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Temperature of mixing system, T = After mixing, NaOH Percentage = Here solid NaOH is mixed with H₂0 45.1

Temperature of mixing system, T = After mixing, NaOH Percentage = Here solid NaOH is mixed with H₂0 45.1

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Temperature of mixing system, T = 70°F After miring, NaOH Perseen tage = 45.1. Here solid NaOH is mixed with H₂O Take the basis of 1lb of product. mass of NaOH in Product = Midling Process How chart boso Here m₁ = H₂ m₂ H2 H₂0 (70°F) might = = m3= H3 Apply NapH balance, = mass of m2 ↑NaOH (70°F) M₂, H₂ mass of H₂O added. Enthalpy of H₂0 added. mass of NaOH added. Enthalpy of NaoH added. Product m2 = Q ↓ Mixing m₂ = 0.45x11b 0.45 lb 0.45x m3 0.45X 1 lb 0.45 ль = Enthalpy of Product Product (45%. NaOH) H3&M3. 1 lb (assumed)
Transcribed Image Text:Temperature of mixing system, T = 70°F After miring, NaOH Perseen tage = 45.1. Here solid NaOH is mixed with H₂O Take the basis of 1lb of product. mass of NaOH in Product = Midling Process How chart boso Here m₁ = H₂ m₂ H2 H₂0 (70°F) might = = m3= H3 Apply NapH balance, = mass of m2 ↑NaOH (70°F) M₂, H₂ mass of H₂O added. Enthalpy of H₂0 added. mass of NaOH added. Enthalpy of NaoH added. Product m2 = Q ↓ Mixing m₂ = 0.45x11b 0.45 lb 0.45x m3 0.45X 1 lb 0.45 ль = Enthalpy of Product Product (45%. NaOH) H3&M3. 1 lb (assumed)
Apply overall mass balance mi + m₂ M3 m3-m₂ Solution 100% H₂O 100%. NaOH 45% NaOH ⇒ mi -mi g = Note the enthalpy value from thermodynamic properties table Enthalpy H₁ = 40 Btu/lbm = 1ль - 0.45 ль 0.55 16 - 2m, = Temperature 70°F 70°F 350 в. т / льт На = 93 Btu / льт Now applying energy balance for the given miding System, m₁ H₁ + M₂ M₂ + g M3H3 → [(0.55 16m) x (t0 Btu)] + [0.45 Sbm x 350 Btu ] +9 H2 70°F - 1 lbm x 93 B tu Ibm (93Btu)-(22 Btu) - (157.5 Btu) - 86.5 Btu Hence 86.5 Bto heat should be transferred for each Pound of mass of Solution formed.
Transcribed Image Text:Apply overall mass balance mi + m₂ M3 m3-m₂ Solution 100% H₂O 100%. NaOH 45% NaOH ⇒ mi -mi g = Note the enthalpy value from thermodynamic properties table Enthalpy H₁ = 40 Btu/lbm = 1ль - 0.45 ль 0.55 16 - 2m, = Temperature 70°F 70°F 350 в. т / льт На = 93 Btu / льт Now applying energy balance for the given miding System, m₁ H₁ + M₂ M₂ + g M3H3 → [(0.55 16m) x (t0 Btu)] + [0.45 Sbm x 350 Btu ] +9 H2 70°F - 1 lbm x 93 B tu Ibm (93Btu)-(22 Btu) - (157.5 Btu) - 86.5 Btu Hence 86.5 Bto heat should be transferred for each Pound of mass of Solution formed.
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