A pitcher with `1-mm` thick porous walls contain 10kg of water. ,Water comes to its outer surface and evaporates at the rate of `0.1gs^(-1)` . The surface area of the pitcher (one side) `=200cm^(2)` . The room temperature `=42^(@)C` , latent heat of vaporization `=2.27xx10^(6)Jkg^(-1)` , and the thermal conductivity of the porous walls `=0.80js^(-1)m^(-1)`^(@)C^(-1)` . Calculate the temperature of water in the pitcher when it attains a constant value.

Answer

B, C

Explanation

`l=1mm =1xx10^(-3)m, M=10kg`
`A=200 cm^(2) =2xx10^(-2)m^(2)`
`L_(vapour)=2.27xx10^(-2) J//kg`
`k=0.08 j//m-s-^(@)C, dQ=2.27xx10^(6)xx10,`
`(dQ/(t)=(2.27xx10^(7))/(10^(5))`
`=2.27xx10^(2) J/s.`
Again we know
`(dQ)/t=(0.80xx2xx10^(-2)xx(42xxT))/(10^(-3))`
`so, (0.80xx2xx10^(2)xx(42-T))/(10^(-3))=2.27xx10^(2)`
`implies16xx42-16T=227`
`impliesT=27.8=28^(@)C.`