Water at `506(@)C` is filled in a closed cylindercal vessel of height `10cm` and cross sectional area `10cm^(2)` The walls of the vessel are adiabatic but the flat parts are made of `1-mm` thick aluminium `(K=200Js^(-2)m^(-1)`^(@)C^(-1)` . Assume that the outside temperature is `20^(@)C` . The density of water is `1000kgm^(-3)` and the specify heat capacity of water `=4200Jkg^(-1)`^(@)C^(-1)` .Estimate the time taken for the temperature to fall by `1.0^(@)C` . Make any simplifying assumotion you need but speify them.

Answer

C

Explanation

`A=10 cm^(2): H=10 cm`
`( Delta Q)/( Delta T)=(kA(T_1-T_2))/(l)`
`=(200xx10xx10^(-4)xx30)/(1xx10^(-3))`
`6000 J//s`
Since heat goes out form both surfaces hence net heat coming out
`( Delta Q)/( Delta T)=6000xx2=12000`
`( Delta Q)/( Delta T)=ms ( Delta T)/( Delta l)
`implies6000xx2=10^(-3)xx10^(-1)xx1000`
`xx4200xx( Delta Q)/( Delta T).`
`implies( Delta Q)/( Delta T)=12000/420=28.57`
so in one sec 28.57^(@)C is dropped
Hence time to drop of `1^(@)C=(1)/(28.57)sec.`
`=0.035 sec is required .