8- 118.29 a. 75 liters C H OH789 g1 molliter46.07 gmol C H OH2523llbg=1284()..CTpCH OHo3kJ / (mol C)=+×⋅−01031 0557 103ej(fitting the two values in Table B.2) 55 L H O1000 g1 molliter18.01 gmol H O22ll=3054.CpHO2kJ mol C=⋅°007541284 mol C2H5OH(l) (70.0oC) 3054 mol H2O(l) (20.0o1284 mol C2H5OH (l) (ToC) 3054 mol H2O(l) (ToC) 37025oIntegrate, solve quadratic equation0 12840.1031 0.557 1030540.0754liquids0adiabaticT=44.3 CTTTdTdTQUHQ−×+=Δ ≅Δ⎫⎪⇒⇓⎬=⎪⎭∫∫b. 1.Heat of mixing could affect the final temperature. 2.Heat loss to the outside (not adiabatic) 3.Heat absorbed by the flask wall & thermometer 4.Evaporation of the liquids will affect the final temperature. 5.Heat capacity of ethanol may not be linear; heat capacity of water may not be
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This note was uploaded on 01/07/2012 for the course CH E 210 at Penn State.