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 = 45°C, latent heat of vaporization = 2.27 × 106 J kg–1, and the thermal conductivity of the porous walls = 0.80 J s–1m–1 °C–1. Calculate the temperature of water in the pitcher when it attains a constant value.
Given:
Thickness of the pitcher: x = 1 mm = 0.001 m
Mass of the water in the pitcher : m = 10 kg
Rate at which water evaporates at it’s outer surface:
= 0.1 gs-1 = 0.1× 10-3 kg s-1.
The surface area of the pitcher : A = 200 cm2= 0.02 m2.
Room temperature : T1 = 42 ° C
Latent heat of vaporization: L = 2.27 × 106 J kg–1.
Thermal conductivity of the porous walls:
K = 0.80 J s–1m–1 °C–1.
Let the constant temperature of the water in the pitcher be T2.
Formula used:
Rate of amount of heat flowing is given as:
Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
Now, 0.1× 10-3 kg of water evaporates in 1 second. Thus by unitary method, 10 kg of water will evaporate in 105 seconds.
Δt = 105 seconds.
Substituting we get,
∴ 42 – T2 = 14.18
∴ T2 = 42 - 14.18
∴ T2 = 27.82 °C
Hence, the temperature of water in the pitcher when it attains a constant value is 27.82 °C.