Find the total entropy of the system?

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The discussion focuses on calculating the total change in entropy when 2.00 m³ of water at 0°C is frozen and cooled to -25°C. The relevant equations include ΔS=Q/T, Q=mL, and Q=mcΔT. The calculation provided yields a total entropy change of 265854.8 J/K. The use of natural logarithm in the calculation is explained as necessary due to the changing temperature during the cooling process, requiring integration of the entropy change. Understanding the integration process is crucial for accurately determining entropy in variable temperature scenarios.
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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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