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

Options

A. equal to 0.002 kJ
B. greater than 0.148 kJ
C. between 0.148 kJ and 0.028 kJ
D. less than 0.028 kJ

Answer

C

Explanation

Heat required to change the temperature of vessel by a small amount dT,
`-dQ=mC_(p)dT`
Total heat required
`-Q=m int_(20)^(4) 32 ((T)/(400))^3dT`
`=(100xx10^(3)xx32)/((4oo)^3)[(T_4)/(4)]_(20)^(4) implies Q= 0.001996 KJ`
Work done required to maintain the temperature os sink to `T_2`,
`W=Q_1-Q_2= (Q_1-Q_2)/(Q_2)Q_2= ((T_1)/(T_2)-1)Q_2 implies W= ((T_1-T_2)/(T_2))Q_2`
For `T_2=20K`,
`W_1= (300-20)/(20)xx1.001996= 0.028KJ`
For `T_2=4K`,
`W_2= (300-4)/(4)xx0.001996= 0.148KJ`
As temperature is changing from 20 K to 4 K, work done required will be more than `W_1` but less than `W_2`.