Answer : The heat required is, 10.3178 KJ
Solution :
The conversions involved in this process are :
[tex](1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(85^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat required = ?
m = mass of water = 15 g
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{15g}{18g/mole}=0.83mole[/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=0.83mole\times 6010J/mole+[15g\times 4.18J/g^oC\times (85-0)^oC][/tex]
[tex]\Delta H=10317.8J=10.3178kJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change is, 10.3178 KJ
