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 one mole of ice at 0°C is melted by heat supplied from surroundings at 276 K,what is the entropy change in the surroundings,in J/K?

Question

Quizplus · Accepted Answer

The entropy change in the surroundings can be calculated using the formula ΔS = -q/T, where q is the heat absorbed by the system (ice) and T is the temperature of the surroundings in Kelvin. Here, q = 6020 J (since it's the enthalpy of fusion for melting one mole of ice) and T = 276 K. Thus, ΔS = -6020 J / 276 K = -21.8 J/K.

Assume That the Enthalpy of Fusion of Ice Is 6020