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 is

Explanation

Heat required to change the temperature of vessel by a small amount dT
dQ=`mC_(P)dT`. Total heat required
`Q=mint_(20)^(4)32((T)/(400))^(3)dT=0.001996kJ`
Work done required to maintain the temperature of sink to `T_(2)`
`W=Q_(1)-Q_(2)=((Q_(1)-Q_(2))/(Q_(2)))Q_(2)` For `T_(2)=20K,`
`W_(1)=((300-20)/(20))0.001996=0.028kJ`
For `T_(2)=4K`
`W_(2)=((300-4)/(4))0.001996=0.148kJ`
`therefore` The work required to cool the vessel from 20K to 4K is `W_(2)-W_(1)=0.148-0.028=0.12K`
As temperature is changing from 20K to 4K work done required will be more than `W_(1)` but less than `W_(2)`.