Question

A swimmer coming out from a pool is covered with a film of water weighing about 80 g. How much heat must be supplied to evaporate this water? ( $\triangle \textdegree _{VapH}$ for water = +40.79 kJ$mol^{-1}$)

Answer 1

The formula for heat (q) is given by

                                q = n Δ$_{vap}$ H

Given data:

We know the molar mass of water is = 18 g/mol

Therefore moles of water in 80g  = (80/18) mol = 4.44mol

Heat required for evaporating 1 mol of water = 40.18 kJ

Therefore,

                                q = n Δ$_{vap}$ H

                             q = (4.44 mol) ( 40.79 kJ/mol)

                                $q = 1.81\times 10^2 kJ$

Hence, the amount of heat released from the swimmer is $1.81\times 10^2 kJ$ that evaporates the water over his body.

Answer 2

$\huge\pink{1.81×10^2\:KJ}$

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