ES 5 (phy 40). a) Wrte the zeroth law o thermodynamcs. b) What s thermal conductvty? c) Identyng all es, draw schematcally a P dagram o the arnot cycle. d) What s the ecency o an engne and what s the coecent o perormance o a rergerator. You wll be gong through an mportant phase change very soon. Show that the volume coecent o thermal epanson or an deal gas at constant pressure s temperature dependent and gven by b( ) where s the temperature o the gas epressed n the elvn scale. (Hnt. Use the denton o the coecent o thermal epanson and the equaton o state or an deal gas: P nr). One mole o an deal gas, wth an ntal temperature o, epands sothermally rom to (> ). a) How much work does the gas perorm n the? b) How much heat s delvered to the gas? c) Determne the change n the entropy o the gas. R 8. mol 4. You add 0g o 5 cream to 00g o 60 coee. onsderng both lquds practcally beng water, ther specc heat s 4. /g and s constant n the consdered temperature range. a) Show that the nal temperature o the mture s 55? (Ignore the cup and the atmosphere.) b) How much heat was transerred between the two lquds? c) alculate the amount o entropy produced n the.
- - a) I two systems A and B are separately n thermal equlbrum wth a thrd system, then they are n thermal equlbrum wth each other. b) he rate at whch energy lows by conducton through a slab o crosssectonal surace s proportonal to the area o the surace and the temperature gradent dt d ka d he proportonalty coecent n the above relaton s called the thermal conductvty o the materal (o the slab). c) P adabatc sothermal Q h adabatc sothermal Q c d) Both numbers reer to the rato o the harvested energy to the necessary energy delvered rom the avalable energy source. he ecency o an engne s dened as the amount o work perormed by the engne per unt energy used rom the energy source W e he coecent o perormance o a rergerator s the amount o energy removed rom the cooled space per unt work necessary to perorm the task Q OP c W Q h
- - he volume coecent o thermal epanson relates the change n temperature o a system wth the change n ts volume. By denton () d βd From the equaton o state or an deal gas, n an sobarc nr () ( ) P. We can relate the (derental) change n volume o the gas, at a constant pressure (and temperature ), n terms o a (derental) change n temperature () d nr P d. he rest s math. Solvng () or the volume coecent o thermal epanson, and substtutng equatons () and () or the volume and the change n volume, we nd the answer β( ) d P nr d d nr P d
- - a) Usng the equaton o state or an deal gas, the pressure s nversely proportonal to volume o the gas nr P he work perormed by the gas s thereore W nr mol 8. d nr mol d nr 7 ln 0 ln.7 0 b) Accordng to the rst law o thermodynamcs, the change n the nternal energy s equal to the heat delvered to the system mnus work done by the system. Snce or an deal gas, the nternal energy depends only on temperature, ts value does not change durng the. It means that the heat delvered to the gas s equal to the work done by the gas Q U + W n v + W 0 +.7 0 c) Accordng to the denton o entropy one has to consder a quasstatc leadng rom the ntal to the nal (equlbrum) state. In ths problem, the consdered s quas-statc. We can determne the change n entropy drectly n the actual Q.7 0 7 0
- 4 - a) Snce the specc heat o each part (cream and coee) s constant n the, the ntegraton o heat delvered to each system s smple ) Q mcd mc (, ), As suggested n the tet o the problem, we can assume that durng the mng, heat was transerred between the cream () and coee only. (he total heat delvered to the mture s zero.) ) m c ( ) + m c ( ) 0,, Solvng or the nal temperature we obtan mc, + mc, m, + m, 0g 5 + 00g 60 55 m c + m c m + m 0g + 00g b) From equaton (), we can nd the heat Q, transerred rom the coee to the cream, ether calculatng the heat delvered to the cream Q Q mc(, ) 0g 4. ( 55 5 ) 4.k g (or the heat released by the coee Q Q mc(, ) 00g 4. ( 55 60 ) 4.k ) g c) From the denton o entropy, each system changes ts entropy durng the transer o heat ds, c md c he change n the entropy o the Unverse s thereore S c m ln c m ln + + 4. 0g ln g, ( 55 + 7) ( 5+ 7), m, m + 4. 00g ln g ln, m, ( 55 + 7). ( 60 + 7)