Respuesta :
Answer : The energy required is, 574.2055 KJ
Solution :
The conversions involved in this process are :
[tex](1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(95^oC)[/tex]
Now we have to calculate the enthalpy change or energy.
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = energy required = ?
m = mass of ice = 1 kg = 1000 g
[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.09J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
n = number of moles of ice = [tex]\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC][/tex]
[tex]\Delta H=574205.5J=574.2055kJ[/tex] (1 KJ = 1000 J)
Therefore, the energy required is, 574.2055 KJ
