A coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperature 80 C. A salt weighing 5.852 g is quickly added.

Question

A coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperature 80 C. A salt weighing 5.852 g is quickly added. The salt has a molar mass of 333.799 g/mol. The final temperature of the solution is 27.973 C.

Assume no heat loss to the surroundings. Assume the specific heat capacity of the solution is equal to that of pure water, and that the mass of the solution is equal to the mass of the solid plus the mass of water in the calorimeter.

What is the molar heat of solution for the salt in kJ/mol?

Please show steps.

Answer 1

q = mC∆T

heat = mass x specific heat x change in temperature

q = ?

m = 100.000 g water + 5.852 g salt = 105.852 g

C = 4.184 J/g-deg

∆T = 52.027 degrees

q = (105.852 g(4.184 J/g-deg)(52.027 deg)

q = 23,042 J = 23.042 kJ (since the temperature went down from 80º to ~28º, the reaction is endothermic)

To find molar heat of solution, we now must find moles of salt that dissolved:

5.852 g x 1 mol/333.799 g = 0.01753 moles

∆H_soln = 23.042 kJ/0.01753 moles = 1314 kJ/mole