Question

How much heat is released when 105g of steam at 100.0C is cooled to ice at -15C? Enthalpy of vaporization of water is 40.67kj/mol, the enthalpy of fusion of water is 6.01kj/mol, the molar heat capacity of water is 75.4J/(mol.C) and the molar heat capacity of ice is 36.4J/(mol.C)

Answer 1

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 2

Answer : The heat released is, 319.28 kJ

Solution :

The conversions involved in this process are :

$(1):H_2O(g)(100^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(s)(100^oC)\\\\(4):H_2O(s)(0^oC)\rightarrow H_2O(s)(-15^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=n\times \Delta H_{condensation}+[n\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{freezing}+[n\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

Mass of water = 105 g

Moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{105g}{18g/mole}=5.83mole$

$c_{p,s}$ = specific heat of solid water = $36.4J/(mol.^oC)$

$c_{p,l}$ = specific heat of liquid water = $75.4J/(mol.^oC)$

$\Delta H_{freezing}$ = enthalpy change for freezing = enthalpy change for fusion = - 6.01 KJ/mole = - 6010 J/mole

$\Delta H_{condensation}$ = 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

$\Delta H=5.83mole\times -40670J/mole+[5.83mole\times 75.4J/(mol.^oC)\times (0-100)^oC]+5.83mole\times -6010J/mole+[5.83mole\times 36.4J/(mol.^oC)\times (-15-0)^oC]$

$\Delta H=-319285.78J=-319.28KJ$     (1 KJ = 1000 J)

Negative sign indicates that the heat is released during the process.

Therefore, the heat released is, 319.28 KJ