Respuesta :
Answer : The amount of heat changes is, 56.463 KJ
Solution :
The conversions involved in this process are :
[tex](1):H_2O(s)(-25^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)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(125^oC)[/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 or heat changes = ?
n = number of moles of water = 1 mole
[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]
[tex]c_{p,g}[/tex] = specific heat of liquid water = [tex]1.84J/g^oC[/tex]
m = mass of water
[tex]\text{Mass of water}=\text{Moles of water}\times \text{Molar mass of water}=1mole\times 18g/mole=18g[/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=[18g\times 4.18J/gK\times (0-(-25))^oC]+1mole\times 6010J/mole+[18g\times 2.09J/gK\times (100-0)^oC]+1mole\times 40670J/mole+[18g\times 1.84J/gK\times (125-100)^oC][/tex]
[tex]\Delta H=56463J=56.463KJ[/tex] (1 KJ = 1000 J)
Therefore, the amount of heat changes is, 56.463 KJ
