Answer : The enthalpy change is, 6.2315 KJ
Solution :
The conversions involved in this process are :
\((1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(90^oC)\)
Now we have to calculate the enthalpy change.
\(\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]\)
where,
\(\Delta H\) = enthalpy change = ?
m = mass of ice = 10.0 g
\(c_{p,s}\) = specific heat of solid water = \(2.09J/g^oC\)
\(c_{p,l}\) = specific heat of liquid water = \(4.18J/g^oC\)
n = number of moles of water = \(\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{10g}{18g/mole}=0.55mole\)
\(\Delta H_{fusion}\) = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
\(\Delta H=[10g\times 4.18J/gK\times (0-(-25))^oC]+0.55mole\times 6010J/mole+[10g\times 2.09J/gK\times (90-0)^oC]\)
\(\Delta H=6231.5J=6.2315kJ\) (1 KJ = 1000 J)
Therefore, the enthalpy change is, 6.2315 KJ
