An open bowl, $0\mathrm{.}5$ m in diameter, contains water at $360$ K which is evaporating into

atmosphere. If the air currents are sufficient strong to remove the water vapor as it formed and if

the resistance to its mass transfer in air is equivalent to that of $2$ mm layer for condition of

molecular diffusion, what will be the rate of cooling due to evaporation. Water can be considered

as well mixed and water equivalent of the system can be taken as $20$ kg$.$ Diffusivity of water

vapor in air can be taken as $0\mathrm{.}2$ cm$2$

$/$s$.$ Vapor pressure of water at $360$ K is $62\mathrm{.}5$ kN$/$m$2$

$.$ Also find

out falling rate of temperature.