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 $$Q=(10)(0.031)(327-400)-(5.5)(10)+(10)(0.033)(25-327)=-177.29 cal(\frac{4.184 J}{1 cal})=-741.8 J$$

  • 1
    $\begingroup$ One of your values is suspiciously close to what one would get if forgetting to convert C -> K... $\endgroup$ – user10851 Sep 8 '13 at 3:58
  • $\begingroup$ I added an edit of my actual calculations. So please let me know what I did wrong $\endgroup$ – Greg Harrington Sep 8 '13 at 6:17
  • $\begingroup$ I changed the temperature to Kelvin in the natural log arguments so hopefully it is correct now $\endgroup$ – Greg Harrington Sep 8 '13 at 16:08
  • $\begingroup$ Though in this case the answers are correct one should stress that entropy always increases for CLOSED systems. It happened that your problem could be considered as a closed system. For example entropy decreases in DNA formation and all of life's processes whose closed system is the earth environment itself. $\endgroup$ – anna v Sep 9 '13 at 4:27

You carried out your calculation using degrees centrigrades instead of kelvins. That will lead to wrong results.

  • $\begingroup$ So I should use Kelvin in the natural log arguments for the entropy changes of the lead? $\endgroup$ – Greg Harrington Sep 8 '13 at 16:01

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.

  • $\begingroup$ I added complete calculations of my problem $\endgroup$ – Greg Harrington Sep 8 '13 at 6:15
  • $\begingroup$ You can use Celsius for changes in temperature, like when calculating the heat. But change in entropy is deltaQ/T where T has to be an absolute temperature. You have to use Kelvin for that. Confession: I did the same thing when I first tried to solve the problem ;) $\endgroup$ – jdj081 Sep 8 '13 at 12:01
  • $\begingroup$ I did use Kelvin when calculating $\Delta Q/T$. I think I messed up by not using Kelvin in the natural log arguments for the lead. $\endgroup$ – Greg Harrington Sep 8 '13 at 16:02
  • $\begingroup$ When in doubt, you can ALWAYS use kelvin and then you don't have to worry about it. $\endgroup$ – jdj081 Sep 9 '13 at 4:46

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.

  • $\begingroup$ I just changed the temperature to Kelvin in the natural log arguments so everything should be right now I hope $\endgroup$ – Greg Harrington Sep 8 '13 at 16:08

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