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April 1, 2015

April 1, 2015

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

- College Physics -
**drwls**, Sunday, June 12, 2011 at 8:14amFor 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 8:33amCheck 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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