Respuesta :
Answer: 357705.6 J
Explanation:
The conversions involved in this process are :
[tex](1):H_2O(s)(-11^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(g)(100^0C)\\\\(4):H_2O(l)(100^0C)\rightarrow H_2O(g)(100^0C)\\\\(5):H_2O(g)(100^0C)\rightarrow H_2O(g)(165^0C)[/tex]
Now we have to calculate the enthalpy change.
[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})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 120 g
[tex]c_{p,s}[/tex] = specific heat of solid water = 2.09 J/gK
[tex]c_{p,l}[/tex] = specific heat of liquid water = 4.18 J/gK
[tex]c_{p,g}[/tex] = specific heat of gaseous water = 1.84 J/gK
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{120g}{18g/mole}=6.7mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[120g\times 4.18J/gK\times (0-(-11))K]+6.7mole\times 6010J/mole+[120g\times 2.09J/gK\times (100-0)K]+6.7mole\times 40670J/mole+[120g\times 1.84J/gK\times (165-100)K][/tex]
[tex]\Delta H=357705.6J[/tex]
Therefore, the energy required to change a 120 g ice cube from ice at -11°C to steam at 165°C is, 357705.6 J
