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A 229.0 g piece of lead is heated to 84.0oC and then dropped into a calorimeter...

A 229.0 g piece of lead is heated to 84.0oC and then dropped into a calorimeter containing 526.0 g of water that initally is at 15.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water.

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Answer #1

specific heat lead = 0.127 J/g°C
specific heat water = 4.18 J/g°C

229.0 * 0.127 ( T - 84.0) + 526 * 4.18 ( T - 15.0)=0

=> T = 15.9°C

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