A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K.
`Delta_("vap")H^(Ө)` for water
at 298K =44.01 kJ `"mol"^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of water
We know the mass of water is given as 18 grams. The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \, \text{g}}{18 \, \text{g/mol}} = 1 \, \text{mol} \]
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