Answer: The total amount of energy needed overall in kilojoules is, 21.8
Explanation:
The conversions involved in this process are :
[tex](1):H_2O(s)(-58^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of ice = 47.7 g = 0.0477 kg (1kg=1000g)
[tex]c_{s}[/tex] = specific heat of solid water = [tex]2.108kJ/kg^0C[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{47.7g}{18g/mole}=2.65mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 kJ/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[0.0477kg\times 2.108kJ/kg^0C\times (0-(-58))^0C]+2.65mole\times 6.01kJ/mole[/tex]
[tex]\Delta H=21.8kJ[/tex]
Therefore, the enthalpy change is, 21.8 kJ
