1.

The specific heat capacity of a metal at low temperature (T) is given as `C_(p)(kJK^(-1) kg^(-1)) =32((T)/(400))^(3)` A 100 gram vessel of this metal is to be cooled from `20^(@)K` to `4^(@)K` by a special refrigerator operating at room temperaturte `(27^(@)C)` . The amount of work required to cool the vessel isA. equal to `0.002` kJB. greater than `0.148` kJC. between `0.148` kJ and `0.028` kJD. less than `0.028` kJ

Answer» Correct Answer - A
Here,
`C_(p)(kJ K^(-1) kg^(-1)) =32 ((T)/(400))^(3) =32xx(T^(3))/((400)^(3))` ,
`m=100 gram =0.1 kg` ,
Workdone to cool the vessel = amount of heat taken out from metal while cooling it from 20 K to 4 K.
`:. W = int_(20)^(4) mC_(p) dT = int_(20)^(4) 0.1xx(32(T^(3))/(400^(3)))dT`
`=0.002 kJ`


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