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What would be the final temperature of the system if 16.6 g of lead at 138 ◦C is dropped into 23.4 g of water at 8.68 ◦C in an insulated container? The specific heat of lead is 0.128 J/g◦C.
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heat lost by lead + heat gained by water = 0 [mass Pb x specific heat Pb x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute all the number and solve for Tfinal. That's the only unknown in that long string.
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