second mole of copper which, initially, is at a temperature the temperature of the two-mole system, which is contained in an adiabatic enclosure, when thermal equilibrium is attained. Why is the common uniform temperature not exactly 50°C? How much heat is transferred and how much entropy is produced by the transfer? The constant pressure molar heat capacity of solid copper varies with temperature as
second mole of copper which, initially, is at a temperature the temperature of the two-mole system, which is contained in an adiabatic enclosure, when thermal equilibrium is attained. Why is the common uniform temperature not exactly 50°C? How much heat is transferred and how much entropy is produced by the transfer? The constant pressure molar heat capacity of solid copper varies with temperature as
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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Transcribed Image Text:3.5 One mole of copper at a uniform temperature of 0°C is placed in thermal contact with a
second mole of copper which, initially, is at a uniform temperature of 100°C. Calculate
the temperature of the two-mole system, which is contained in an adiabatic enclosure,
when thermal equilibrium is attained. Why is the common uniform temperature not
exactly 50°C? How much heat is transferred and how much entropy is produced by the
transfer? The constant pressure molar heat capacity of solid copper varies with
temperature as
Cp = 22.64 +6.28 x 10-³T J/mole.K
(Note: the volume of two copper samples does not change with temperature in this problem. The heat
absorbed by the cold Cu and released by the hot Cu are the same (but have different signs, though),
which can be calculated from the integration of Cp*dT. The entropy change can be calculated by
integration of dS = Cp*dT/T).
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