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Question

# Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for is taken from the water itself and the water is cooled down. If pitcher has 10 kg of water, find the time in which the temperature of water falls by 5∘C. The approximate rate of vaporisation is 5 g min−1 (Cw=1 cal/g-∘C, Lv=540 cal/g)

A
20 min 10 s
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B
18 min 20 s
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C
14 min 12 s
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D
none of these
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Solution

## The correct option is B 18 min 20 sLet dmdt be the rate of evaporation and M=M0−tdmdt is the amount of water left in the pitcher at any instant. Heat lost by left over water=Heat gained for evaporation MCw dθ=(dmdtt)Lv (M0−dmdt.t)Cwdθ=Lvtdmdt (10000−560.t)1×5=560t(540) or 50000=45t+512t or t=50,000×12545=18 min 20 s

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