Question

The enthalpy of vaporization of liquid water at 100°C is 2257 kJ/kg. Determine the enthalpy for apordato of iuod eeat capacity of liquid water is 4 154 kJl(kg °C) and for water vapor is 1.859 kJ/(kg °C)

Answer 1

Step-by-step explanation:

The given data is as follows.

`T_(1) = 100^(o)C`,
`T_(2) = 10^(o)C`

`\Delta H_(vap1)` = 2257 kJ/kg,
`\Delta H_(vap2)` = ?

For water,
`C_(p)` = 4.184
`kJ/kg ^(o)C`

Formula to calculate heat of vaporization is as follows.

`\Delta H_(vap1) - \Delta H_(vap2)` =
`C_(p)(T_(1) - T_(2))`

Hence, putting the values into the above formula as follows.

`\Delta H_(vap1) - \Delta H_(vap2)` =
`C_(p)(T_(1) - T_(2))`

`2257 kJ/kg - \Delta H_(vap2)` =
`4.184 kJ/kg ^(o)C (100 - 10)^(o)C`

`\Delta H_(vap2)` = 2257 kJ/kg - 376.56 kJ/kg

= 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at
`10^(o)C` is 1880.44 kJ/kg.