Question

2) From the lecture (compare also text page 193-194) we learn that the maximum amount of work which can be extracted from a system interacting with a single heat reservoir is given by W =-AF where AF = F,-F, is the change of the Helmholtz free energy of the system. Specifically for water at initial temperature To and constant heat capacity Cp interacting with a heat reservoir at temperature Tr this maximum amount of work reads W =-AF =C (T-TR)+CT, In(TR/T%). Perform now an explicit calculation to confirm the To=To(t) with above expression. Let a Carnot cycle run between the lim 7,)=T, water and the reservoir (see sketch). The Carnot engine is small taking in only infinitesimal small amounts of heat per cycle. The water of constant heat capacity Cp is initially at temperature To. The reservoir is at constant temperature Tr. The Carnot cycle runs until equilibrium is reached (hint: take into Carnot account that the temperature of the reservoir initially at To is changing and thus the changes the Carnot efficiency continuously). Calculate the total amount of work produced by the Carnot cycle. Express your result in terms of To, Tr and Cp and compare with W... from above. TR=const (5points) w

Answer 1

This question is based on Carnot engine.

I am giving solution here in image form.

The relation between carnot efficiency and work is as follows:

Carnot efficiency = work done / heat

η = w / q.

(Page no.-ol) Carnot Cycles of flip othetical & wel Noflung issual. This cycle us a theoretical TiD cycle. r bypoflictical) Ceiven by sadio not meal. Cenot. at provides an upper limit for the efficiency of a engine is the conversion of heat into work on the efficiency of a refrigerator in creating the temp. difference by the application of work to the System. n = marinum. l'Heat - 100% work at rid of in real, & engine of soll | lengine = COP “พฤษไอฯ Each seetwens vidabecho Compuer for - coefficient - WORK 100% Temp of I Heat aliff. of performance • Single T.D system exist in a particular state , when a system is taken through a series of dit states & finally it Hesults to the initial state; Then a TiD cycle is said to have occured. Isothernial Expansion od 2yetem wlicke undoyoel Curmot circle is called Adliabatic in Carnot heat engine" expansion . The Carnot leat engine Consist of twor reservoirs Compoussion To HOT RESERVOIR :- at lugh temp. TI (TA) 2) COLD RESERVOIR - at low temp. 2 (Tc) These meservaines have such a large canut of thermal capacity that Huir temperokwees puuetically seemeis constant, during a particulas (single) T.D cycle. That at the COLD 1. HOT end RES ERUDIRL cold out ELT2) The (T) isoleermat RESERVOIR System Conducting (IDEAL GAS) Conducting ( 4. 1 peel off! csalt concepts bypothetic

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Points: theoretically page noos Since the cycle is reversible, therefore there is no change in eutropy in the single T. D cycle since there is no volume change in either reservoir , therefore the work done by any reservoir is nero . A definite ant of entropy (AS) is extracted from hot reservoir & same ant is transferred to the cold als ervoice. so you I Net AS =o7 or 152-5, =o) or (3₂=S, or AS, - AS2] A definite cand of heat Thas (or T, AS) is extracted from hot (48=&or & sies ervoise.. But a smaller amt of heat. TCAS FORTAS) is trouste As) is teous ferred to cold desenvois. COLD Fe d ot the work & 2 c = TEAS 124 TAASI the difference diw these two quantities in equal to done by the system, ut - Те, as - Тодо wa te-Te) As or 10 = (T; - (2) AS) Th=Tigan E = T2 J

Stages of Carnot's cycles - Disothermal Expansion for Act this stage, heat is released 2. hot reservoir & it is absorb by the system. & the boundaries are conducting. • Therefore, the temp. of the system ises. This high temp. causes the gas to expand, Pusting the piston upside & doing work on Surrounding. o although in this step, pressure uses; but temp of gas cloesn't fall & therefore the expansion is Isothermal pour od de finite art of heat (2) ils absorbed in this step, therefore the entropy of system will increase by the cent of AS, = 2h , 2, . In this, there is no change in internal enery. DU=nCvdT [dT=o) Lou=0 Work done by system on Surrounding Pd, r, s , Teoust. Conducting Boundary. Th T

by ho decreasing pressure Pagendo 2 Adiabatic Expansion - ( 2 ) o for this step, the gas in the ter engine is thermally insulated from both reservoirs. pun • The to expand by do decreasing system (sas) continues cling work on the sworounding boosing an amt of internal du cataw is equal to work done du = ordw fou dwl Tau=-vellut, Pr, vr, Temp. abse S (means - entropy-conet INSULATING BOUNDARY rau = nCuci vel become -ve OLEDROS 3) Isothermal Compression (3-):- CONDUCTING BOUNDARY 08 / so thermal Heat Rejection) In this step, heat is transferred to the cold reservoir at const, temp. (To). In this step, Swerounding do work on system. Gw - the in vel , wu=CAT, 15 decrease (by the art of ! IT>, AU=0) U $S2 -223) T dediabatie Compressiou (4-1):- | INSULATING BOUNDARY Work to swerrounding do wosk an system. A w = tue vi Paz AU=Aatw Tire AU=tve [S ] Lunge Efficiency of Carnot cugine: - n output - Wook done - Th Tc) AS I T input input heat input Te AS #S, =S2