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

4909.486Kj/mol

Step-by-step 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* \Delta H_(condensation)+[n* c_(p,l)* (T_(final)-T_(initial))]+n* \Delta H_(freezing)+[n* c_(p,s)* (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}}=(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* -40670J/mole+[5.83mole* 75.4J/(mol.^oC)* (0-100)^oC]+5.83mole* -6010J/mole+[5.83mole* 36.4J/(mol.^oC)* (-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