Calorimeter Experiment to Determine the Specific Heat of Aluminum

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Calorimeter Experiment to Determine the Specific Heat of Aluminum

Presentation transcript:

Calorimeter Experiment to Determine the Specific Heat of Aluminum Assume calorimeter is perfect insulator

Sign Conventions Heat into system (endothermic): +q Heat out of system (exothermic): -q DT = TFinal - TInitial

Heat gained by the water: Heat lost by the aluminum: Solve for C Compare this result with the value provided on your periodic table.

Determination of Calorimeter Constant (Calibration of Calorimeter) In this case, we will not assume that the calorimeter is a perfect insulator. Some of the heat from the hot water will flow into the calorimeter. The hot water will cool off more than the cold water heats up. The difference represents how much heat was required to change the temperature of the calorimeter to TF .

The two volumes of water are combined. Heat gained by cold water: + q = m C DT = (49.539 g) (4.184 J/g oC) (28.9 oC) = + 5990.14 J Heat lost by hot water: - q = m C DT = (51.742 g) (4.184 J/g oC) (-33.0 oC) = - 7144.12 J