Assume that the enthalpy of fusion of ice is 6020 J/mol and does not vary appreciably over the temperature range 270-290 K. If 1.20 mol of ice at 0°C is melted by heat supplied from surroundings at 284 K, what is the entropy change in the surroundings in J/K?

Question

Quizplus · Accepted Answer

The entropy change in the surroundings is calculated using the formula ΔS_surroundings = -ΔH/T_surroundings. Here, ΔH = 6020 J/mol * 1.20 mol = 7224 J, and T_surroundings = 284 K. Thus, ΔS_surroundings = -7224 J / 284 K = -25.4 J/K, which is closest to -23.5 J/K.

Assume That the Enthalpy of Fusion of Ice Is 6020