Question

A student has calibrated his/her calorimeter and finds the heat capacity to be 16.0 J/°C. S/he then determines the molar heat capacity of aluminum. The data are: 25.4 g Al at 100.0°C are put into the calorimeter, which contains 99.1 g H₂O at 19.9°C. The final temperature comes to 23.9°C. Calculate the heat capacity of Al in J/mol·°C.

Answer 1

m(water) = 99.1 g

T(water) = 19.9 oC

C(water) = 4.184 J/goC

m(aluminium) = 25.4 g

T(aluminium) = 100.0 oC

C(aluminium) = to be calculated

C3 = 16.0 J/oC

We will be using heat conservation equation

use:

heat lost by aluminium = heat gained by water and 3

m(aluminium)*C(aluminium)*(T(aluminium)-T) = m(water)*C(water)*(T-T(water)) + C3*(T-T(water))

25.4*C(aluminium)*(100.0-23.9) = 99.1*4.184*(23.9-19.9) + 16.0*(23.9-19.9)

1932.94*C(aluminium) = 1658.5376 + 64

C(aluminium)= 0.8911 J/goC

Molar mass of Al = 26.98 g/mol

use:

molar heat capacity = specific heat capacity * molar mass

molar heat capacity = 0.8911 J/g.oC * 26.98 g/mol

molar heat capacity = 24.04 J/mol.oC

Answer: 24.0 J/mol.oC