Find the total entropy of the system?

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SUMMARY

The total change in entropy for the process of freezing 2.00 m³ of water at 0°C and cooling it to -25°C is calculated to be 265854.8 J/K. The formula used includes ΔS=Q/T, where Q is determined using Q=mL for phase change and Q=mcΔT for temperature change. The natural logarithm is employed in the calculation due to the variable temperature during the cooling process, necessitating integration to accurately determine ΔS.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with the concepts of heat transfer and phase changes.
  • Knowledge of the specific heat capacity of water (4190 J/kg·K) and the heat of fusion (336000 J/kg).
  • Basic calculus, particularly integration and natural logarithms.
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  • Study the derivation and application of the entropy formula ΔS=\int {dQ \over T}.
  • Learn about the specific heat capacities of different substances and their impact on entropy calculations.
  • Explore the principles of thermodynamic cycles and their relation to entropy changes.
  • Investigate the role of temperature in phase transitions and its effect on entropy.
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Students and professionals in physics, chemistry, and engineering fields who are studying thermodynamics and entropy calculations, particularly in relation to phase changes and heat transfer processes.

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Homework Statement


If 2.00 m3 of water at 0°C is frozen and cooled to -25°C by being in contact with a great deal of ice at -25°C, what would be the total change in entropy of the process?


Homework Equations


ΔS=Q/T Q=mL Q=mcΔT


The Attempt at a Solution


2000 * 4190 * ln (273/(273-25) + 2000* 336000/273 = 265854.8 J/K
Why is natural log being used?
 
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Hi gibson101! :smile:

The natural log is used because ΔS depends on the temperature, which is changing during cooling.
This means that ΔS needs to be integrated (\Delta S=\int {dQ \over T}) yielding the natural log.
 

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