Answer:
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