Chapter 20 - 20.1: a) %. 8 . 33 338 . b) J. 6500 J 4300 J...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 20.1: a) %. 8 . 33 338 . b) J. 6500 J 4300 J 2200 6500 2200 = = = + 20.2: a) %. 9 . 28 289 . b) J. 2600 J 6400 J 9000 J 9000 J 2600 = = =- 20.3: a) %. . 23 230 . 100 , 16 3700 = = b) J. 12,400 J 3700 J 100 , 16 =- c) . g 350 . kg J 10 60 . 4 J 100 , 16 4 = d) hp. 298 kW 222 s) 60.0 J)( 3700 ( = = 20.4: a) J. 10 43 . 6 5 ) 280 . ( s) W)(1.00 10 180 ( 1 3 = = = Pt Q e b) J. 10 4.63 s) W)(1.00 10 180 ( J 10 43 . 6 5 3 5 = - =- Pt Q 20.5: a) MW. 970 MW 330 MW 1300 b) %. 25 25 . MW 1300 MW 330 =- = = = e 20.6: Solving , for (20.6) Eq. r or ) 1 ln( ln ) 1 ( e r - =- . 8 . 13 ) 350 . ( ) 1 ( 5 . 2 1 1 = =- =-- e r If the first equation is used (for instance, using a calculator without the y x function), note that the symbol e is the ideal efficiency , not the base of natural logarithms. 20.7: a) C. 453 K 726 K)(9.5) 15 . 295 ( 0.40 1 = = = =- a b r T T b) Pa. 10 99 . 1 Pa)(9.50) 10 50 . 8 ( 6 4 = = = a b r p p 20.8: a) From %. 58 58 . ) 8 . 8 ( 1 1 (20.6), Eq. 40 . 1 = =- =- =-- r e b) %, 60 ) 6 . 9 ( 1 40 . =-- an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%. 20.9: a) J. 10 62 . 1 4 10 . 2 J 10 40 . 3 4 = = = K Q C W b) J. 10 02 . 5 ) 1 ( 4 1 C C H = + = + = K Q W Q Q 20.10: ) ( 1 f C T c L t m K t K Q t W P p + = = = ( 29 W. 128 K) K)(2.5 kg J (485 kg) J 10 60 . 1 ( s 3600 kg . 8 8 . 2 1 5 = + = 20.11: a) , b) W. 767 s . 60 J 10 80 . 9 J 10 44 . 1 4 5 P H EER = = - or . 27 . 7 ) 413 . 3 ( W 767 W 1633 ) 413 . 3 ( ] s) (60 J) 10 8 . 9 ( s) (60 J) 10 44 . 1 ( [ s) (60 J) 10 8 . 9 ( 4 5 4 = = - = EER 20.12: a) ) ( water water ice ice C T c T c L m Q f + + = ( 29 J. 10 90 . 8 K) K)(25.0 kg J 4190 ( K) K)(5.0 kg J 2100 ( kg J 10 334 kg) 80 . 1 ( 5 3 = + + = b) J. 10 37 . 3 5 40 . 2 J 10 08 . 8 | | 5 C = = = K Q W c) = = + = + = | | that (note J 10 1.14 J 10 8.08 J 10 37 . 3 | | | | H 6 5 5 C H Q Q W Q ).) 1 ( | | 1 C K Q + 20.13: a) J. 215 J 335 J 550 | | | | C H =- =- Q Q b) K. 378 J) 550 J K)(335 620 ( |) | | | ( H C H C = = = Q Q T T c) %. 39 J) 550 J 335 ( 1 ) | | | | ( 1 H C =- =- Q Q 20.14: a) From Eq. (20.13), the rejected heat is J. 10 3.72 J) 6450 )( ( 3 K 520 K 300 = b) J. 10 2.73 J 10 3.72 J 6450 3 3 = - c) From either Eq. (20.4) or Eq. (20.14), e= 0.423=42.3%. 20.15: a) C H f C H C H | | | | T T mL T T Q Q = = J, 10 088 . 3 K) (273.15 K) (287.15 kg) J 10 kg)(334 . 85 ( 7 3 = = or =- =- = )) ( 1 ( | | | | | | | | b) figures. two to J 10 09 . 3 H H C H 7 T T Q Q Q W C J) 10 09 . 3 ( 7 J. 10 49 . 2 )) 15 . 297 15 . 273 ( 1 ( 6 =- 20.16: a) From Eq. (20.13), J. 492 J) 415 )( ( K 270 K 320 = b) The work per cycle is J, 77 J 415 J 492 =- and W, 212 ) 75 . 2 ( s 1.00 J 77 = = P keeping an extra figure....
View Full Document

This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

Page1 / 25

Chapter 20 - 20.1: a) %. 8 . 33 338 . b) J. 6500 J 4300 J...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online