Answer : The heat required is, 10.3178 KJ
Solution :
The conversions involved in this process are :
\((1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(85^oC)\)
Now we have to calculate the enthalpy change.
\(\Delta H=n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]\)
where,
\(\Delta H\) = enthalpy change or heat required = ?
m = mass of water = 15 g
\(c_{p,l}\) = specific heat of liquid water = \(4.18J/g^oC\)
n = number of moles of water = \(\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{15g}{18g/mole}=0.83mole\)
\(\Delta H_{fusion}\) = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get:
\(\Delta H=0.83mole\times 6010J/mole+[15g\times 4.18J/g^oC\times (85-0)^oC]\)
\(\Delta H=10317.8J=10.3178kJ\) (1 KJ = 1000 J)
Therefore, the enthalpy change is, 10.3178 KJ
