The entropy change during heating of 50 g water (sp. heat = 4200 Jkg-1 K-1) from 20°C to 40°C will  be: ( given log313 = 2.4955 and log293 =2.4669)

Mass (m) of water                     =100g×1kg1000g=110kg; sp. heat                     =4200Jkg-1K−1,T1=20+273=293K;T2=40+273=313K.We know that :ΔS=2.303×m×Cplog⁡T2T1                           ∴ ΔS=2.303×110kg×4200Jkg−1K−1log=967.26[log⁡313−log⁡293]=967.26[2.4955−2.4669]=967.26×0.0286 K−1=27.66JK−1So, the correct answer is (3)