How much energy {in kJ} must be removed when 15 mL of liquid (density= 1.0 g/mol) freezes at 0ºC. Atomic Mass: H= 1.01, O=16.0, molar enthalpy of fusion is 6.009 kJ/mole.

Question

Process and answer is needed!

Answer 1

When a liquid freezes, heat energy is released (removed) and the amount is based on the mass of the liquid and the enthalpy of vaporization for that liquid. The equation is q = m∆Hvap where q is heat, m is mass and ∆Hvap is the enthalpy of vaporization.

In the current problem, 15 mls is equivalent to 15 g (since density is 1 g / ml;, not 1 g / mol) = mass

∆Hvap is given as kJ/mol, so we'll convert 15 g of water to moles using the molar mass of water of 18 g/mol

q = ?

m = 15 g x 1 mol / 18 g = 0.833 mols

∆Hvap = 6.009 kJ / mol

Solving for q we have ...

q = (0.833 mols)(6.009 kJ/mol+

q = 5.0 kJ (2 sig. figs.)