Solucionario cap 22 serway

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Chapter 22 Solutions
W 25.0 J = = 0.0694 Qh 360 J

22.1

(a) (b)

e=

or

6.94%

Qc = Qh – W = 360 J – 25.0 J = 335 J e= W W 1 = = = 0.333 Q h 3W 3 or 33.3%

22.2

(a) (b)

Qc = Qh – W = 3W – W = 2W Therefore, Q c 2W 2 = = Q h 3W 3 Qh – Qc Qc W = =1– = 0.250, Qh Qh Qh

22.3

(a)

We have e =

with Qc = 8000 J, we have Qh = 10.7 kJ (b) W = Qh – Qc = 2667 J and from ℘ =22.4 W = Qh – Qc = 200 J e= Qc W =1– = 0.300 Qh Qh W W 2667 J , we have t = = = 0.533 s t ℘ 5000 J/s (1) (2)

From (2), Qc = 0.700Qh (3) Solving (3) and (1) simultaneously, we have Q h = 667 J 22.5 and Qc = 467 J

It is easiest to solve part (b) first: (b) ∆Eint = nCV ∆T and since the temperature is held constant during the compression, ∆Eint = 0 . (a) From the first law of thermodynamics,∆Eint = Q – W. Since ∆Eint = 0, this gives: W = Q = 1000 J = 1.00 kJ

© 2000 by Harcourt College Publishers. All rights reserved.

2

Chapter 22 Solutions Qc = heat to melt 15.0 g of Hg = mLf = (15.0 × 10–3 kg)(1.18 × 104 J/kg) = 177 J Qh = heat absorbed to freeze 1.00 g of aluminum = mLf = (10–3 kg)(3.97 × 105 J/kg) = 397 J and the work output = W = Qh – Qc = 220 J e= W 220 J = = 0.554, or 55.4%Q h 397 J

22.6

 Theoretical Eff (Carnot) =  Th  = 933 K – 243.1 K = 0.749 = 74.9% 933 K   T h – T c 
22.7 Tc = 703 K, (a) (b) eC = Th = 2143 K ∆T 1440 = = 67.2% Th 2143 W = 0.420Qh

Qh = 1.40 × 105 J, ℘=

W 5.88 × 104 J = = 58.8 kW t 1s

*22.8

The Carnot efficiency of the engine is eC = ∆T 120 K = = 0.253 Th 473 K

At 20.0% of this maximum efficiency, e = (0.200)(0.253)= 0.0506 From Equation 22.2, W = Qh e and Qh = W 10.0 kJ = = 197 kJ e 0.0506

22.9

When e = eC ,

Tc t Tc W 1– = , and Q =1– Th Q h Th h  

W

t

(a)

Qh =

(W /t)t (1.50 × 105 W)(3600 s) = 1 – (Tc /Th) 1 – (293/773)

Qh = 8.69 × 108 J = 869 MJ Qc = Qh –  W t = 8.69 × 108 – (1.50 × 105)(3600) = 3.30 × 108 J = 330 MJ t 

(c)

© 2000 by Harcourt College Publishers.All rights reserved.

Chapter 22 Solutions
22.10 From Equation 22.4, (a) ∆T 100 eC = = = 0.268 = 26.8% Th 373 eC = ∆T 200 = = 0.423 = 42.3% Th 473

3

P A Qh B W D Qc C Tc V Th

(b) *22.11

Isothermal expansion at Th = 523 K Isothermal compression at Tc = 323 K Gas absorbs 1200 J during expansion. (a) Qc = Qh Tc 323 = (1200 J)  = 741 J Th 523

(b)

W = Qh – Qc = (1200 – 741) J= 459 J Tc Th 573 K Th

*22.12

We use eC = 1 –

as, 0.300 = 1 –

From which, Th = 819 K = 546°C 22.13 The Carnot summer efficiency is eC,s = 1 – Tc (273 + 20)K =1– = 0.530 Th (273 + 350)K 283 = 0.546 623

And in winter, eC,w = 1 –

Then the actual winter efficiency is 0.320  0.546 = 0.330 0.530 or 33.0%

© 2000 by Harcourt College Publishers. All rights reserved.

4Chapter 22 Solutions P fV f γ  P iV i γ =  Tf   Ti 

*22.14

(a)

In an adiabatic process, PfVf = PiVi . Also, 
γ γ

Pf (γ – 1) /γ Dividing the second equation by the first yields Tf = Ti   Pi Since γ = 5 γ–1 2 for Argon, = = 0.400 and we have 3 γ 5 300 × 103 Pa  0.400 = 564 K 1.50 × 106 Pa

Tf = (1073 K)  (b)

∆Eint = nCV ∆T = Q – W = 0 – W, so W = –nCV ∆T, and the poweroutput is ℘= W –nCV ∆T = t t = or

(–80.0 kg)(1.00 mol/0.0399 kg)(3/2)(8.315 J/mol ⋅ K)(564 – 1073) K 60.0 s

℘ = 2.12 × 105 W = 212 kW Tc 564 K =1– = 0.475 Th 1073 K

(c)

eC = 1 –

or

47.5%

22.15

(a)

emax = 1 –

Tc 278 =1– = 5.12 × 10–2 = 5.12% Th 293

(b)

℘=

W = 75.0 × 106 J/s t W = (75.0 × 106 J/s)(3600 s/h) = 2.70 × 1011 J/h

Therefore, From e =

W , we findQh W 2.70 × 1011 J/h = = 5.27 × 1012 J/h = 5.27 TJ/h e 5.12 × 10–2

Qh = (c)

As fossil-fuel prices rise, this way to use solar energy will become a good buy.

© 2000 by Harcourt College Publishers. All rights reserved.

Chapter 22 Solutions
1 m (5.00 m/s)2 2 train

5

22.16

The work output is W = W Qh

We are told e =

0.200 =

1 m (5.00 m/s)2/Qh 2 t 300 K 1 = mt (6.50...
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