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`        Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross-sectional area 10 cm2. The walls of the vessel are adiabatic but the flat parts are made of 1 mm thick aluminum (K = 200 J/m-s-°C). Assume that the outside temperature is 20°C. The density of water is 1000 kg/m°3, and the specific heat capacity of water = 4200 J/kg-°C. Estimate the time taken for the temperature to fall by 1.0°C. Make any simplifying assumptions you need but specify them.`
3 years ago

396 Points
```										Sol. A = 10cm^2, h = 10 cm
∆Q/∆t = KA(θ base 1 - θ base 2)ℓ = 200 * 10^-3 * 30/1 * 10^-3 = 6000
Since heat goes out form both surfaces. Hence net heat coming out.
= ∆Q/∆t = 6000 * 2 = 12000, ∆Q/∆t = MS ∆θ/∆t
⇒ 6000 * 2 * 10^-3 * 10^-1 * 1000 * 4200 * ∆θ/∆t
⇒ ∆θ/∆t = 72000/420 = 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop 1°C 1/28.57 sec. = 0.0035 sec. is required

```
3 years ago
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