A swimmer coming out from a pool is covered with a film of water weighing about `80g`. How much heat must be supplied to evaporate this water ? If latent heat of evaporation for `H_(2)O` is `40.79 kJ mol^(-1)` at `100^(@)C`.

Explanation

We can represent the process of evaportion as

`Delta_(vap)H^(Θ)` for water is `40.79 kJ mol^(-1)` and the molar mass of water is `18.0 g mol^(-1)`.
`80 g` of water is, `(80 g)/(18.0 g mol^(-1))=4.44 mol`
Therefore, heat that must be supplied is as follows:
`q_(p)=nDelta_(vap)H^(Θ)=(4.44 mol)(40.79 kJ mol^(-1))`
`=1.81xx10^(2) kJ`