An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in the pitcher to cool to `28^@C` from its original temperature of `30^@C` Neglect radiation effect. Latent heat of vapourization of water in this range of temperature is 580 cal/g and specific heat of water is `1 kcal//gC^(@)`

Explanation

Here water at the surface is evaporated at the cost of the water in the vessel losing heat.
Heat lost by the water in the vessel
`=(9.5+0.5)xx1000xx(30-20)=10^5`cal
Let t be the required time in seconds. Heat gained by the water at the surface
`=(txx10^-3)xx540xx10^3`
`(L=540(cal)/(g)=540xx10^3cal//kg)`
`:. 10^5=540t` or `t=185s=3min5s`