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Question
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 aluminium (K = 200 J s−1 m−1°C−1). Assume that the outside temperature is 20°C. The density of water is 100 kg m−3, and the specific heat capacity of water = 4200 J k−1g °C−1. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.
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Solution
Area of cross section, A = 10 cm2 = 10 × 10–4 m2
Thermal conductivity, K = 200 Js–1 m–1 °C–1
Height, H = 10 cm
Length, l = 1 mm =10–3 m
Temperature inside the cylindrical vessel, T1 = 50°C
temperature outside the vessel, T2 = 30°C
Rate of flow of heat from 1 flat surface will be given by
`(DeltaQ)/(Deltat) = (T_1 -T_2)/( l /(KA)`
`(DeltaQ)/(Deltat) = ((50 - 30 ) xx 200 xx 10^-3)/ 10^-3`
`(DeltaQ)/(Deltat)` = 6000 J/s
Heat escapes out from both the flat surfaces.
Net rate of heat flow = 2 × 6000 = 12000 J/sec
`(DeltaQ)/(Deltat) = (m.s.DeltaT)/ (Deltat)`
Mass = Volume density
`⇒10^-3 xx 0.1 xx 1000 =0.1g`
Using this in the above formula for finding the rate of flow of heat, we get
`12000 = 0.1 xx 4200 xx (DeltaT)/(t)`
`⇒ (DeltaT)/(Deltat) = 12000/420 = 28.57 `
As `DeltaT = 1^circ C`
⇒ `1/t = 28.57`
⇒ `t = 1/28.57 = 0.035 sec`
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