Respuesta :
Answer : The enthalpy change is, 7.205 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)(80^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})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 10 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
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{10g}{18g/mole}=\frac{10}{18}mole[/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=[10g\times 2.09J/gK\times (273-248)k]+\frac{10}{18}mole\times 6010J/mole+[10g\times 4.18J/gK\times (353-273)k][/tex]
[tex]\Delta H=7205.39J=7.205KJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change is, 7.205 KJ
