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 $0.1g{s}^{-1}$. The surface are aof the pitcher (one side) = 200 $c{m}^2$. The room temperature = ${42}^{\circ }$, latent heat of vaporization = $2.27\times {10}^{6}Jk{g}^{-1}$, and the thermal conductivity of the porous walls = $0.80J{s}^{-1}{m}^{-1\circ }{C}^{-1}$. Calculate the temperature of water in the pitcher when it attains a constant value.

Answer 1

l = 1 mm = $1\times {10}^{-3}m,$ M = 10 kg

A = 200 $c{m}^2=2\times {10}^{-2}{m}^2$

$L_{vapour}=2.27\times {10}^{-2}J/kg$

$K=0.80J/m-s{-}^{\circ }C,dQ=2.27\times {10}^6\times 10$;

$\frac{dQ}{t}=\frac{2.27\times {10}^7}{{10}^5}$

$=2.27\times {10}^{2}J/s$

Again we know

$\frac{dQ}{t}=\frac{0.80\times 2\times {10}^{-2}\times (42-T)}{{10}^{-3}}$

So, $\frac{0.80\times 2\times {10}^{-2}\times (42-T)}{{10}^{-3}}=2.27\times {10}^2$

$\Rightarrow 16\times 42-16T=227$

$\Rightarrow T=27.8={28}^{\circ }C$