e•& • 'i-yLea & Burke Physics: The Nature of Things

= - (2 aim) (20 L) = 60 L • atm

CD: Isochoric

22.5 GIveu^rT^N

Tc =^"410 K, TH = 750 K, QH = 3.0 x 103 J

Find: -WOT\<., W and the heat rejected, Qc

, Tc n 410

= -(40 L )(-! atm) = -60 L • atm

Wen = 0&UCD = QCD - WCD

~ QCD = -60 L • atm

DA: isobaric compression

QDA = |Pi(Vi - V2) - |(1 atm)(20 L - 40 :

= —50 L • atm

WDA - Pi(Vi - V2) = 1 atm(-20 L )= -20 L -atm

The work done per cycle is:

W =^&oQn ~ (0.453)(3.0 x 103 J)('= 1.4 x lO3^; <^.

The waste heat exhausted is:

\QC\ =_ [1._-£e)Qg = (0.547)(3.0 x 103 J)fl~= 1.6xl03J\ ^'

-—H __—^As a check on the answers, note that

W + \QC\ jr. 1.4 x 103 J + 1.6 x 103 J= 3.0 x 103 J = QH/

= —30 L • atm

Table: Since 1 L atm = 101 J, we have:

Process:

AB

BC

CD

DA

Q(J)3030

10100

-6060

-5050

W(J)

0

4040

0

-2020

A£/(J)

3030

6060

-6060

-3030

As a check, note that the sum of the third col-umn gives:

A[7net - 3030 J + 6060 J - 6060 J - 3030 J = 0

As required for a cyclic set of processes,

b) The efficiency is denned by:

_ net work obtained Wheat input Q-in

from the table in 3a),

Waei = 4040 J-2020 J =+2020J

Qin = 3030 J + 10100 J = 13130 J

therefore,

Wnei 2020 JQ-m 13130 J

-0.154

22.4 e = 44%

as required by the 1st Law of Thermodynamics.

22.6 4.96

22.7 No. For example, an irreversible processes car-ried out inside an insulated container involvesno exchange of energy with the environment.Therefore the only entropy change possible isthat of the system. When an ice cube melts,heat is transferred to the cube from the environ-ment. The entropy change of the environment isnegative, while the entropy change of the cubeis positive and sufficiently large that the total

,- entropy change (of cube plus environment) isalso positive.

22.8 11 J/K

22.9 The Carnot cycle is the most .efficient heat en-gine, and has the maximum efficiency possible,

e < ec = I ~ 7j=r

for Tc = 300 K, TH = 600 K

.<.-^-l-M-«K

The engine can be no more than 50% efficient.

22.10 137 K

22.11

El = -5.4 x HT19 J; <?2 - 8

£2 = -2.2 x 10~18 J; 9-2 = 2

T = 104K

/<w

1

c-

" I

\

^>Ai-*— |

|] ^-

22.14 The number of states is 52!. Thus the entropy is:

5 = Jfeln52! = Jfelne156 = 156fc= 156 (1.38 x ID"23 J/K) = 2. 15 x 10~21 J/K

22.16 See solutions manual

22.18 In both cases the gas undergoes a complete cycle of processes and then returns toits initial state. Since internal energy is a state variable, its value is the same at the end ofthe cycle as at the beginning - the change in internal energy is zero independent of whetherthe processes are reversible. Of course the irreversible cycle does produce a. net change ofentropy while the reversible cycle does not

22.20 ̂ vvfe consider each process in detail to find the heat input and net work done."22?: isobaric compression. Point B is on a lower isotherm than point .A, so heat transfer

is out of the system during this leg. The work done by the system is negative:

WAB = PA (Vs - VA] = PA (\VA - VA] = -\PAVA\& J 4

BC: Constant volume process. Point C is on a higher isotherm than point B, so heattransfer is into the system:

= XR (Tc - TB)

= | (1PB - PB) \VA = \PBVA = \

0 i

>lCv-

^

QBC =

Then, using the ideal gas law:

QBC = \(Pc- PB

Since the volume does not change, no work is done:

\J WBC = 0CA: Along the isotherm, there is no change in internal energy, so Q = W. Both are

positive. Using equation 19.13:

QCA = WCA = PGVC In ?± = In In2

Thus the efficiency of the cycle is:

net work done WAB + W3C + WCA _ ~\_heat input QBC + QCA %PAVA + PAVA}n2

~l + 2hi23 + 41n2 ^

2122 We consider each leg of the cycle to compute the heat input and the net workdone. Since Q = 0 along both adiabats, the calculations are most easily done in terms ofheat transfers. For a complete cycle, there is no change of internal energy. Thus from thefirst law of thermodynamics,

= 0 = QMt - Wnet =>

Q» Qin

2

Lea &: Burke Physics: The Nature of Thingi

from Figure 22.25,

r2 =

8JL

8_L2L

22.27 a) The coefficient of performance for a Carnol"T"i\—"**^ refrigerator is:^•^? >

eDiesel _ 1 — —

for r = T2-For diatomic gas, 7 — | and so 7 — 1 = |.

Then

eDiese! = 0.328

eotto = 0.426

(For the numbers in the Figure, 7 = |.)

52.24 A. large volume of water may approximate anideal heat reservoir through its very great heatcapacity (C^o ~ 4184 J/K • kg) However, thethermal conductivity of water is extremely low,being only about 0.6 ^5- While the thermalconductivity of copper is extraordinarily high(fa 400 TJ-rg), it's usefulness as an ideal heatreservoir is limited by the small value of its spe-cific heat capacity (393 J/K • kg)

22.25 By increasing the hot temperature while keep-ing VA constant, the entire cycle moves upwardin the P-V diagram.

P 4A\\

D

PA increases, and since AD is an adiabat PjyV^also increases. The product PD^D is fixed bythe value of the temperature TC of the reservoir,thus VD must increase. The pressure and vol-ume throughout the cycle increase. Similarly ifTC is decreased, pressure and volume through-out the cycle decrease.

= ^(36°F-32°F)+273K

= 275.2 K

- ^(82°F-32°F)+273K

= 300.8 K275.2 K

(300.8 K) - (275.2 K)-11

b)

QdQdt

= Kc\W\

dt= (10.8)(220 W)= 2400W

22.28 a) 260 K

b) The efficiency increases by 1%

c) decreases by 1%

22.29 According to Newton's law of cooling, the rat<at which heat leaks into the freezer is propor-tional to the temperature difference between in-side and outside.

The freezer uses net power P to remove thifheat:

p _ Wnet per cycleAt per cycle

For the Carnot cycle (§22.3)

Wnet - Nk(TH

to remove heat

each cycle. To remove heat faster, the refrigerator must cycle more rapidly.

A J Qcoid Tc

H

Thus the power used is

(Tn-12.26 540 K

P-Ai

oc oc (TH ~ Tcf

3MUUIC3

both the first and second laws.After sufficient water is absorbed, the bird becomes mechanically unbalanced. When

the torque arising from the weight of absorbed water exceeds the torques due to friction inthe axle and due to our inventor's generating equipment, the bird will move to its "wet"state. In this process, ordered gravitational energy is transformed into ordered electri-cal energy plus some thermal energy due to friction. Only the latter leads to entropyincrease.

NUW, exposed to ory air, water molecules evaporate, their entropy increasing as theybecome still less well known in position. Evaporation reducesthe water temperature,resulting in heat flow from the air into the bird and further production of entropy. Even-tually the bird becomes unbalanced again and its head drops to the water for another drink.Again ordered gravitational energy is transformed to electrical energy via the inventor'sgenerator.

Energy transfer to the system during the cycle occurs as heat transfer during the wetstate, accounting for frictional dissipation and any electrical energy produced. Net en-tropy is produced in each phase of the cycle. So both laws are satisifed overall.

\Vater in contact with dry air is not in equilibrium, and evaporation results in energy flowand entropy production. The "dippy bird" is an example of how devices can exploit non-equilibrium to carry on organized activity. Life on earth is another example, which takesadvantage of entropy production as the earth converts sunlight to infrared radiation.

(̂ 22.38 First the water is heated from room temperature to the boiling point, at constant^StiOMSBS*551

pressure:

= (4.0 x 10-3 m3) (1.0 x 103 kg/m3) (4186 J/kg - K) In — =*' 4.0 x 103 J/K^

The change in entropy due to conversion of water to steam at 100°C = 373 K is given by:

= 9. = H^i = (4-0 * 10"3 m3) (1-0 x 1(}3 k8fa3) (2272 x 1Q3 J/kg) =JI^373 K

So the total entropy change is me sum of the two amounts we have calculated:

A5totai = 2.8 x 104 J/K

22.40 The rate of heat transfer through the roof is:

where A is the roof's area, R/ is the thermal resistance of the roof material, TH is the (hot)temperature inside the building and Tc is the (cold) temperature outside. From Example22.5, we have :

dS dQc/dt dQH/dtdt ~~ Tc TH .

Heat flow is into the cold exterior and out of the hot interior, so:

dS H H A . I l\ A(TH-TcfRf TCTS

Putting in the numbers:

TH = 69°F = - (69 - 32) + 273 K = 293. 56 K

Tc = 44°F = | (44 - 32) + 273 K = 279. 67K

dS 510 m2 (293. 56 K - 279. 67 K)2

dt ~0.20m2-K/W (293. 56 KU279. 67 K]

22. Entropy and the Second Law of Thermodynamics 409

Then

~ - x w

dt T 6300 K(See front endpapers of text)

~ = -6.2 x 1022 W/Kdt

At least as much entropy is generated elsewherein the universe by the absorption of solar radiation.

22.38 AS = 2.8 x 104 J/K

22.39 Given

m = 1 kg

Ti = 15° C = 288 KT2 = 80° C = 353 K

The entropy increase of the water isTi mcdT

AS = ^ = = me inf rp

- mcln | T

= (1.0 x 103 g)(4.2

= 850 J/K

22.40 6.0 W/K

1j2.4rShow:

-IM f353"H288,

t

for an ideal gas.

, Solution: for an ideal gas

_ PV = NkT

dU = -

1st Law:

CD

(2)

dQ =dQT

r^

du + PdVdU dVT ~f~

mcv—±P-

mcv^ + N.T

fdQ

^>

dV( PV\Nk,

dVfc~

-22,42 0.22

22.43JThe measured efficiency, e, may be expressed="™"""ln terms of the Carnot efficiency, &c'-

e = (0.55)ec = 0.17

thereforeec = 0.31

But we know ec = 1 — Tc/Tn with

Tc = (273 + 5) K = 278 K

so

TH =278 K

= 402.9 K = 4.0 x 102 Kv_22.44 74 K

22.45 The required work is:

TC

differentiating with respect to time yields

W_=L_Tc\ dQndt \ TH I dt

with TH in Kelvin. Thus

35 W(268 K) = 153.6 K

Tc = 150 K (to 2 sig fig.)

22.46 Only the fast molecules near the surface ofa liquid can overcome the attraction of the

' molecules in the surface and escape by evapora-tion. The average kinetic energy of the remain-ing molecules drops correspondingly, leaving theliquid at a lower temperature.

22.47 The Maxwellian velocity distribution (Eqn.22.13)

m

,27r/bTVexp{ —mv2 ^

2kT J

—V

gives the fraction of molecules in a speed rangev to v + dv. For a fixed value of v and dv, twothings happen as T increases by a factor of 2:

as reauired.

i) The normalization factor

by a factor 2§ = 2.83 and[2irkT

decreases