Calculating Entropy of Water During Melting Process: Ice Cube on Kitchen Table

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SUMMARY

The discussion focuses on calculating the entropy change of water as an ice cube melts and its temperature rises from 0 degrees Celsius to 25 degrees Celsius. The specific heat capacities used are 2.0 J/g°C for ice and 4.18 J/g°C for water. The formula applied for entropy change is delta(S) = Q/T, with Q calculated as Q = C * m * delta(T). The final entropy change is determined to be 10.52 J/K, emphasizing the need for integration due to the non-constant temperature during the melting process.

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Homework Statement



An ice cube (mass 30 g) at 0 degrees celsuis is left sitting on the kitchen table, where it graduallyy melts. The temperature in the kitchen is 25 degrees celsuis. Calculate the entropy of the water (from melted ice ) as its temperature rises from 0 degrees celsuis to 25 degrees celsuis.

Homework Equations


c(ice)=2.0 J/g*C
c(water)=4.18 J/g*C
delta(S)= Q/T; Q=C*delta(T)=c*m*delta(T)/T

The Attempt at a Solution



delta(S)= Q/T=c*m*delta(T)/T= ((4.18)*(30 g)*(25 Kelvins))/298 Kelvins= 10.52 J/K
 
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[itex]\Delta S = \frac{Q}{T}[/itex] isn't going to give the right answer, because T isn't constant. Try integrating

[tex]dS=c\,m\,\frac{dT}{T}[/itex]<br /> <br /> and plugging the initial and final temperatures into the resulting equation.[/tex]
 
You simply have to integrate (sum over the transformation path)
[tex]dS=\frac{dU}{T} + \frac{PdV}{T} - \sum_i \frac{\mu_i dN_i}{T}[/tex]
Fourier's law tells you that
[tex]dU=\rho c dT[/tex]
hence what is written above.
You could also ask yourself what has happened with the second term (the last one is irrelevant here)
[tex]\frac{PdV}{T}[/tex]
Hint: linked to the ice coefficient of thermal expansion
[tex]\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P[/tex]
 
Last edited:

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