Posted by **Chris** on Sunday, June 12, 2011 at 4:38am.

100 grams of molten lead (600oC) is used to make musket balls. If the lead shot is allowed to cool to room temperature (21oC), what is the change in entropy (in J/K) of the lead? (For the specific heat of molten and solid lead use 1.29 J/g oC; the latent heat of fusion and the melting point of lead are 2.45 x 104 J/kg and 327oC.)

- College Physics -
**drwls**, Sunday, June 12, 2011 at 8:14am
For each incremental heat loss dQ, there is an entropy loss of the lead that is equal to dQ/T, where T must be in Kelvin. Compute the total entropy loss in three steps:

(1) liquid lead cooling from 873 to 600 C

(2) molten lead turning to solid at a contant temperature of 600 K while the heat of fusion is transferred away, and

(3) solid lead cooling from 600 K to 294 K.

You are supposed to use the same specific heats for the liquid and solid lead in this case; this is an approximation. Call the specific heat of both C.

Entropy loss is:

(1) Integral of dQ/T from T = 873 to 600 = M C ln(873/600), where M is the mass and C is the specific heat,PLUS

(2) Integral of dQ/T as the lead freezes, M*Qf/600, where Qf is the latent heat of fusion, PLUS

(3) Integral of dQ/T from T=600 to T = 294 K, = M C ln(600/400)

Now just add up the numbers. Use numeric values for M, Qf and C.

- College Physics -
**drwls**, Sunday, June 12, 2011 at 8:33am
Check your given specific heat of lead. Yours appears to be off by a factor of about ten. I get 0.127 J/g*K for the solid and 0.120 J/g*K for the liquid, from various web sites.

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