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`        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. The surface area of the pitcher (one side) = 200 cm2. The room temperature = 42°C, latent heat of vaporization = 2.27 × 106 J/kg, and the thermal conductivity of the porous walls = 0.80 J/m-s-°C. Calculate the temperature of water in the pitcher when it attains a constant value.`
3 years ago

Jitender Pal
365 Points
```										Sol. ℓ = 1 mm = 10^–3 m m = 10 kg
A = 200 cm2 = 2 × 10^–2 m^2
L base vap = 2.27 × 10^6 J/kg
K = 0.80 J/m-s-°C
dQ = 2.27 × 10^6 × 10,
dQ/st = 2.27 * 10^7/10^5 = 2.27 * 10^2 J/s
again we know
dQ/dt = 0.80 * 2 * 10^2 * (42 - T)/1 * 10^-3
So, 8 * 2 * 10^-3(42 -T)/10^-3 = 2.27 * 10^2
⇒ 16 * 42 – 16T = 227 ⇒ T = 27.8 = 28°C

```
3 years ago
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