ChBE 2100 Exam 3 Solution

ChBE 2100 Exam 3 Solution - Problem #1 (10 pts) A) (5 pts)...

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Problem #1 (10 pts) A) (5 pts) Vaporization of water at 37°C 1) Latent heat of vaporization interpolated from data at 36°C and 38°C is 2414 kJ/kg. V = (2414 J/g)(1 cal/4.184 J) = 577 cal/g 2) Consider a hypothetical path: Liquid at 37°C Vapor at 37°C Overall path 1 3 Warm liquid Cool vapor 2 Liquid at 100°C Vapor at 100°C Vaporization 1 = C p (T 2 – T 1 ) = (75.4 J/mol·°C)(1 mol/18 g)(1 cal/ 4.184 J) (100°C – 37°C) = 63.1 cal/g 2 = (2257 J/g)(1 cal/4.184 J) = 539 cal/g 3 = = = -28.4 cal/g 1 + 2 + 3 = 574 cal/g
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B) (5 pts) Cold pack temperature Heat needed to dissolve 60 g NH 4 NO 3 : = n S n = 60 g × (1 mol/80 g) = 0.75 mol S = 25.69 kJ/mol = n S = 19.24 kJ This is the heat that will be lost from H 2 O = H2O = -19.24 kJ Using the available information on water: H2O = nC p ( T) -19.24 kJ = (11.11 mol)(75.4 J/mol·°C)( T) ( T) = -22.95°C Cold pack initially at 25°C: T f = T o + T = 25°C - 22.5°C = 2.05°C C) (10 pts)Vaporization of ammonia From
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ChBE 2100 Exam 3 Solution - Problem #1 (10 pts) A) (5 pts)...

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