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Derive an expression for the temperature at the center of a planet of radius \(a\) with uniform density \(\rho\) and internal heat generation \(H\). Heat transfer in the planet is by conduction only in the lithosphere, which extends from \(r=b\) to \(r=a\). For \(0 \leq r \leq b\) heat transfer is by convection, which maintains the temperature gradient \(d T / d r\) constant at the adiabatic value \(-\Gamma\). The surface temperature is \(T_{0}\). To solve for \(T(r)\), you need to assume that \(T\) and the heat flux are continuous at \(r=b\).

Short Answer

Expert verified
Temperature at the center, \(T_c\), requires using boundary conditions and solving for the derived equations.

Step by step solution

01

Understand the Regions of Heat Transfer

In this problem, there are two regions to consider: the convective region from the center to radius \(b\), and the conductive lithosphere from \(b\) to \(a\). We need to develop temperature expressions for each region.
02

Solve for Temperature in the Convective Region (0 to b)

In the convective region, the temperature gradient \(\frac{dT}{dr}\) is constant and equal to \(-\Gamma\). Therefore, the temperature distribution can be expressed as:\[T(r) = T_c + \Gamma \cdot r\]where \(T_c\) is the temperature at the center of the planet (\(r=0\)).
03

Boundary Condition at b: Continuity of Temperature

At \(r = b\), the temperature must be continuous. Thus, the temperature at the boundary from the convective region must equal that of the conductive region. We call this temperature \(T(b)\). From Step 2, the temperature at \(r = b\) is:\[T(b) = T_c + \Gamma \cdot b\]
04

Solve for Heat Flux and Temperature in the Conductive Region (b to a)

In the conductive region, the heat equation describes how temperature changes due to conduction:\[-q = k \frac{dT}{dr}\]where \(q\) is the heat flux and \(k\) is the thermal conductivity.The general form for the temperature is:\[T(r) = T_s - \frac{H}{6k} (r^2 - a^2)\]and heat flux can be calculated using continuity conditions for flux at the boundary.
05

Determine Surface Condition at a

At \(r = a\), the temperature is equal to the surface temperature \(T_0\). Using the temperature expression from Step 4 for \(r = a\), substitute \(T(a) = T_0\) to solve for constants.\[T_0 = T_s - \frac{H}{6k}(a^2 - a^2)\]After simplification, match constants with boundary conditions.
06

Equation for Central Temperature T_c

We find \(T_c\) by solving the system from boundary conditions and simplified equations:From the continuity of temperature and flux at \(r = b\), solve for \(T_c\).The continuity condition:\[T_c + \Gamma \cdot b = T_s - \frac{H}{6k} (b^2 - a^2)\]Solve this equation to find \(T_c\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convective Heat Transfer
Convective heat transfer occurs in the region ranging from the center of the planet out to radius \( b \). This type of heat transfer involves the movement of fluid-like material within the planet, enabling the transfer of heat through the motion of the material itself. Within this region, the temperature gradient \( \frac{dT}{dr} \) is maintained constant due to convective action.

This means that at any point in this region, the rate of temperature change across distance is uniform and equal to \(-\Gamma\). The constant negative gradient \(-\Gamma\) implies that the temperature decreases at a steady rate as one moves from the center towards radius \( b \).
  • The temperature expression is given by:\[ T(r) = T_c + \Gamma \cdot r \]where \( T_c \) is the central temperature.
  • This formula shows how temperature is projected linearly from the center to radius \( b \).
Convective heat transfer is essential for maintaining the planet's inner temperature without it reaching extreme temperatures rapidly, thanks to the redistribution of heat through the circulating material.
Conductive Heat Transfer
Conductive heat transfer is predominantly active in the lithosphere, ranging from \( r = b \) to \( r = a \). This process involves the transfer of heat through material without the involvement of mass movement.

Here, heat is conducted due to the temperature differences within the planet's material, moving from higher to lower temperature regions through direct contact. The heat equation helps us describe how temperature varies as it moves through this region. This relation is typically expressed through:
  • The heat conduction equation: \( -q = k \frac{dT}{dr} \)
  • The general temperature equation: \[ T(r) = T_s - \frac{H}{6k} (r^2 - a^2) \]
This relation signifies that the temperature distribution within the conductive region depends on factors such as internal heat generation \( H \), thermal conductivity \( k \), and the square of the radius. The conduction mechanism is crucial in regulating the planet's outer temperature, ensuring a gradient from the deeper to the shallower regions, ultimately reaching the surface temperature \( T_0 \).
This process efficiently enables heat flux through solid layers, maintaining thermal equilibrium over longer periods.
Temperature Distribution
Understanding temperature distribution in a planet provides insights into its thermal dynamics and stabilization. The temperature within the planet is a result of both convective and conductive processes.

Initially, in the convective core region, the temperature is linearly distributed due to constant convective heat transfer:
  • Formula: \( T(r) = T_c + \Gamma \cdot r \)
  • Defined by the adiabatic \(-\Gamma\), maintaining a consistent temperature gradient.
Subsequently, as one transitions into the lithosphere, the temperature distribution follows a more complex pattern due to conduction:
  • Formula: \[ T(r) = T_s - \frac{H}{6k} (r^2 - a^2) \]
  • Reveals how temperature changes with radius squared (due to conduction).
At the interface \( r = b \), where convective and conductive regions meet, temperature and heat flux are continuous, ensuring a smooth transition. This boundary condition helps determine unknowns such as the central temperature \( T_c \).
Ultimately, the planet's surface temperature \( T_0 \) serves as a stabilizer, indicating weather at a larger scale. Temperature distribution helps in understanding geothermal gradients, impacting geological activity and planetary behavior over time.

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Most popular questions from this chapter

The mantle rocks of the asthenosphere from which the lithosphere forms are expected to contain a small amount of magma. If the mass fraction of magma is 0.05 , determine the depth of the lithosphere-asthenosphere boundary for oceanic lithosphere with an age of 60 Ma. Assume \(L=400 \mathrm{~kJ}\) \(\mathrm{kg}^{-1}, c=1 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, T_{m}=1600 \mathrm{~K}, T_{0}=275 \mathrm{~K},\) and \(\kappa=1 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).

Estimate the effects of variations in bottom water temperature on measurements of oceanic heat flow by using the model of a semi-infinite half-space subjected to periodic surface temperature fluctuations. Such water temperature variations at a specific location on the ocean floor can be due to, for example, the transport of water with variable temperature past the site by deep ocean currents. Find the amplitude of water temperature variations that cause surface heat flux variations of \(40 \mathrm{~mW} \mathrm{~m}^{-2}\) above and below the mean on a time scale of 1 day. Assume that the thermal conductivity of sediments is \(0.8 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) and the sediment thermal diffusivity is \(0.2 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).

Assume that the radioactive elements in the Earth are uniformly distributed through a nearsurface layer. The surface heat flow is \(70 \mathrm{~mW} \mathrm{~m}^{-2}\), and there is no heat flow into the base of the layer. If \(k=4 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, T_{0}=0^{\circ} \mathrm{C},\) and the temperature at the base of the layer is \(1200^{\circ} \mathrm{C}\), determine the thickness of the layer and the volumetric heat production.

Scientists believe that early in its evolution, the Moon was covered by a magma ocean with a depth of \(50 \mathrm{~km}\). Assuming that the magma was at its melt temperature of \(1500 \mathrm{~K}\) and that the surface of the Moon was maintained at \(500 \mathrm{~K}\), how long did it take for the magma ocean to solidify if it was cooled from the surface? Take \(L=320 \mathrm{~kJ} \mathrm{~kg}^{-1}, \kappa=1 \mathrm{~mm}^{2}\) \(\mathrm{s}^{-1}\), and \(c=1 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).

A body of water at \(0^{\circ} \mathrm{C}\) is subjected to a constant surface temperature of \(-10^{\circ} \mathrm{C}\) for 10 days. How thick is the surface layer of ice? Use \(L=320 \mathrm{~kJ}\) \(\mathrm{kg}^{-1}, k=2 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \mathrm{~K}^{-1}, c=4 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, \rho=\) \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

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