Given that the heat of fusion of water is +6.02 kJ/mol that the heat capacity of H2O(l)H2O(l) is 75.2 J/mol⋅KJ/mol⋅K and that the heat capacity of H2O(s)H2O(s) is 37.7 J/mol⋅KJ/mol⋅K, calculate the heat of fusion of water at -15 ∘C∘C.

The answer to the question is:

The heat of fusion (heat required to melt a quantity of solid water) at -15 ∘C is 6.586 kJ .

Explanation:

To solve the question, we note that you have

H₂O(s) at-15 °C and you are asked find the heat of fusion of the water (ice)

The heat of fusion = Heat required to melt the solid H₂O (ice)

The total heat required to melt the ice is

(Heat required to raise the temperature of the solid water, H₂O(s) from -15 °C to 0 °C) + (Heat required to melt the ice to liquid at 0 °C)

The heat required to raise the temperature of the solid water, H₂O(s) from -15 °C to 0 °C is given by;

Heat required or transferred ΔQ = n·c·ΔT

Where:

ΔQ = Heat transfer required

n = Number of moles to be heated

c = Specific heat capacity in J/mol⋅K

ΔT = Temperature change

We however note that the heat capacity, C = n×c

Therefore we have for the raising of the temperature of the ice from -15 °C to 0 °C is given by

ΔQ = n·c·ΔT = C·(T₂ - T₁)

Where

T₁ = Initial temperature = -15 °C

T₂ = Final temperature = 0 °C

C·= 37.7 J/mol⋅K

Therefore

ΔQ = n·c·ΔT = 37.7 J/mol⋅K×(0 °C - (-15 °C))

=  565.5 J

Also we find the heat required to melt the ice to liquid at 0 °C as follows

q = n ×Δ

Where;

Δ = heat of fusion of water is +6.02 kJ/mol

n Number of moles = 1

q = 1 × +6.02 kJ/mol  = 6.02 kJ

The heat required to raise the temperature of the solid water, H₂O(s) from -15 °C to 0 °C is 565. 5 J + 6.02 kJ

= 6.586 kJ

The heat of fusion of water at -15 ∘C is 6.586 kJ.