How Does Quenching Copper Affect Entropy Changes in Water?

[kalbuskj31]

Homework Statement

A piece of Copper, consisting of 1 mole, is quenched rapidly from 525^oC into a water bath at 25^oC. Calculate $$\Delta$$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 25^oC.

Homework Equations

$$\Delta$$S = n*Cp/T * dT (when temperature varies)

$$\Delta$$S = $$\Delta$$H / T (when temperature is constant.

The Attempt at a Solution

I think I figured out the change in entropy in the Copper.

$$\Delta$$S_Cu = 1 mole * 25 J/K*mole * $$\int$$ dT/T
$$\Delta$$S = 25 J/K ln T (from 798K to 298K)
$$\Delta$$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 $$\Delta$$H_Cu = -$$\Delta$$H_H2O
$$\Delta$$H_Cu = n*Cp*dT = 1 mole * 25 J/K *mole * (298-798K)
$$\Delta$$H_Cu = -12500J

$$\Delta$$H_H2O = - (-12500J) = 12500J

$$\Delta$$S_H2O = $$\Delta$$H * (1/T) = 12500J / 298K
= 41.95 J/K

$$\Delta$$S_Total = $$\Delta$$S_Cu + $$\Delta$$S_H2O

$$\Delta$$S_Total = -24.63 J/K + 41.95 J/K = 17.32 J/K

 

[Andrew Mason]

Your method and answers are correct.

AM