Specific heat capacity

Question

5.21 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70 °C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.184 J/g･ °C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.

Answer 1

Molar mass MgSO4 = 120.4 g / mol

q = mC∆T

q = heat = ?

m = mass = 100.0 ml H2O x 1 g / ml = 100 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 6.70º

q = (100 g)(4.184 J/gº)(6.7º)

q = 2803 J .. this is the heat generated from 5.21 g of MgSO4

To find ∆H in kJ/mol, convert 2803 J to kJ and convert 5.21 g to moles:

2803 J x 1 kJ / 1000 J = 2.803 kJ

5.21 g MgSO4 x 1 mol / 120.4 g = 0.0433 mols

∆H = 2.803 kJ / 0.0433 mols = 64.7 kJ / mole