Student has calibrated his/her calorimeter and finds the heat capacity to be 14.2 j/°c. s/he then determines the molar heat capacity of aluminum. the data are: 25.5 g al at 100.0°c are put into the calorimeter, which contains 99.0 g h2o at 18.6°c. the final temperature comes to 22.7°c. calculate the heat capacity of al in j/mol·°c.

24.03 J/mol.ºC

Explanation:

For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.

-Qal = Qw + Qc (minus signal represents that the heat is lost)

-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc

Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC

-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)

1971.15Cal = 1699.10 + 58.22

1971.15Cal = 1757.32

Cal = 0.89 J/g.ºC

The molar mass of Al is 27 g/mol

Cal = 0.89 J/g.ºC * 27 g/mol

Cal = 24.03 J/mol.ºC

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