A feed solution of 10,000 lb at 130 °F containing 47.0 lb FeSO4 per 100 lb total water is cooled to 80 °F, where FeSO4·7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 per 100 lb total water. The average heat capacity of the feed solution is 0.70 Btu/lb-°F. The heat of solution at 18 °C is -4.4 kcal/gmol FeSO4·7H2O. Calculate the yield of crystals and the total heat to be removed from the system.
A feed solution of 10,000 lb at 130 °F containing 47.0 lb FeSO4 per 100 lb total water is cooled to 80 °F, where FeSO4·7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 per 100 lb total water. The average heat capacity of the feed solution is 0.70 Btu/lb-°F. The heat of solution at 18 °C is -4.4 kcal/gmol FeSO4·7H2O. Calculate the yield of crystals and the total heat to be removed from the system.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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A feed solution of 10,000 lb at 130 °F containing 47.0 lb FeSO4 per 100 lb total water is
cooled to 80 °F, where FeSO4·7H2O crystals are removed. The solubility of the salt is 30.5
lb FeSO4 per 100 lb total water. The average heat capacity of the feed solution is 0.70
Btu/lb-°F. The heat of solution at 18 °C is -4.4 kcal/gmol FeSO4·7H2O. Calculate the yield
of crystals and the total heat to be removed from the system.
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