A 44.0 gram sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter containing 80.0 g of water at 24.0 oC. The final temperature of the system was found to be 28.4 oC. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 12.4 J/oC.)
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heat lost by metal = heat gained by water + heat gained by calorimeter
44g x c x (99 - 28.4) = 80g x 4.184J/gºC x (28.4 - 24) + (12.4J/ºC x (28.4 - 24))
since deltaT water = deltaT calorimeter:
44g x c x 70.6ºC = 4.4ºC(80g x 4.184J/g-ºC + 12.4J/ºC)
3106.4c = 1527.33
c = 0.49J/gºC
44g x c x (99 - 28.4) = 80g x 4.184J/gºC x (28.4 - 24) + (12.4J/ºC x (28.4 - 24))
since deltaT water = deltaT calorimeter:
44g x c x 70.6ºC = 4.4ºC(80g x 4.184J/g-ºC + 12.4J/ºC)
3106.4c = 1527.33
c = 0.49J/gºC