Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(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.

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

For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of ε(see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains significantly less than zero as long as T is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must ε have in order for there to be an abrupt Bose-Einstein condensation?

The speed of sound in copper is 3560m/s. Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure 7.28, and compare.

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.)

For a system of particles at room temperature, how large must ϵ-μbe before the Fermi-Dirac, Bose-Einstein, and Boltzmann distributions agree within 1%? Is this condition ever violated for the gases in our atmosphere? Explain.

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have ben conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

TTν32π445+u(T)+f(T)=constant

as the universe expands and cools. Evaluate the constant by assuming that T=Tv when the temperatures are very high.

(c) Calculate the ratio T/Tv, in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/Tv, as a function of T, for kT/mc2ranging from 0 to 3.*

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