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Entropy and heat

by Benzoate
Tags: entropy, heat
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Benzoate
#1
Mar15-08, 09:22 PM
P: 569
1. The problem statement, all variables and given/known data

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.

2. Relevant 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
3. 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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Mapes
#2
Mar16-08, 08:26 AM
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P: 2,532
[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]

and plugging the initial and final temperatures into the resulting equation.
SeniorTotor
#3
Mar21-08, 11:16 AM
P: 15
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]


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