Exp. (c)Heat required to change the temperature of vessel by a small amount dT,−dQ=mCp​dTTotal heat required−Q=m∫204​32(400T​)3dT=(400)3100×10−3×32​[4T4​]204​⇒Q=0.001996 kJWork done required to maintain the temperature of sink to T2​,W=Q1​−Q2​=Q2​Q1​−Q2​​Q2​=(T2​T1​​−1)Q2​⇒W=(T2​T1​−T2​​)Q2​For T2​=20 K,W1​=20300−20​×0.001996=0.028 kJFor T2​=4 K,W2​=4300−4​×0.001996=0.148 kJ