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# Calorimetry Pre Lab

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Assume you use calorimetry to calculate the specific heat capacity of a 125.24 g piece of unknown metal. You intially heat the metal to 100.0 Â°C in boiling water. You then drop the chunk of metal into a calorimeter containing 47.22 g of water at 19.7 Â°C. After closing and stiring the calorimeter thoroughly, the metal and water both come to equilibrium at a temperature of 27.6 Â°C.
Part A
What is the temperature change of the water?
100.0 Â°C
19.7 Â°C
7.9 Â°C

Part B
What is the temperature change of the metal?
-72.4 Â°C
125.2 Â°C
-27.6 Â°C

Part C
How much heat was gained by the water? (calculate the qwater)
33.1 J
1561 J
418.4 J

Part D
Knowing the above, what must qmetal?
-33.1 J
-1561 J
-125240 J

Part E
Then what must the the Specific Heat of the metal be?
0.4184 J/gÂ°C
25.00 J/gÂ°C
0.1721 J/gÂ°C