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Chapter 1: Q. 1.61 (page 40)

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 by20∘Cper kilometer of depth and the thermal conductivity of the rock is 2.5W/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 the earth's surface, at approximately what rate is the earth losing heat via conduction? (The radius of the earth is 6400km.)

Short Answer

Expert verified

Rate of heat conduction and the rate at which the earth loses heat via conduction

QΔt=0.05W

QtotalΔt=2.573×1013

Step by step solution

01

Calculation of ratio

Because the rock that makes up the Earth's crust has a thermal conductivity and there is a temperature differential between a point underground and the Earth's surface, the Earth loses energy through heat conduction. We've calculated the following using Schroeder's values:ΔT=20∘Kper Δx=1000mand kt=2.5W∘K−1, the rate of heat conduction in an area of 1m2is, therefore:

localid="1651746825640" QΔt=ktAΔTΔx=2.5×1×201000=0.05W/m2

localid="1651746839506" QΔt=0.05W/m2

02

Calculation of heat

Even if the rate of heat loss for a square metre is fairly low, if we assume that this figure applies to the entire Earth, the total heat loss is:

QtotalΔt=Loss per meter square×Total area of earth

QtotalΔt=0.05×4πr2=0.05×4π6400×1032

QtotalΔt=2.573×1013W

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Most popular questions from this chapter

Even at low density, real gases don’t quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial

expansion,

PV−nRT(1+B(T)(V/n)+C(T)(V/n)2+⋯)

where the functions B(T), C(T), and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations, it’s sufficient to omit the third term and concentrate on the second, whose coefficient B(T)is called the second virial coefficient (the first coefficient is 1). Here are some measured values of the second virial coefficient for nitrogen (N2):

T(K)
B(cm3/mol)
100–160
200–35
300–4.2
4009.0
50016.9
60021.3
  1. For each temperature in the table, compute the second term in the virial equation, B(T)/(V/n), for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.
  2. Think about the forces between molecules, and explain why we might expect B(T)to be negative at low temperatures but positive at high temperatures.
  3. Any proposed relation between P, V, andT, like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,
    (P+an2V2)(V−nb)=nRT
    where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients (Band C) for a gas obeying the van der Waals equation, in terms of aand b. (Hint: The binomial expansion says that (1+x)p≈1+px+12p(p−1)x2, provided that |px|≪1. Apply this approximation to the quantity [1−(nb/V)]−1.)
  4. Plot a graph of the van der Waals prediction for B(T), choosing aand bso as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.)

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000 . The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β :
β≡ΔV/VΔT
(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β =1 / 550,000 K-1=1.81 x 10-4 K-1. (The exact value varies with temperature, but between 0oC and 200oC the variation is less than 1 %.)
(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 -4 K-1 at 100oC, but decreases as the temperature is lowered until it becomes zero at 4oC. Below 4oC it is slightly negative, reaching a value of -0.68 x 10-4K-1 at 0oC. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

By applying a pressure of 200atm, you can compress water to 99%of its usual volume. Sketch this process (not necessarily to scale) on aPV diagram, and estimate the work required to compress a liter of water by this amount. Does the result surprise you?

An ideal gas is made to undergo the cyclic process shown in the given figure. For each of the steps A, B, and C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas; (b) the change in the energy content of the gas; (c) the heat added to the gas.

Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish?

Derive the equation 1.40 from the equation 1.39
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