Respuesta :
Answer:
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[tex]_s[/tex]·(T₂ - T₁)
Where
T₁ = Initial temperature = -15 °C
T₂ = Final temperature = 0 °C
C[tex]_s[/tex]·= 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[tex]H_f[/tex]
Where;
Δ[tex]H_f[/tex] = 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.
