#### Question

Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm^{2}. 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^{−1}g °C^{−1}. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.

#### Solution

Area of cross section*, A* = 10 cm^{2} = 10 × 10^{–4} m^{2}

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*, **T*_{1} = 50°C

temperature outside the vessel, *T*_{2} = 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`