- #1
kalbuskj31
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Homework Statement
A piece of Copper, consisting of 1 mole, is quenched rapidly from 525oC into a water bath at 25oC. Calculate [tex]\Delta[/tex]S for the Copper and the water bath. Assume that Copper has a constant Cp of 25 J/K and that the water bath is so large (infinitely), that its temperature remains essentially unchanged at 25oC.
Homework Equations
[tex]\Delta[/tex]S = n*Cp/T * dT (when temperature varies)
[tex]\Delta[/tex]S = [tex]\Delta[/tex]H / T (when temperature is constant.
The Attempt at a Solution
I think I figured out the change in entropy in the Copper.
[tex]\Delta[/tex]SCu = 1 mole * 25 J/K*mole * [tex]\int[/tex] dT/T
[tex]\Delta[/tex]S = 25 J/K ln T (from 798K to 298K)
[tex]\Delta[/tex]S = -24.63 J/KThe entropy for the water bath is where I'm having trouble. The change in the enthalpy and entropy in the water bath comes from the heat absorbed from the water quenching of the copper. Can someone please double check this for me?
so [tex]\Delta[/tex]HCu = -[tex]\Delta[/tex]HH2O
[tex]\Delta[/tex]HCu = n*Cp*dT = 1 mole * 25 J/K *mole * (298-798K)
[tex]\Delta[/tex]HCu = -12500J
[tex]\Delta[/tex]HH2O = - (-12500J) = 12500J
[tex]\Delta[/tex]SH2O = [tex]\Delta[/tex]H * (1/T) = 12500J / 298K
= 41.95 J/K
[tex]\Delta[/tex]STotal = [tex]\Delta[/tex]SCu + [tex]\Delta[/tex]SH2O
[tex]\Delta[/tex]STotal = -24.63 J/K + 41.95 J/K = 17.32 J/K