Entropy of molten lead freezing

[jmm5872]

Entropy of molten lead "freezing"

Lead melts at 327.5 C.° The latent heat of melting of lead is 24.1 J/g, and the heat capacity of solid lead is 0.14 J/g °C. You take 100 grams of molten lead at a temperature of 327.5 C° and pour it on the sidewalk. The lead freezes and then comes into thermal equilibrium with the sidewalk. The heat capacity of the sidewalk is so large that its temperature stays at 20 °C at all times. How much new entropy have you created by pouring the molten lead on the sidewalk and allowing it to freeze and cool to 20 °C?

Attempt:

This must be split up into two parts...the entropy created during the freezing process at constant temperature

Q = mL = (100 g)(24.1 J/g) = 2410 J

S = Q/T = (2410 J)/(600.65 K) = 4 J/K

The second part is the cooling of the solid lead from 600.65 K to 293.15 K...

ΔS = m$\int$$^{Tf}_{Ti}$(C_v/T)dT

ΔS = mC_vln(Tf/Ti) = -10 J/K

I'm not sure about the signs...it seems like the entropy in freezing a liquid to a solid will decrease (get more ordered).

My other question is about the total entropy created...
Since the sidewalk can absorb all the heat energy that the lead releases does the entropy of the sidewalk technically increase, even though the sidewalk is large enough for the temp change to be negligable? In other words, is the amount of total entropy created the same as the decrease in entropy of the lead?

Thanks

 

[jmm5872]

Sorry, I just realized I need to take the Latent heat of melting and make it negative for freezing...this solves my first problem.

 

[Andrew Mason]

My other question is about the total entropy created...
Since the sidewalk can absorb all the heat energy that the lead releases does the entropy of the sidewalk technically increase, even though the sidewalk is large enough for the temp change to be negligable? In other words, is the amount of total entropy created the same as the decrease in entropy of the lead?

The temperature of the sidewalk does not change appreciably. But the sidewalk does absorb heat, so at the end of the process it is not in the same state that it was before the process started - it has absorbed the heat lost by the lead. It has absorbed this heat at essentially constant temperature. It is in the same state it would be in if this heat flow had occurred reversibly. Since there is reversible heat flow into the sidewalk, what can you say about the change in entropy of the sidewalk?

AM