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Theoretical Final Temperature: Qlost+Qgained =0 = -Qwater Quilver Q = mcAT (50g)(0.235 J/g °C)(T₂ - 80°C) = -(200g)(4.184 J/g °C)(T₂-20°C) T₂ = 19.14550633295

Theoretical Final Temperature: Qlost+Qgained =0 = -Qwater Quilver Q = mcAT (50g)(0.235 J/g °C)(T₂ - 80°C) = -(200g)(4.184 J/g °C)(T₂-20°C) T₂ = 19.14550633295

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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How did this arrive from that value? Because when I tried to solve it manually, the value I get is 20.83...
Theoretical Final Temperature: Qlost+Qgained =0 = -Qwater Quilver Q = mcAT (50g)(0.235 J/g °C)(T₂ - 80°C) = -(200g)(4.184 J/g °C)(T₂-20°C) T₂ = 19.14550633295
Transcribed Image Text:Theoretical Final Temperature: Qlost+Qgained =0 = -Qwater Quilver Q = mcAT (50g)(0.235 J/g °C)(T₂ - 80°C) = -(200g)(4.184 J/g °C)(T₂-20°C) T₂ = 19.14550633295
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