Question

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 1.40 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

Answer 1

The entropy change in the surroundings when 1.40 mol of ice at 0°C melts is 29.68 J/K, which is calculated using the enthalpy of fusion and the temperature of the surroundings.

Step-by-step explanation:

The change in entropy of the surroundings when 1.40 mol of ice at 0°C is melted can be calculated using the given enthalpy of fusion and the temperature of the surroundings. The equation ΔS = Q/T is used, where Q is the heat absorbed (6.02 kJ/mol × 1.40 mol) and T is the temperature of the surroundings in kelvins (284 K). We must convert Q to joules by multiplying kJ by 1000 (since 1 kJ = 1000 J).

So the calculation would be:

ΔS = (6.02 kJ/mol × 1.40 mol × 1000 J/kJ) / 284 K

ΔS = (8,428 J) / 284 K

ΔS = 29.68 J/K

Therefore, the entropy change in the surroundings is 29.68 J/K. This indicates that when the ice melts, the surroundings lose this amount of entropy due to the heat transfer.

Answer 2

29.7 J/K

Step-by-step explanation:

To solve this problem we can use the equation

• ΔS = Δq/T

Where ΔS is the entropy change, Δq is the heat change, and T is the temperature.

• Δq= ΔH*mol

We put the data given by the problem and solve for ΔS:

• ΔS = 6020J/mol * 1.40 mol / 284 K

• ΔS = 29.7 J/K