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Problem 1.61. Geologists measure conductive heat flow out of the earth by drilling holes (a few hundred meters deep) and measuring the temperature as a function of depth. Suppose that in a certain location the temperature increases by \(20^{\circ} \mathrm{C}\) per kilometer of depth and the thermal conductivity of the rock is \(2.5 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) What is the rate of heat conduction per square meter in this location? Assuming that this value is typical of other locations over all of earth's surface, at approximately what rate is the earth losing heat via conduction? (The radius of the earth is \(6400 \mathrm{km} .\) )

Short Answer

Expert verified
The Earth's heat loss via conduction is \( 2.575 \times 10^{13} \mathrm{W} \).

Step by step solution

01

Identify the given data

We are provided with the following information: the temperature gradient is \( 20^{\circ} \mathrm{C/km} \) and the thermal conductivity of the rock is \( k = 2.5 \mathrm{W/m \cdot K} \). The radius of the Earth is \( R = 6400 \mathrm{km} \).
02

Use Fourier's Law of Heat Conduction

Fourier's Law of Heat Conduction is used to find the rate of heat flow per unit area, given by the formula \( q = -k \frac{dT}{dz} \), where \( q \) is the heat flux, \( k \) is the thermal conductivity, and \( \frac{dT}{dz} \) is the temperature gradient. Substituting the given values, we have \( q = -2.5 \mathrm{W/m \, K} \times \left(-20 \mathrm{K/km}\right) \).
03

Convert units appropriately

The temperature gradient is given in \( \mathrm{C/km} \), which can be directly used as \( \mathrm{K/km} \) since the change is the same in both Celsius and Kelvin. Convert kilometers to meters by dividing by 1000: \( 20 \mathrm{K/km} = 20 \times 10^{-3} \mathrm{K/m} \).
04

Calculate the heat flux

Now calculate the heat flux using the identified values: \( q = 2.5 \times 20 \times 10^{-3} = 0.05 \mathrm{W/m^2} \).
05

Calculate Earth's surface area

The surface area \( A \) of a sphere is calculated using the formula \( A = 4 \pi R^2 \). Substitute in the Earth's radius to find \( A = 4 \pi (6400 \times 10^3 \mathrm{m})^2 \).
06

Determine Earth's total heat loss

Using the heat flux and Earth's surface area, calculate the total heat loss rate \( Q \) using \( Q = q \times A \). Substitute the values: \( Q = 0.05 \mathrm{W/m^2} \times 4 \pi (6400 \times 10^3)^2 \).
07

Perform the calculation

Carrying out the calculation for \( Q \): \( 4 \pi (6400 \times 10^3)^2 \approx 5.15 \times 10^{14} \mathrm{m^2} \). So, \( Q = 0.05 \mathrm{W/m^2} \times 5.15 \times 10^{14} = 2.575 \times 10^{13} \mathrm{W} \).

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Key Concepts

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

Fourier's law of heat conduction
Fourier's law of heat conduction is a fundamental principle that describes how heat energy is transferred through a material. Simply put, it states that the rate at which heat flows through a substance is proportional to the temperature gradient and the material's thermal conductivity. The mathematical expression for this law is:
\[ q = -k \frac{dT}{dz} \]
Here, \( q \) represents the heat flux, \( k \) is the thermal conductivity, and \( \frac{dT}{dz} \) is the temperature gradient. The negative sign indicates that heat flows from high to low temperature, aligning with the second law of thermodynamics. Fourier's law is applicable to many heat transfer scenarios, such as understanding geothermal heat flow, which helps in predicting how much heat is released from the Earth's interior to its surface.
Thermal conductivity
Thermal conductivity is a property that defines how well a material conducts heat. It's denoted by \( k \) and measured in watts per meter per degree Kelvin (W/m·K). High thermal conductivity means a material can effectively transfer heat, while low thermal conductivity indicates good thermal insulation.
Factors affecting thermal conductivity:
  • Material composition: Metals typically have high thermal conductivity, while gases and plastics have lower values.
  • Temperature: Some materials show significant changes in thermal conductivity with temperature variations.
  • Phase of the material: Solids, liquids, and gases may have different thermal conductivities.
In the given problem, the thermal conductivity of rock is 2.5 W/m·K, suggesting that it is not an excellent conductor, but it can still transfer some heat from the Earth's interior towards the surface.
Temperature gradient
The temperature gradient is a measure of how rapidly temperature changes over a distance. It's represented by the term \( \frac{dT}{dz} \) in Fourier's law of heat conduction. In our problem, it means that for every kilometer into the Earth's crust, the temperature increases by \( 20^{\circ} \) Celsius.
Key Characteristics of Temperature Gradient:
  • Direction: Indicates the path along which temperature changes, usually moving from hot to cold regions.
  • Magnitude: A higher gradient means a steeper temperature change across a given distance.
  • Units: Generally expressed in degrees per unit length, such as Celsius per kilometer or Kelvin per meter.
Understanding the temperature gradient is crucial for determining how much heat is conducted, especially in geological processes such as the one described here.
Earth's surface area
Calculating the Earth's surface area is important for determining total heat loss via conduction. The formula for the surface area of a sphere is:
\[ A = 4 \pi R^2 \]
Here, \( A \) is the surface area, and \( R \) is the radius of the sphere—in this context, the Earth's radius is used. By knowing the surface area, one can find the total heat lost to space from the Earth's surface using the heat flux calculated from Fourier's law.
Steps to calculate the surface area:
  • Use the radius: For Earth, \( R = 6400 \) km, or \( 6400 \times 10^3 \) meters.
  • Apply the formula: Calculate \( A = 4 \pi (6400 \times 10^3)^2 \).
  • Result: Approximately \( 5.15 \times 10^{14} \) square meters, illustrating the vastness of Earth's surface area aiding in heat conduction calculations.
This understanding helps estimate Earth's thermal energy loss, a key factor in studies of geothermal systems and global heat balance.

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