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Question
A pitcher with 1-mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of 0.1 g s−1. The surface area of the pitcher (one side) = 200 cm2. The room temperature = 42°C, latent heat of vaporization = 2.27 × 106 J kg−1, and the thermal conductivity of the porous walls = 0.80 J s−1 m−1°C−1. Calculate the temperature of water in the pitcher when it attains a constant value.
Solution
Thickness of porous walls, l = 1mm = 10-3 m
mass, m =10 kg
Latent heat of vapourisation, Lv = 2.27 × 106 J/kg
Thermal conductivity, K = 0.80 J/m s °C
ΔQ = 2.27 × 106 × 10 J
0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 105 s
`⇒ (DeltaQ)/(Deltat) = (2.27 xx 107)/10^5`
`⇒ (DeltaQ)/(Deltat) = 2.27 xx 10^2 ` J/s
`⇒ (DeltaQ)/(Deltat)=(DeltaT)/(l/(kA))`
`⇒ (DeltaQ)/( Deltat) = ((42 - T)/ 10^-3) . 0.80 xx 2 xx 10^-2`
⇒ `2.27xx10^2=(42-"T")/10^-3xx0.80xx2xx10^-2`
⇒ T = 27.8° C
⇒ T = 28° C
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