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A container of negligible heat capacity contains `1kg` of water. It is connected by a steel rod of length `10m` and area of cross-section `10cm^(2)` to a large steam chamber which is maintained at `100^(@)C`. If initial temperature of water is `0^(@)C`, find the time after which it beomes `50^(@)C`. (Neglect heat capacity of steel rod and assume no loss of heat to surroundings) (use table `3.1`, take specific heat of water `=4180 J//kg.^(@)C)`

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Let temperature of water at time t be `T`, then thermal current at time t,
`I = ((100 -T)/(R ))`
This increases the temperature of water form `T to T +dT`
`rArr i = (dH)/(dt) = ms (dT)/(dt)`
`rArr (100-T)/(R ) = ms(dT)/(dt) rArr overset(50)underset(0)int (dT)/(100-T) = overset(t)underset(0)int (dT)/(Rms)`
`rArr -ln ((1)/(2)) = (t)/(Rms)`
or `t = Rms ln2 sec`
`= (L)/(KA) ms ln2 sec`
`= ((10m)(1kg)(4180J//kg-^(@)C))/(46(w//m^(@)C)xx(10xx10^(-4)m^(2)))ln2`
`= (418)/(46) (0.69) xx 10^(5) = 6.27 xx 10^(5) sec = 174.16 hours`

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