Explanation

(c) Heat required to change the temperature of vessel by a small amount dT
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - dQ = mC_{P}dT$
Total heat required
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - Q = m\int_{20}^{4}{32\left( \frac{T}{400} \right)^{3}dT}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{100 \times 10^{- 3} \times 32}{(400)^{3}}\left\lbrack \frac{T^{4}}{4} \right\rbrack_{20}^{4}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q = 0.001996\ kJ$
Work done required to maintain the temperature of sink to $T_{2}$
$W = Q_{1} - Q_{2}$
$= \frac{Q_{1} - Q_{2}}{Q_{2}}Q_{2}$
$= \left( \frac{T_{1}}{T_{2}} - 1 \right)Q_{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W = \left( \frac{T_{1} - T_{2}}{T_{2}} \right)Q_{2}$
$For\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = 20\ K$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W_{1} = \frac{300 - 20}{20} \times 0.001996$
=0.028 kJ
For $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = 4\ K$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W}_{2} = \frac{300 - 4}{4} \times 0.001996$
=0.148 kJ
As temperature is changing from 20 K to 4 K, work done required will be more than $W_{1}\ but\ less\ than\ W_{2}$.