Physics 41 Chapter 22 HW

Transcription

1 Pysis 41 apter 22 H 1. eat ine performs 200 J of work in ea yle and as an effiieny of 30.0%. For ea yle, ow mu energy is (a) taken in and (b) expelled as eat? = = 200 J (1) e = = = (2) From (2), = (3) Solving (3) and (1) simultaneously, we ave (a) = 667 J and (b) = 467 J. 2. refrigerator as a oeffiient of performane of Te ie tray ompartment is at 20.0, and te room temperature is Te refrigerator an onvert 30.0 g of water at 22.0 to 30.0 g of ie at 20.0 ea minute. at input power is required? Give your answer in watts. OP 3.00 = =. Terefore, = Te eat removed ea minute is 5 = kg J kg kg J kg t kg J kg 20.0 = J min ( )( )( ) ( )( ) ( )( )( ) or, t = 233 J s Tus, te work done per seond is 233 J s P = = n eletri power plant tat would make use of te temperature gradient in te oean as been proposed. Te system is to operate between 20.0 (surfae water temperature) and 5.00 (water temperature at a dept of about 1 km). (a) at is te maximum effiieny of su a system? (b) If te useful power output of te plant is 75.0 M, ow mu energy is taken in from te warm reservoir per our? () In view of your answer to part (a), do you tink su a system is wortwile? Note tat te fuel is free. a) e max T = 1 = 1 = = 5.12% T 293 (b) 6 P = = J s t 6 11 Terefore, ( ) ( ) = J s s = J From e= we find J 12 = = = J = 5.27 TJ 2 e () s fossil-fuel pries rise, tis way to use solar energy will beome a good buy.

2 4. Te temperature at te surfae of te Sun is approximately K, and te temperature at te surfae of te art is approximately 290 K. at entropy ange ours wen J of energy is transferred by radiation from te Sun to te art? S T T = = J K = J K L ontainer as a enter partition tat divides it into two equal parts, as sown. Te left side ontains H2 gas, and te rigt side ontains O2 gas. ot gases are at room temperature and at atmosperi pressure. Te partition is removed and te gases are allowed to mix. at is te entropy inrease of te system? f S = nrln = ( )( 2) Rln 2 Vi ( ) S = ln 2 = J K 6. t point in a arnot yle, 2.34 mol of a monatomi ideal gas as a pressure of kpa, a volume of 10.0 L, and a temperature of 720 K. It expands isotermally to point, and ten expands adiabatially to point were its volume is 24.0 L. n isotermal ompression brings it to point, were its volume is 15.0 L. n adiabati proess returns te gas to point. (a) raw te diagram. (b) etermine all te unknown pressures, volumes and temperatures as you fill in te following table. () Find te energy added by eat, te work done by te ine, and te ange in internal energy for ea of te steps,,, and. ut values in te table. (d) alulate te effiieny net/. Sow tat it is equal to 1 T/T, te arnot effiieny. Results: State P(kPa) V(L) T(K) Proess (kj) (kj) int (kj)

3 #6. alulations: (a) First, onsider te adiabati proess : L = so P = P = kpa = 712 kpa 15.0 L nrt nrt lso V = V Now, onsider te isotermal proess : T = T = 549 K or T = T = 720 K = 549 K V V kpa ( 10.0 L ) P = P = P = P 1 = = 23 V V V VV 24.0 L ( 15.0 L ) Next, onsider te adiabati proess : = 445 kpa V ut, P = from above. lso onsidering te isotermal proess, P 1 = P VV VV 10.0 L ( 24.0 L ) Hene, P V = V 1 wi redues to V = = = 16.0 L V VV V 15.0 L Finally, 10.0 L P = P = kpa = 875 kpa 16.0 L (b) For te isotermal proess : int = n T = 0 V 16.0 = = ln = 2.34 mol J mol K 720 K ln = kj V 10.0 so nrt ( )( ) For te adiabati proess : = 0 3 int = nv ( T T) = 2.34 mol ( J mol K ) ( ) K = 4.98 kj 2 and = + int = 0 + ( 4.98 kj) = 4.98 kj For te isotermal proess : int = n T = and = = nrt ln = 2.34 mol ( J mol K )( 549 K ) ln = 5.02 kj V 24.0 Finally, for te adiabati proess : = 0 3 int = nv ( T T) = 2.34 mol ( J mol K ) ( ) K = kj 2 and = + int = kj= kj V Te work done by te ine is te negative of te work input. Te output work is given by te work olumn in te table wit all signs reversed. () 1.56 kj e= = = = or 23.7% 6.58 kj T 549 e = 1 = 1 = or 23.7% T 720

4 7. eat ine wit a diatomi gas as te working substane uses te losed yle sown. How mu work does tis ine do per yle and wat is its termal effiieny? ut all your values in te tables as usual. P(atm) V(m3) T(K) (kj) (kj) int (kj) Net

5 8 Otto yle: pproximates te proess ourring in an internal ombustion ine. Te Otto yle represents a four-stroke yle onsisting of two upstrokes and two downstrokes. Te ompression ratio (V1/V2)=8 and gas is diatomi: 1 T T Otto yle: e = 1 = 1 = 1, 1 ( V / V ) T T 1 2 Te tables look like: State T(K) P (kpa) 3 V ( m ) int (J) Proess (J) output (J) int (J) (a), (b) Te quantity of gas is ( Pa)( m ) n = = = mol RT ( J mol K )( 293 K ) int, ( Pa )( = nrt = = m ) = 125 J P = P = = In proess, ( )( ) 1.40 T int, nr Pa Pa ( Pa)( m 8.00) ( mol )( J mol K ) = = = 673 K 5 5 = nrt = ( mol )( J mol K )( 673 K ) = 287 J 2 2 so int, 287 J 125 J 162 J out 0 out = = = = = 162 J Proess takes us to: int, int, ( mol )( J mol K )( K ) nrt P = = = 6 3 V m out Pa 5 5 = nrt = ( mol )( J mol K )( K ) = 436 J 2 2 = 436 J 287 J = 149 J = = 0 = 149 J

6 In proess : P = P = ( Pa) = Pa 8.00 T int, nr ( Pa)( m ) ( mol )( J mol K ) = = = 445 K 5 5 = nrt = ( mol )( J mol K )( 445 K ) = 190 J 2 2 int, = 190 J 436 J= 246 J = out = 0 out = 246 J and int, int, int, out = = 125 J 190 J = 65.0 J = = 0 = 65.0 J For te entire yle, int, net = 162 J = 0. Te net work is net = 162 J J+ 0 = 84.3 J = J J= 84.3 J () Te input energy is = 149 J, te waste is = 65.0 J, and = 84.3 J. (d) Te effiieny is: 84.3 J e= = = 149 J (e) Let f represent te angular speed of te ranksaft. Ten 2 f obtain work in te amount of 84.3 J/yle: is te frequeny at wi we f J s = ( 84.3 J yle ) J s f = = 84.3 J yle = rev s rev min