Calculate the heat energy released when 26.5 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.

Question

Constants for mercury at 1 atm
heat capacity of Hg(l)Hg(l)	28.0 J/(mol⋅K)28.0 J/(mol⋅K)
melting point	234.32 K
enthalpy of fusion	2.29 kJ/mol

Answer 1

Let us first convert 26.5 g Hg to moles since the constants are given in units per mole

26.5 g Hg x 1 mol Hg / 200.59 g = 0.1321 mols Hg

Step 1: heat released when 26.5 g Hg(l) @ 25ºC (298.15K) goes to 234.32K

q = mC∆T = (0.1321 mol)(28.0 J / mol)(-64K) = -237 J

Step 2: heat released when 26.5 g Hg changes phases from liquid to solid @ 234K:

q = ∆Hf x m = 2.29 kJ / mol x 0.1321 mol = 0.303 kJ = -303 J

Step 3: total up the heat released in each step:

237 J + 303 J = 540 J of heat are released