PS5_2011Ans

# PS5_2011Ans - Name Section Rec TA Ch 1b Solution Set for...

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Name: Ch 1b, Solution Set for Problem Set Five Section: Sid e 1 of 12 Due Friday, Feb. 12, 2010 Rec TA: at 4 PM in the Drop Box 1 . En t halpy Chang e & Pha s e Tran s i t ion s (24 poin ts ) Calculate ' H (units: kJ) for this process: Process Start: 2.00 mol of steam at 130 q C Process End: 2.00 mol of ice at ± 15.0 q C Use ǻ ƨ f u s ion = 6.03 kJ mol -1 and ǻ ƨ vap = 40.7 kJ mol -1 for the specific molar heats of fusion and vaporization, respectively, at their transition temperatures. (NOTE: 6RPHWLPHV \RX¶OO VHH D FDUDW RYHU WKH V\PERO IRU HQWKDOS\ ZKHQ LW KDV XQLWV RI HQHUJ\ SHU XQLW PROH DQG LW¶V FDOOHG VSHFLILF PRODU HQWKDOS\² 2WKHU WLPHV SHRSOH ZL ll just use the same symbol, H (without the carat), for enthalpy, specific molar enthalpy and specific mass enthalpy ± watch the units c lo s e ly to determine if you have to multiply by the number of moles or the mass.) You may assume that specific molar heat capacity (see Table below) is independent of temperature for each phase. H e at c apa c ity ( c P ) J mol -1 K -1 Ice 38.0 Liquid water 74.5 Steam 36.0 You may treat steam as an ideal gas, and you may assume the process is isobaric. Solu t ion Break the process into five steps, all occurring at 1.00 atm. ( 5 poin ts for clearly stating these states: picture/graph or words are OK) State A o State B: cool steam from 130 q C to 100 q C State B o State C: condense steam at 100 q C State C o State D: coolliquid water from 100 q C to 0.00 q C State D o State E: freeze water at 0.00 q C State E o State F: cool ice from 0.00 q C to ± 15.0 q C A o B: cool steam from 130 q C to 100 q C H A o B = nc P , steam T ( 2 poin ts ) ' o B A H (2.00 mol) ( 36.0 J/K/mol )( 100 q C - 130 q C) ' o B A H -2160 J ( 1 poin t )

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Name: Ch 1b, Solution Set for Problem Set Five Section: Sid e 2 of 12 Due Friday, Feb. 12, 2010 Rec TA: at 4 PM in the Drop Box B o C: condense steam at 100 q C H B o C = n Ö H vap ( 2 poin ts ) ' o C B H (2.00 mol)(-40.7 kJ/mol) ' o C B H -81400 J ( 1 poin t ) C o D: cool liquid water from 100 q C to 0.00 q C T n c H liquid P D C ' ' o , ( 2 poin ts ) ' o D C H (2.00 mol) ( 74.5 J/K/mol )( 0.0-100.0 q C) ' o D C H -14900 J ( 1 poin t ) D o E: freeze water at 0.00 q C H D o E = n Ö H fus ( 2 poin ts ) ' o E D H (2.00 mol)(-6.03 kJ/mol) NO T E: s hould b e 6 . 01 kJ / mol ' o E D H -12100 J ( 1 poin t ) E o F: cool ice from 0.00 q C to ± 15.0 q C H E o F = nc P , ice T ( 2 poin ts ) ' o F E H (2.00 mol)(38.0 J/K/mol)( ± 15.0-0.0 q C) ' o F E H -1140 J ( 1 poin t ) Overall Process (A o F) F E E D D C C B B A F A H H H H H H H o o o o o o ' ± ' ± ' ± ' ± ' ' ' ( 1 poin t ) ' H -2160 J + -81400 J + -14900 J + -12100 J + -1140 J ' H -111700 J H = 111.7 kJ ( 3 poin ts t o t al : 2 points for value, 1 point for sig figs) NO T E: fu s ion e nthalpy for wat e r = 6 . 01 kJ / mol a s s tat e d in th e probl e m , whi c h giv e s th e final an s w e r a s -111 . 6 kJ . A note about sig figs: There are four significant figures in the answer because when added, the least accurate value is significant to the hundreds place.
Name: Ch 1b, Solution Set for Problem Set Five Section: Sid e 3 of 12 Due Friday, Feb. 12, 2010 Rec TA: at 4 PM in the Drop Box 1RWH³ , GLGQ¶W FRQYHUW WHPSHUDWXUH IURP &HOVLXV WR .HOYLQ HYHQ WKRXJK WK e denominator for the heat capacity has units of Kelvin. As long as you deal with t e mp e ra t ur e diff e r e n ce s

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