Question

1.18 (a) Two fluids F1, F2 of fixed volumes and constant heat capaci- ties C1, C, are initially at temperatures Ti, T, (T; > Tı), respectively. They are adiabatically insulated from each other. A quasistatically acting Carnot engine E uses F, as heat source and F2 as heat sink, and acts between the systems until they reach a common temperature, To say. Obtain an expression for T, and for the work done by the Carnot engine. (b) If a common temperature is established by allowing direct heat flow between F, and F2, what is the final temperature and what is the change in entropy? (c) Show that for all positive C1, C2 this change is an increase.

Answer 1

a) from of the mo dynamics, and law as do ds=dQrev .: Considering 9. Tds partial different V wie 7 of), as a co osis-ch at For flund, the temperature is common kemp. attained – So entropy lost s = flows into it For fluid 2 B The fluid 2 gains entropy as heat So the temperature rises from Ta to To. So entispy gained, Sea Caldt s. -cifen ) Tz asen To - den ty] Car Sea & Cala ( 7 ) cannot engine undergoes a cyclic process

So S₂-S, = 0 w Czen (s) + Gino). † G+C2 = ln 1 Cit2 † (2 a Tx T, Tz CitC2 To = Tie (T) = 1 = † Taxi I law. Since for a I dg+dw=do] dr fork done can be obtained from cydic process du=0 do--dw Heat lost by F, . o - c. = - G (to -7) . Q = C, (T, - To) caldt T2 Heat gerned by Fr, Q = Q = C (700-tz) work done & Q. 9, - Q2 - GT GT-C, J + G To - Toffth)

Work done is 8 - Q. wa C, T, -C, To - C. Tot (₂ Tz W = C, T, & C T - To (C+C2). (6) In the conservation absence of perfomance of work, energy gields c, (T.-To) - Ca (To-Tz) = 0 exchange is direct / As the C.T. - C. To = Ca To - Ca Tz To (CI+ (₂) & c.T, + C₂ Tz To - CT + C2 Tz Entropy lost by & Fi is a Sana la = c.la I Entropy gained by F is S- & les at = cala (2) :AS= S2 - S, - Entropy change C, A( ) - Gen czin s) + Cien E) T2