$100 g$ of water at $353 K$ is mixed with $200 g$ of water at $293 K$ in an insulated container. Take $c_{P} = 4.18 J g^{- 1} K^{- 1}$ (constant) and ignore the container. The total entropy change of the system in $J K^{- 1}$ (two decimals) is

Numerical answer type, answer in J/K, tolerance 0.02.

Answer: 4.93 J/K

Final temperature:

T_{f} = (100 \times 353 + 200 \times 293) /300 = 313.0 K

Δ S = 418 ln (T_{f} /353) + 836 ln (T_{f} /293) = - 50.271 + 55.202 = 4.931 J K^{- 1}

. The positive value reflects irreversible heat flow.