathrm{K}$	0.2
Total			10.0

Mathematical hint for the theoretical competition

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;

$\int \frac{dx}{x} = \ln|x| + C$ , where $C$ stands for an arbitrary constant;

$\int \frac{dx}{(a^2+x^2)^{3/2}} = \frac{x}{a^2\sqrt{a^2+x^2}} + C$ , where a – constant, $C$ – arbitrary constant;

$(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2$ , for $|x| \ll 1$ and any value of $\gamma$ ;

$\tan x \approx \sin x \approx x$ , for $|x| \ll 1$ ;

$\ln(1+x) \approx x$ , for $|x| \ll 1$ .

List of physical constants

Speed of light in vacuum $c = 3.00 \cdot 10^{8}$ m/s;

Reduced Planck constant $\hbar = 1.05 \cdot 10^{-34}$ J·s;

Magnetic constant $\mu_0 = 4\pi \cdot 10^{-7}$ H·m;

Electric constant $\varepsilon_0 = 8.85 \cdot 10^{-12}$ F/m;

Stefan-Boltzmann constant $\sigma = 5.67 \cdot 10^{-8}\ W/(m^2 \cdot K^4)$ ;

Elementary charge $e = 1.60 \cdot 10^{-19}$ C;

Electron mass $m_e = 9.11 \cdot 10^{-31}$ kg.

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