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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sometimes it is useful to know the free energy of a photon gas.

(a) Calculate the (Helmholtz) free energy directly from the definition

(Express the answer in terms of T' and V.)

(b) Check the formula S=-(F/T)Vfor this system.

(c) Differentiate F with respect to V to obtain the pressure of a photon gas. Check that your result agrees with that of the previous problem.

(d) A more interesting way to calculate F is to apply the formula F=-kTlnZ separately to each mode (that is, each effective oscillator), then sum over all modes. Carry out this calculation, to obtain

F=8πV(kT)4(hc)30x2ln1-e-xdx

Integrate by parts, and check that your answer agrees with part (a).

When the attractive forces of the ions in a crystal are taken into account, the allowed electron energies are no longer given by the simple formula 7.36; instead, the allowed energies are grouped into bands, separated by gaps where there are no allowed energies. In a conductor the Fermi energy lies within one of the bands; in this section we have treated the electrons in this band as "free" particles confined to a fixed volume. In an insulator, on the other hand, the Fermi energy lies within a gap, so that at T = 0 the band below the gap is completely occupied while the band above the gap is unoccupied. Because there are no empty states close in energy to those that are occupied, the electrons are "stuck in place" and the material does not conduct electricity. A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature. Figure 7 .17 shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor, and defines some terminology and notation to be used in this problem.

(a) As a first approximation, let us model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas, with an appropriate zero-point: g(ϵ)=g0ϵ-ϵc, where go is the same constant as in equation 7.51. Let us also model the density of states near the top

Figure 7.17. The periodic potential of a crystal lattice results in a densityof-states function consisting of "bands" (with many states) and "gaps" (with no states). For an insulator or a semiconductor, the Fermi energy lies in the middle of a gap so that at T = 0, the "valence band" is completely full while the-"conduction band" is completely empty. of the valence band as a mirror image of this function. Explain why, in this approximation, the chemical potential must always lie precisely in the middle of the gap, regardless of temperature.

(b) Normally the width of the gap is much greater than kT. Working in this limit, derive an expression for the number of conduction electrons per unit volume, in terms of the temperature and the width of the gap.

(c) For silicon near room temperature, the gap between the valence and conduction bands is approximately 1.11 eV. Roughly how many conduction electrons are there in a cubic centimeter of silicon at room temperature? How does this compare to the number of conduction electrons in a similar amount of copper?

( d) Explain why a semiconductor conducts electricity much better at higher temperatures. Back up your explanation with some numbers. (Ordinary conductors like copper, on the other hand, conduct better at low temperatures.) (e) Very roughly, how wide would the gap between the valence and conduction bands have to be in order to consider a material an insulator rather than a semiconductor?


Fill in the steps to derive equations 7.112and7.117.

Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles.

(a) Describe the ground state of this system, for each of these three cases.

(b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case?

(c) Repeat part (b) for two units of energy and for three units of energy.

(d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.

Change variables in equation 7.83 to λ=hc/ϵ and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and find a numerical formula for the wavelength where the spectrum peaks, in terms of hc/kT. Explain why the peak does not occur at hc/(2.82kT).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free