Respuesta :
Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :
[tex](1):H_2O(s)(-18^0C)\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(l)(25^0C)[/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}+[m\times c_{l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 72.0 g
[tex]c_{s}[/tex] = specific heat of ice = [tex]2.09J/g^0C[/tex]
[tex]c_{l}[/tex] = specific heat of liquid water = [tex]4.184J/g^0C[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C][/tex][tex]\Delta H=34279.8J=34.3kJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
