Take Specific Heat (or Cp) of Water as 4.1813 Kilojoules-Per-Kilogram-Dot-Kelvin
and write 19°Celsius and 21°Celsius as 292.15 Kelvin and 294.15 Kelvin.
Equate Heat Transfer Q to mCp(T2 − T1) which evaluates to
(0.495 kg)(4.1813 kJ/[kg • K]){294.15 − 292.15} K or 4.139487 kJ.
Then set down ΔSnet as mCp • ln[T2/T1] + {Q/T2} equal here to
(0.495 kg)(4.1813 kJ/[kg • K]) • ln[294.15 K/292.15 K] + {4.139487 kJ/294.15 K}
which reduces to 0.02819347552 kJ/K (or 28.19347552 J/K).