Split_00062 - 8.87 (contd) Air @ 200C: Table B.8 H = 515 kJ...

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8-62 8.87 (cont’d) Air @ 200 ° C: Table B.8 ⇒= ± . H 515 kJ mol Air (dry) @ 50 ° C: Table B.8 ± . H 073 kJ mol HO , 50C 2 v ° bg : Table B.5 = ± . H 2592 104.8 kJ 1 kg 18.0 g kg 10 g 1 mol kJ mol 3 44 81 substance NaOH in mol min HO in k Jmo l Dry air in in out out 2 ² ± ² ± .. . . ² . ± nH n H aq n vH 015 4247 285 147 44 81 1061 515 1061 0 73 −− neglect out in Energy balance: 1900 kJ min transferred to unit Δ Δ E ii n Q H ²²±± == = ∑∑ 8.88 a. Basis : 1 L 4.00 molar H 2 SO 4 solution (S.G. = 1.231) 1 L 1231 g L g 4.00 mol H SO 392.3 g H SO 1231 g H O 46 mol H O mol H O/mol H SO kJ/mol 24 2 2 22 4 Table B.11 s2 4 =⇒ = −= = ⎯⎯ ⎯= 1231 392 3 838 7 57 1164 67 6 . . ± . rH Δ Ref :H O , 25 C 2 l ° ,H SO C 25 ° substance i n m o l HSO k l C , in in out out 2 n H lT n lH n ±± . ± . b g 4657 0 0754 25 400 0 25 4 00 67 6 °= QH T T = = ° Δ 0 4 00 67 6 4657 0 0754 25 52 ... . b g b g C ( The water would not be liquid at this temperature impossible alternative! ) b. Ref : H O , 25 C 2 l
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