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Physics / An Introduction to Thermal Physics 1 / Chapter 1 / Problem 61P

An Introduction to Thermal Physics | 1st Edition | ISBN: 9780201380279 | Authors: Daniel V. Schroeder

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Geologists measure conductive heat flow out of the earth

Chapter 1, Problem 61P

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QUESTION:

Problem 61P

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°C per kilometer of depth and the thermal conductivity of the rock is 2.5 W/m∙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 km.)

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QUESTION:

Problem 61P

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°C per kilometer of depth and the thermal conductivity of the rock is 2.5 W/m∙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 km.)

ANSWER:

Solution 61P

To solve this question, we shall have to consider the Fourier heat conduction law. The mathematical equation for Fourier heat conduction law is written as $\frac{Q}{\Delta{}t}=\ -kA\frac{dT}{dx}$ …..(1)

Here, $Q=$ amount of heat

$\Delta{}t=$ time taken for the heat to flow

$k=$ thermal conductivity

$A=$ area of cross-section

$dT=$ change in temperature

$dx=$ change in length

We need to calculate $\frac{\frac{Q}{\Delta{}t}}{A}$ .

Step 1

Given, $\frac{dT}{dx}=\frac{20}{1}\frac{{\ }^{0}C}{km}=\frac{20\ {\ }^{0}C}{1000\ m}=0.02\ \frac{{\ }^{0}C}{m}$

Thermal conductivity $k=2.5\ W/m.K$

Considering the magnitude only, from equation (1)

$\frac{\frac{Q}{\Delta{}t}}{A}=2.5\ W/m.K\times{}0.02\ \frac{{\ }^{0}C}{m}$

$\frac{\frac{Q}{\Delta{}t}}{A}=0.05\ W/{m}^{2}$

The rate of heat conduction per square meter in this location is $0.05\ W/{m}^{2}$ .

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