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.
Homework Answers
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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A 229.0 g piece of lead is heated to 84.0oC and then
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A 239.0 g piece of lead is heated to 85.0oC and then dropped into a calorimeter containing 645.0 g of water that initally is at 24.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water. A: 7.89 B: 1.05×101 C: 1.40×101 D: 1.86×101 E: 2.47×101 F: 3.28×101 G: 4.37×101 H: 5.81×10
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