The following information is given for copper at 1atm

Question

boiling point = 2567 °C	Hvap(2567 °C) = 305 kJ/mol
melting point = 1083 °C	Hfus(1083 °C) = 13.0 kJ/mol
specific heat solid= 0.385 J/g°C
specific heat liquid = 0.494 J/g°C

What is H in kJ for the process of freezing a 35.0 g sample of liquid copper at its normal melting point of 1083 °C.

Answer 1

The copper is at its melting point. The melting point is the temperature a solid melts into a liquid. It is also the temperature that the liquid freezes into a solid. The problem statement wants to know the delta H for freezing copper starting at its melting point. Since the melting point and freezing point are the same point (same temperature), no temperature change takes place. The only things we need to know for this question are the H_fus(1083 °C) = 13.0 kJ/mol, the molecular mass of Cu=63.55, and the number of grams.

Grams to moles:

35.0gX1mol/63.55g=0.55075mol

Then the delta H for freezing:

0.55075molX(-13.0)KJ/mol=-7.15975KJ = -7.16Kj to 3 significant figures.

Note that freezing is the negative of melting.

Answer 2

ΔH = mc_f = (35.0 g) (1 mol/63.5 g) (13.0 kJ/mol = 7.17 kJ