Find ∆H for converting 100.0 g of solid mercury at−75.0 o C to liquid mercury at 25.0 o C at a constant pressure of 1.000 atm

Question

The enthalpy change of fusion of mercury is 2331 J mol−1 . Find ∆H for converting 100.0 g of solid mercury at−75.0 o C to liquid mercury at 25.0 o C at a constant pressure of 1.000 atm. Heat capacity of Hg(l) is 28.0 J/mol*K

Answer 1

Taking the melting point of mercury to be -38.8ºC (Wikipedia) and the specific heat of solid mercury to be 0.14 J/gºC...

Step 1: heat needed to convert 100.0 g solid Hg @ -75ºC to solid Hg @ -38.8ºC:

q = mC∆T = (100.0 g)(0.14 J/gº)(40º) = 560 J

Step 2: heat needed to melt 100.0 g Hg @ -38.8º:

q = m∆Hfusion = (100.0 g)(2331 J/mol)(1 mol/200.6 g) = 1162 J

Step 3: heat needed to raise temp of 100.0 g liquid Hg @-38.8º to 25º:

q = mC∆T = (100.0 g)(28.0 J/mol)(1 mol/200.6g)(63.8º) = 891 J

Step 4: Sum the heats from the previous 3 steps:

560 J + 1162 J + 891 J = 2613 J