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A reversible melting of $32.0\ \mathrm g$ of ice at $0\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$. Solve for $\Delta G, \Delta S_\text{universe}$, and $\Delta A$

I'm also given latent heat of fusion/vaporization, heat capacity at constant pressure, ($C_p$) and the densities of water and ice at the appropriate temperature and pressure,

Solving for $\Delta S_\text{universe}$

from equation sheet: $$\Delta S_\text{vap} = \frac {q_{p,\text{vap}}}{T_\mathrm b}$$ Which is for vaporization and I assume I can transform that into $$\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$ So solving for $\Delta S_\text{system}$ I get $$\Delta S_\text{system}=\Delta S_\text{melt}= \frac {q_{p,\text{fus}}}{T_\mathrm m}$$

But the question asked for $\Delta S_\text{universe}$ not $\Delta S_\text{system}$. I know that $$S_\text{universe}=S_\text{surroundings}+S_\text{system}$$, so $\Delta S_\text{universe}$ must be $$\Delta S_{universe}=\Delta S_\text{surroundings}+\Delta S_\text{system}$$ But I'm given no information on the surroundings. How can I solve this? Is it possible that the change in entropy in the surroundings constant such that $\Delta S_\text{universe}=\Delta S_\text{system}$?

Solving for $\Delta A$ $$A=U-TS$$ $$\Delta A = \Delta U -T\Delta S_\text{sys}$$ $$\Delta A = (q-P\Delta V) -T\Delta S_\text{sys}$$ $$\Delta A = (q-P[\rho_\text{water}m-\rho_\text{ice}m])-T\Delta S_\text{sys}$$

But I don't know how to solve $q$ for this problem. Additionally, I'm not sure if the ice melted completely, so perhaps the change in Volume is less that what I wrote in the equation.

Solving for $\Delta G$ $$\Delta G = \Delta H-T\Delta S_\text{sys}$$ $$\Delta G = \Delta U+P\Delta V-TS_\text{sys}$$ $$\Delta G = (q-P\Delta V)+P\Delta V-TS_\text{sys}$$ $$\Delta G = q-TS_\text{sys}$$

But same as before I don't know how to solve for $q$.

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  • $\begingroup$ What are the initial and final thermodynamic equilibrium states of the system? $\endgroup$ – Chet Miller Oct 15 '16 at 14:12
  • $\begingroup$ Um I know that S(universe) becomes maximum @ equilibrium. I guess gibbs would need to be negative because the melting happened spontaneously. I don't have much intuitive sense with Thermo. $\endgroup$ – Jess L Oct 15 '16 at 23:27
  • $\begingroup$ $$q_{fus}=T_m\Delta S_{syst}$$What does that tell you about $\Delta G$? $\endgroup$ – Chet Miller Oct 16 '16 at 14:56
  • $\begingroup$ dG is Zero?! But the melting is spontaneous! how did you get the equation you just wrote?! what kinds of situations can I say dG is zero? $\endgroup$ – Jess L Oct 17 '16 at 0:14
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    $\begingroup$ I can't make sense of this, but, anyway, $$\Delta H=\Delta U+\Delta(PV)$$. So $$\Delta A=\Delta G-\Delta (PV)$$For constant pressure, $$\Delta A=\Delta G-P\Delta V=-P\Delta V$$where $\Delta V$ is the difference in volume between 32 g of liquid water and 32 g of water ice. $\endgroup$ – Chet Miller Oct 18 '16 at 0:26

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