Question

How much heat to change 1 mole of ice at -25 °C to steam at +125 °C? • heat to warm ice - heat to melt ice (no temperature change) - heat to warm water • heat to boil water (no temperature change) - heat to warm steam

Answer 1

Answer : The amount of heat changes is, 56.463 KJ

Solution :

The conversions involved in this process are :

`(1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(125^oC)`

Now we have to calculate the enthalpy change.

`\Delta H=[m* c_(p,s)* (T_(final)-T_(initial))]+n* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]+n* \Delta H_(vap)+[m* c_(p,g)* (T_(final)-T_(initial))]`

where,

`\Delta H` = enthalpy change or heat changes = ?

n = number of moles of water = 1 mole

`c_(p,s)` = specific heat of solid water =
`2.09J/g^oC`

`c_(p,l)` = specific heat of liquid water =
`4.18J/g^oC`

`c_(p,g)` = specific heat of liquid water =
`1.84J/g^oC`

m = mass of water

`\text{Mass of water}=\text{Moles of water}* \text{Molar mass of water}=1mole* 18g/mole=18g`

`\Delta H_(fusion)` = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

`\Delta H_(vap)` = 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=[18g* 4.18J/gK* (0-(-25))^oC]+1mole* 6010J/mole+[18g* 2.09J/gK* (100-0)^oC]+1mole* 40670J/mole+[18g* 1.84J/gK* (125-100)^oC]`

`\Delta H=56463J=56.463KJ` (1 KJ = 1000 J)

Therefore, the amount of heat changes is, 56.463 KJ