/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 3.7 Use the result of Problem 2.42 t... [FREE SOLUTION] | 91影视

91影视

Use the result of Problem 2.42 to calculate the temperature of a black hole, in terms of its mass M. (The energy is Mc2. ) Evaluate the resulting expression for a one-solar-mass black hole. Also sketch the entropy as a function of energy, and discuss the implications of the shape of the graph.

Short Answer

Expert verified

The required expression is T=hc3162GMc2kand for one-solar-mass of a black hole, the temperature can be calculated be 6.1510-8K.

The graph below depicts the entropy as a function of energy which is a concave up graph.

Step by step solution

01

Given

The expression for the entropy of a black hole is given as:

S=82GM2hck..........(1)

Where,

Gis the gravitational constant, Mis mass, his Planck's constant, cis the speed of light, and kis the Boltzmann's constant

The energy of the black hole is given by Einstein's relation as:

U=Mc2..........(2)

02

Calculation for Temperature

Mathematically, temperature can be defined as:

1T=SU..........(3)

Where,

Sis the change in entropy and Uis the change in the internal energy of the body.

Equation (1) can be modified as:

role="math" localid="1646995322714" S=82GM2hckc4c4S=82G(Mc2)2hc5k

By replacing Mc2as U, we get,

role="math" localid="1646997428892" S=82GU2hc5k..........(4)

Now, by substituting this value of Sin equation (3), we get,

1T=SU=U82GU2hc5k1T=162GUhc5kT=hc5162GUk

By resusbstituting the value of Uin the above equation, we get the desired result in terms of mass,

T=hc3162GMc2k

For a solar mass black hole, M=21030kg.

Also, by substituting 6.6710-11m3kg-1s-2for G, 6.6210-34J.sfor h, 1.3810-23J/Kfor kand 3108ms-1for cin the above equation, we get,

T=6.6210-34310831626.6710-11210301.3810-23T=6.1510-8K

03

Graph of entropy as a function of energy

Consider the equation (4),

S=82GU2hc5k

Here,

G,K,h,care all constants

Hence, it can be modified as:

SU2

Therefore, the graph of entropy as a function of energy can be sketched as follows:

It can be observed that the graph is a concave up graph. Objects exhibiting such behavior would have a negative heat capacity.

04

Final answer

The required expression is T=hc3162GMc2kand the temperature can be calculated to be 6.1510-8K. Also the graph of entropy as a function of energy which is a concave up graph, can be sketched as follows:

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 91影视!

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

Use a computer to reproduce Table 3.2 and the associated graphs of entropy, temperature, heat capacity, and magnetization. (The graphs in this section are actually drawn from the analytic formulas derived below, so your numerical graphs won't be quite as smooth.)

Use a computer to study the entropy, temperature, and heat capacity of an Einstein solid, as follows. Let the solid contain 50 oscillators (initially), and from 0 to 100 units of energy. Make a table, analogous to Table 3.2, in which each row represents a different value for the energy. Use separate columns for the energy, multiplicity, entropy, temperature, and heat capacity. To calculate the temperature, evaluate U/Sfor two nearby rows in the table. (Recall that U=qfor some constant .) The heat capacity (U/T)can be computed in a similar way. The first few rows of the table should look something like this:

(In this table I have computed derivatives using a "centered-difference" approximation. For example, the temperature .28is computed as 2/(7.15-0).) Make a graph of entropy vs. energy and a graph of heat capacity vs. temperature. Then change the number of oscillators to 5000 (to "dilute" the system and look at lower temperatures), and again make a graph of heat capacity vs. temperature. Discuss your prediction for the heat capacity, and compare it to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the numerical value of in electron-volts, for each of those real solids.

Consider an ideal two-state electronic paramagnet such as DPPH, with =B. In the experiment described above, the magnetic field strength was 2.06T and the minimum temperature was 2.2K. Calculate the energy, magnetization, and entropy of this system, expressing each quantity as a fraction of its maximum possible value. What would the experimenters have had to do to attain99% of the maximum possible magnetization?

What partial-derivative relation can you derive from the thermodynamic identity by considering a process that takes place at constant entropy? Does the resulting equation agree with what you already knew? Explain.

Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)

CP=a+bT-cT2

where a=16.86J/K,b=4.7710-3J/K2, and c=8.54105JK. Suppose, then, that a mole of graphite is heated at constant pressure from 298Kto 500K. Calculate the increase in its entropy during this process. Add on the tabulated value of S(298K)(from the back of this book) to obtain S(500K).

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.