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fluidistic
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Homework Statement
A kilogram of water at 0°C is put in contact with a reservoir at 100°C. When the water reaches 100°C, what is the change of entropy of the water, the reservoir, and the whole system?
Homework Equations
[tex]\Delta S = \int \frac{dQ}{T}[/tex].
The Attempt at a Solution
I realize that the change of entropy of the water is positive while the change of entropy of the tank must be negative. However the sum of the changes of entropy must be positive since it's an irreversible process.
However I don't reach this. What I did was, for the water : [tex]\Delta S =mc \int_{T_0}^{T_{100}} \frac{dT}{T}=1000g \cdot \frac{1 cal}{g K}\ln \left ( \frac{373}{273} \right )=74.66J/K[/tex].
Now the change of entropy of the tank : [tex]\Delta S = \frac{Q}{T}[/tex] since [tex]T[/tex] is constant.
[tex]Q=-mc100K \Rightarrow \Delta S=-1120.64J/K[/tex].
Sidenote : I considered 1cal=4.18 J.
Where did I go wrong?