Respuesta :
Answer:
4909.486Kj/mol
Explanation:
the heat would be required to change steam at 100°c to water at 100°c then change the water at that temperature to water at 0°c then change water at 0°c to ice at 0°c then ice at 0°c to ice at -15°c
Answer : The heat released is, 319.28 kJ
Solution :
The conversions involved in this process are :
[tex](1):H_2O(g)(100^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(s)(100^oC)\\\\(4):H_2O(s)(0^oC)\rightarrow H_2O(s)(-15^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=n\times \Delta H_{condensation}+[n\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{freezing}+[n\times c_{p,s}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
Mass of water = 105 g
Moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{105g}{18g/mole}=5.83mole[/tex]
[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]36.4J/(mol.^oC)[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]75.4J/(mol.^oC)[/tex]
[tex]\Delta H_{freezing}[/tex] = enthalpy change for freezing = enthalpy change for fusion = - 6.01 KJ/mole = - 6010 J/mole
[tex]\Delta H_{condensation}[/tex] = enthalpy change for condensation = enthalpy change for vaporization = -40.67 KJ/mole = -40670 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=5.83mole\times -40670J/mole+[5.83mole\times 75.4J/(mol.^oC)\times (0-100)^oC]+5.83mole\times -6010J/mole+[5.83mole\times 36.4J/(mol.^oC)\times (-15-0)^oC][/tex]
[tex]\Delta H=-319285.78J=-319.28KJ[/tex] (1 KJ = 1000 J)
Negative sign indicates that the heat is released during the process.
Therefore, the heat released is, 319.28 KJ
