Indian style of cooling drinking water is to kept it is a pitcher having porous walls water comes to the outer surface very alowly and evaporates .Most of the itself and the water is cooles down .Assums that a pitcher containe `10kg` water and `0.2g` of water comes from the itomsphere decrease by ``5^(@)C` specific beat capicity of water` = 4200J kg^(-1)^(@)C^(-1)` and letent head of vaporization of water `= 2.27 xx 10^(6)J kg^(-1)`

Explanation

Energy required to decreases the temp of `10kg` of water to `5^(@)C`
`U = 10 xx 4200J//kg^(@)C xx 5^(@)C`
`= 210.000 = 21 xx 10^(6)J`
`Energy required to for evaporation of water to `0.2kg//sec`
`= 2 xx 10^(-4) xx 2.27 xx 10^(6)`
= 454`
`454J` energy losing system per second
`= (21 xx 10^(4))/(454)`
`21 xx 10^(4)J`energy losing system during one minute
`= (21 xx 10^(4))/(454 xx 60) = 7.7 minute`
The time required to decrease temp by
`5^(@)C is 7.7 minute`