The enthalpy and entropy of fusion of ice at 0.00 $\circ$ C are 6.01 kJ/mol and 22.0 J/mol . K,respectively.What is the entropy change of the universe when one mole of ice melts in a large room maintained at 34.00 $\circ$ C? Assume the final temperature of the water is 0.00 $\circ$ C.

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The entropy change of the universe is the sum of the entropy change of the system (ice melting) and the surroundings (room). The entropy change of the system is ΔS_system = ΔH_fusion / T = 6.01 kJ/mol / 273.15 K = 22.0 J/K. The entropy change of the surroundings is ΔS_surroundings = -ΔH_fusion / T_room = -6.01 kJ/mol / 307.15 K = -19.6 J/K. Therefore, the total entropy change of the universe is ΔS_universe = ΔS_system + ΔS_surroundings = 22.0 J/K - 19.6 J/K = +2.4 J/K.

The Enthalpy and Entropy of Fusion of Ice at 0 ∘\circ∘

The Enthalpy and Entropy of Fusion of Ice at 0 $\circ$