Shouldn't the sign of generated entropy always be positive? I have a process where 10 g of liquid lead at 400 C is dropped into a water bath that is at 25 C. The lead solidifies over time and comes to thermal equilibrium with the water bath. The bath is so large that it stays at nearly 25C. The specific heat of solid lead is $0.031 \frac{cal}{g*C}$ and the specific heat of liquid lead is $0.033 \frac{cal}{g*C}$. The latent heat of fusion is $5.5 \frac{cal}{g}$. I am confused by the sign of the entropy generated in the process. The melting point of lead is 327 C.
I found the change in entropy of the lead as it comes into thermal equilibrium to be $-4.15 \frac{J}{K}$. I also found the entropy change of the water to be $2.488 \frac{J}{K}$ by using $\Delta S_{w}=\frac{\Delta U_{w}}{T_w}$.
I am getting a negative entropy generated in the process because I am using the equation $$\Delta S_{h2o}+\Delta S_{lead}=\sigma$$. This leads to a entropy generated of $-1.662 \frac{J}{K}$
I am confused by this because I have learned that the entropy generated by a process is always positive no matter what and this is going against that. If anyone could help out with this then it would be greatly appreciated
Edit: Here are the equations I used for the cooling of lead from 400C to 327C, the solidification of lead, and then the cooling from 327C to 25C, respectively
Edit: I converted the temperature to Kelvin for the natural log arguments
$$\Delta S=[(10g)[(0.031 \frac{cal}{g*C})Ln(\frac{600.15}{673.15})]](\frac{4.184J}{1cal})=-0.1488\frac{J}{K}$$
$$\Delta S=[(10g)\frac{5.5\frac{cal}{g}}{600.15 K}](\frac{4.184 J}{1 cal})=-0.383\frac{J}{K}$$
$$\Delta S=[(10g)[(0.033 \frac{cal}{g*C})Ln(\frac{298.15}{600.15})]](\frac{4.184J}{1cal})=-0.9659\frac{J}{K}$$
$$\Delta S_{lead} = -0.9659 + -0.261 + -0.1488 = -1.3757 \frac{J}{K}$$
and the total heat lost by the lead was calculated as
$$Q=(10)(0.031)(327-400)-(5.5)(10)+(10)(0.033)(25-327)=-177.29 cal(\frac{4.184 J}{1 cal})=-741.8 J$$
 A: You carried out your calculation using degrees centrigrades instead of kelvins. That will lead to wrong results.
A: You are correct that the total entropy of the lead + water should increase during this process. I worked it out and got different results than you; if necessary you may want to edit and show some work or intermediate results of your calculations.
I took separately the entropy changes of cooling the liquid lead, solidifying the lead, and cooling the solid lead. Being careful with units, I ended up with -0.184, 0.445, and -1.013 J/K respectively, for a total change of -1.643 J/K for the lead. The total heat lost by the lead was ~840 J, which added to the water at 25C gave me +2.816 J/K for the water.
Putting these together gives +1.17 J/K which is positive as expected.
BTW, good job asking the question. That is one great way to evaluate each answer you get; you were right to question the negative answer.
A: My answers are slightly different from the ones posted above. I got a total change of  -1.45 J/K for the lead, +2.416 J/K for the water and an entropy generation of 0.966 J/K. I am curious why our answers are different because we followed the same method. 
