Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 kJ/mol*K and that the heat capacity of H2O(s) is 37.7 kJ/mol*K, calculate the heat of fusion of water at -11 K.

To calculate the heat of fusion of water at -11 K, we need to find the heat transfer when water at -11 K changes from solid to liquid at 0°C.

The heat transfer during a phase change is given by the formula:

Q = n * ΔH

Where:
Q is the heat transfer in Joules or kilojoules (kJ)
n is the number of moles of the substance
ΔH is the heat of fusion in Joules or kilojoules per mole (J/mol or kJ/mol)

First, let's determine the number of moles of water involved in the phase change. We can use the ideal gas law formula:

PV = nRT

Since we know the temperature (-11 K), we can assume the pressure is constant (1 atm, standard pressure) and use the molar volume at standard conditions (22.4 L/mol) for the volume. The gas constant R is 0.0821 L·atm/(mol·K).

Solving for n:

n = PV / RT

n = (1 atm * 22.4 L/mol) / (0.0821 L·atm/(mol·K) * (-11 K))

n ≈ -24.5 mol

Since the number of moles cannot be negative, we take the absolute value, resulting in approximately 24.5 mol.

Now, we can calculate the heat transfer using the formula Q = n * ΔH:

Q = (24.5 mol) * (-6.02 kJ/mol)

Q ≈ -147.19 kJ (rounded to two decimal places)

So, the heat transfer during the phase change of water at -11 K is approximately -147.19 kJ.