(a) Estimate (roughly) the total power radiated by your body, neglecting any energy that is returned by your clothes and environment. (Whatever the color of your skin, its emissivity at infrared wavelengths is quite close to 1; almost any nonmetal is a near-perfect blackbody at these wavelengths.)

(b) Compare the total energy radiated by your body in one day (expressed in kilocalories) to the energy in the food you cat. Why is there such a large discrepancy?

(c) The sun has a mass of 2×1030kgand radiates energy at a rate of 3.9×1026watts. Which puts out more power per units mass-the sun or your body?

Short Answer

Expert verified

(a) . The total power radiated by our body 1kW

(b) .role="math" localid="1647768533942" The energy we would lose in one dayis20,000cal.

(c) .The radiation rate per kilogramis0.0002W/kg

Step by step solution

01

Step 1. Given information

  • The total power radiated by our body we use is given by Power=σeAT4
02

Step 2. Calculating the total power radiated by our body we use .

The power radiated is

Power=σeAT4

T=310K, A=2m2, e=1.

Power=5.67×10-8W/m2·K42m2(310K)4

=1050W

1000W

1.0×103W

1kW

Hence, This is the rate at which we would lose energy, if we were naked in empty space .

03

Step 3. Calculating the energy we would lose in one day .

The rate, the energy we would lose in one day will be

(1050W)(24hr)3600s1hr=9.0×107J1kcal4186J

=20,000kcal

As we can see that it is ten times the numbers of calories that an average person conserves in a day about (2000). The discrepancy is due to the fact that we are not naked in empty space, most of the energy that we radiate is replaced by energy radiated or conducted back to us by our clothes and other surroundings

04

Step 4. Calculating the radiation rate per kilogram.

Now, the radiation rate per kilogram is

1000W75kg=14W/kgin case of seen it is

3.9×1026W2×1030kg=0.0002W/kg

As it is 70,000 times less, have we get a reason that, how is it possible of although the sun is bright it is also very massive. Although it generates energy by nuclear fusion the reaction in its core actually proceeds extremely slowly giving it a ten billion year life time. So, I on the other hand have to replenish my chemical fuel supply on daily basis.

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Most popular questions from this chapter

An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of 0.18fm-3(where 1fm=10-15m). Because nucleons come in two different types (protons and neutrons), each with spin 1/2, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result.

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).

Explain in some detail why the three graphs in Figure 7.28 all intercept the vertical axis in about the same place, whereas their slopes differ considerably.

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).

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic ϵmc2so that their energies are ϵ=pc(where p is the magnitude of the momentum vector).

(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by =

μ=hc(3N/8πV)1/3

(b) Find a formula for the total energy of this system in terms of N and μ.

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