How Does Quenching Copper Affect Entropy Changes in Water?

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SUMMARY

The discussion focuses on calculating the entropy changes (\DeltaS) for a quenching process involving 1 mole of Copper, which is rapidly cooled from 525°C to 25°C in a water bath. The specific heat capacity (Cp) of Copper is given as 25 J/K·mole. The calculated entropy change for Copper (\DeltaSCu) is -24.63 J/K, while the entropy change for the water bath (\DeltaSH2O) is 41.95 J/K. The total entropy change (\DeltaSTotal) for the system is determined to be 17.32 J/K, indicating an overall increase in entropy.

PREREQUISITES
  • Understanding of thermodynamic concepts such as entropy and enthalpy.
  • Familiarity with the equations for calculating entropy changes (\DeltaS = n*Cp/T * dT and \DeltaS = \DeltaH / T).
  • Knowledge of specific heat capacity (Cp) and its application in thermodynamic calculations.
  • Basic proficiency in calculus for evaluating integrals in entropy calculations.
NEXT STEPS
  • Study the principles of thermodynamics, focusing on the laws of entropy and enthalpy.
  • Learn about the heat transfer mechanisms involved in phase changes and temperature changes.
  • Explore advanced topics in thermodynamics, such as Gibbs free energy and its relation to entropy.
  • Investigate the implications of entropy changes in chemical reactions and physical processes.
USEFUL FOR

Students and professionals in chemistry and physics, particularly those studying thermodynamics, heat transfer, and material science. This discussion is beneficial for anyone involved in calculating entropy changes in thermal processes.

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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 \DeltaS 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



\DeltaS = n*Cp/T * dT (when temperature varies)

\DeltaS = \DeltaH / T (when temperature is constant.

The Attempt at a Solution



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

\DeltaSCu = 1 mole * 25 J/K*mole * \int dT/T
\DeltaS = 25 J/K ln T (from 798K to 298K)
\DeltaS = -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 \DeltaHCu = -\DeltaHH2O
\DeltaHCu = n*Cp*dT = 1 mole * 25 J/K *mole * (298-798K)
\DeltaHCu = -12500J

\DeltaHH2O = - (-12500J) = 12500J

\DeltaSH2O = \DeltaH * (1/T) = 12500J / 298K
= 41.95 J/K

\DeltaSTotal = \DeltaSCu + \DeltaSH2O

\DeltaSTotal = -24.63 J/K + 41.95 J/K = 17.32 J/K
 
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Your method and answers are correct.

AM
 

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