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.

Answer 1

Thickness of porous walls, l = 1 mm = 10⁻³ m
Mass, m = 10 kg
Latent heat of vapourisation, L_v = 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.

$$\begin{gathered} \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{2.27\times 10{}^7}{{10}^5} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=2.27\times {10}^2\mathrm{J}/\mathrm{s} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{\Delta }t}=\left(\frac{42-T}{{10}^{-3}}\right)·0.80\times 2\times {10}^{-2} \\ \Rightarrow 2.27\times {10}^2=\frac{\left(42-\mathrm{T}\right)}{{10}^{-3}}\times 0.80\times 2\times {10}^{-2} \\ \Rightarrow T=27.8°\mathrm{C} \\ \Rightarrow T=28°\mathrm{C} \end{gathered}$$