1.

In a pitcher when water is filled some water comes to outer surface slowly through its porous walls and gets evaprated. Most of the latent heat needed for evaporation is taken from water inside and hence this water is cooled down. If 10 kg water is taken in the pitcher and 12 gm water comes out and evaporated per minute. Neglect heat transfer by convection and radiation to surrounding, find the time in which the temperature of water is pitcher decreases by 5^(@)C.

Answer»

Solution :It is given that 12 gm WATER is evaporated per minute, thus heatrequired per minute for it is
`Q=mL_(v)`
`=12xx540=6480` cal/min
After time t, mass of inside water is
`m=10000-12t`
If in further tiem t`dt,dm=12 dt`, mass is vapourized and the temperature of inside water falls by dT we have
`(10000-12t)xx1xxdeT=12 dT 540`
or `dT=12xx540(dt)/(10000-12t)`
Integrating the above expanssion in PROPER limits, we get
`int_(T_(0))^(T_(0)-5)dT=12xx540xxint_(0)^(t)(dt)/(10000-12t)`
`5=540 In (10000/(10000-12t))`
or `E^(5//540)=10000/(10000-12t)`
or `t=10000/12 [e^(1//108)-1]`
`=7.737` minutes As here the amount of wather vaporized is very small, we can assume that hte QUANTITY of water inside remains constant. Thus to reducethe temperature of inside water, amount of heat to be rejected is
`Q=m^(3)DeltaT`
`=10000xx1xx5`
`=50,000` calories
As obtained from the given DATA that 6480 calories of heat is required per minute to vapourize 12 gm water thus if in time t minutes, temperature of inside water falls by `5^(@)C` we have
`50000=6480t`
or `t=50000/6480` Which is approximately same as obtained earlier. But students should note that this method will give the correct answer if the mass of inside water actually remains constant i.e. when the rate of evaporation is very very low.


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