sol to assignment11 - 12.6 The sketches for(a and(hf are...

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Unformatted text preview: 12.6 The sketches for (a) and (hf are shown bdbw' HWWDQ‘C’ S ‘7 Dean M (c)There is more work done in process (a). We recognize this horn the fignres because there is more area under the PV curve in (a). Phymcafly, more work is done because of the higher pressure during the expansion part of the process. _ 12.13 (a) W=PAV= 0,3 10?" 5' “_‘_ =[.15(2j. 3x10 Pa)](3.00x10'3m3—8.00x10'3 1313) (b) AU=:Q— w We are ' given that —- AU=~400J-(-152 J) = -248 J. Q— 40” Tim -‘————_._._,__'____ 12.18 = 1:3: Qo :onstant volume), WCA <0 (A V<0), WAB> o (A V> 0) _ - givesAUBca-Qgc- Mc<0(becau_se In summary QB>O Q I . , , c<0,QjA<O. WAB>O , =0! AUAB>O,AUB(:<O,andAUCA:‘(’)hC WCA<OI 12.23 '— = _& eff 1 Q: 0.300, which gives, Q =0.700Q. (21) Therefore, Q- = 0.700(800J) = 560] (b) For a Carnot C — i ycle, eff-— 1 - 7i: = 0.300,-from which Tc =O.700Th = 0.7(500K) = 350 K. 12.33 Q=va= (1.0 kg)(2.26 x 1 051/kg)=226 1 . . 061 AS=Q__ 2.26 x 106 1 x r 373 K =6.1x103 J/K 12.43 The heat discarded is 7096 of the input energy for the plant. Thus, the energy going in to the river each second is: $2 0.7(25 x 108 W) = 17.5 x 108 J/s. Also Q‘— (E)cA1C or nefi— 17.5 x 10 A = 2.8 “C. _ t = m/ t ( t)c (WMISG filo—c) ...
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