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Suppose that the concentration of infrared-absorbing gases in earth's atmosphere were to double, effectively creating a second "blanket" to warm the surface. Estimate the equilibrium surface temperature of the earth that would result from this catastrophe. (Hint: First show that the lower atmospheric blanket is warmer than the upper one by a factor of 21/4. The surface is warmer than the lower blanket by a smaller factor.)

Short Answer

Expert verified

The temperature of the lower atmosphere blanket is303K.

Step by step solution

01

Step 1. Given information

The below figure shows two blankets of the earth's atmosphere for receiving the energy from the sun.

In the above figure the upper blanket must send one unit of the infrared energy for every unit of energy absorbed from the sun.

02

Step 2. Calculating the temperature of the upper blanket

Let the30%of the solar light is reflected. So, the temperature of the upper blanket is

Tupper=(0.7)solar constant4σ1/4

where, σis the Stefan Boltzmann constant.

Putting1370W/m2for the solar constant and5.67×10-8W/m2K4forσ.

Tupper=(0.70)1370W/m245.67×10-8W/m2K414

=255K

So, the temperature of the upper blanket255K.

03

Step 3. Calculating the temperature of the lower blanket

The temperature of the lower blanket is

Tlower=2(0.7)solar constant4σ1/4

=(2)1/4(0.7)solar constant4σ1/4

PuttingTupperfor(0.7)solar constant4σ1/4we get

Tlower=21/4Tupper

So, the temperature of the lower atmosphere blanket is21/4times of the temperature of the upper atmosphere.

Tlower=21/4(255K)

=303K

Hence, the temperature of the lower atmosphere blanket is303K.

04

Step 4. Calculating the equilibrium surface temperature of the earth 

The temperature of the earth is

Tground=3(0.7)solar constant4σ1/4

=(3)1/4(0.7)solar constant4σ1/4

Putting Tupperfor(0.7)solar constant4σ1/4

Tground=31/4Tupper

Tground=314(255K)

=335.6K

=(335.6-273.15)°C

=62°C

Hence, the equilibrium surface temperature of the earth is62°C.

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Most popular questions from this chapter

At the center of the sun, the temperature is approximately 107K and the concentration of electrons is approximately 1032 per cubic meter. Would it be (approximately) valid to treat these electrons as a "classical" ideal gas (using Boltzmann statistics), or as a degenerate Fermi gas (with T0 ), or neither?

Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.

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(c) Explain why the heat capacity must approach 32Nkin the high- Tlimit.

(d) For T>Tcyou can evaluate the integral using the values of μcalculated in Problem 7.69. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with Figure 7.37.

Figure 7.37. Heat capacity of an ideal Bose gas in a three-dimensional box.

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For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of "background radiation," in addition to the photons and neutrinos (see Figure 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (ea.ch with the same mass), and each has two independent polarization states. During the time period of interest, the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in Figure 7.21. When the temperature was greater than the electron mass times c2k, the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutrinos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles. the previous problem. Recall from special relativity that the energy of a massive particle is ϵ=(pc)2+mc22.

(a) Show that the energy density of electrons and positrons at temperature Tis given by

u(T)=0x2x2+mc2/kT2ex2+mc2/kT2+1dx;whereu(T)=0x2x2+mc2/kT2ex2+mc2/kT2+1dx

(b) Show that u(T)goes to zero when kTmc2, and explain why this is a

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