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 `0.002kJ`
B. greater than `0.148kJ`
C. between `0.148kJ "and" 0.028kJ`
D. less than `0.028kJ`

Answer

C

Explanation

Specific heat at low is
`C_(P) = 32((T)/(400))^(3)`
`Q = intm.c.dT = int_(20)^(4)(100)/(1000)xx 32((T)/(400))^(3)dT`
=`(32)/(10)xx(1)/((400)^(3))((T^(4))/(4))`
=`(32)/(10 xx(400)^(3))xx (1)/(4)(20^(4) - 4^(4))`
=`(32)/(10 xx (400)^(3))xx (1)/(4)xx (160000-256)`
=`0.002W`
`beta = (T_(2))/(T_(1) -T_(2)) = (Q_(3))/(W)`
`implies (20)/(300-20) = (0.2)/(W)impliesW = 0.028kJ`
`implies (4)/(300-4) = (0.002)/(W) implies W = 0.0148kJ`