A student conducts a coffee cup calorimetry experiment to determine the specific

Question

A student conducts a coffee cup calorimetry experiment to determine the specific heat of an unknown metal. First, he collects data to determine the calorimeter constant. He measures 40.0 mL of room temperature water at 20.7 degree C into the calorimeter. He then adds 50.0 mL of water heated to 67.6 degree C to the water in the calorimeter, and through graphical extrapolation, the instantaneous final temperature of the mixture is determined to be 42.1 degree C. He then measures 100.0 mL of water at 20.7 degree C into the calorimeter. He had a 14.54 g sample of the metal in a boiling water bath, which was at 99.0 degree C. He then transferred the metal into the water in the calorimeter, and the final temperature of the water and the metal was 23.2 degree C. Determine the most probable identity of the metal. Determine the quantity of heat energy transferred to the atmosphere when 5.0 ounces of solid calcium carbide (CaC_2) reacts with liquid water to yield solid calcium hydroxide and gaseous acetylene (C_2H_2).

Explanation / Answer

1. heat gained by cold water (qcold) = mCpdT

= 40 x 4.184 x (42.1 - 20.7) = 3581.504 J

heat lost by hot water (qhot) = mCpdT

= 50 x 4.184 x (67.6 - 42.1) = 5334.6 J

The difference is heat gained by calorimeter = 5334.6 - 3581.5 = 1753.1 J

Calorimeter constant Cp(cal) = 1753.1/46.9 = 37.38 J/K

For specific heat of metal,

heat gained by water = 100 x 4.184 x (23.2 - 20.7) = 1046 J

heat lost by metal = heat gained by water

specific heat of metal = 1046/14.54 x (99 - 23.2) = 0.95 J/g.oC

Identity of the metal = Aluminum (Al)

2. For the reaction,

CaC2(s) + 2H2O(l) --> Ca(OH)2 + C2H2

dHo = (-986.6 + 226.73) - (2 x -285.8 - 59.83)

        = -128.44 kJ/mol

moles of CaC2 = 141.75 g/64.1 g/mol = 2.21 mols

So,

Heat energy transferred = -128.44/2.21 = -58.08 kJ

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