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how much of the aluminum will solidify when thermal equilibrium is reached

how much of the aluminum will solidify when thermal equilibrium is reached

Introduction to Chemical Engineering Thermodynamics
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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b) how much of the aluminum will solidify when thermal equilibrium is reached?

 

In a laboratory, 100.0 g of molten aluminum with a temperature of 932.0 K are poured into 1.0 Lof water at a room temperature of 295.2 K found in an ideal calorimeter. If the melting temperature of aluminum is 932.0 K, a) What will be the final temperature of the water-aluminum system when thermal equilibrium is reached? b) how much of the aluminum will solidify when thermal equilibrium is reached? c) How much aluminum should be added to bring the liter of water to its boiling point? d) How would the analysis have changed if instead of 100.0 g of aluminum they indicate that we have 1.0 kg of aluminum? Explain without doing any calculations. Liquid CH,0 = 4190 Vapor CH,0 = 1840 kg:K Solid aluminium = 910- kg ·K . 6y L,CH,0 = 2.26 - 106 kg Lf aluminium = 3.97 · 105 kg kg pwater = 4190
Transcribed Image Text:In a laboratory, 100.0 g of molten aluminum with a temperature of 932.0 K are poured into 1.0 Lof water at a room temperature of 295.2 K found in an ideal calorimeter. If the melting temperature of aluminum is 932.0 K, a) What will be the final temperature of the water-aluminum system when thermal equilibrium is reached? b) how much of the aluminum will solidify when thermal equilibrium is reached? c) How much aluminum should be added to bring the liter of water to its boiling point? d) How would the analysis have changed if instead of 100.0 g of aluminum they indicate that we have 1.0 kg of aluminum? Explain without doing any calculations. Liquid CH,0 = 4190 Vapor CH,0 = 1840 kg:K Solid aluminium = 910- kg ·K . 6y L,CH,0 = 2.26 - 106 kg Lf aluminium = 3.97 · 105 kg kg pwater = 4190
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