Question

5. The heat of fusion of ice is 333.5J/g. The entropy change for the water when freezing 5.0 g of water at 0°C

and 1 atm pressure is:

(a) 6.1 J/K

(b) 1.2 J/K

(C) OJ/K

(d) -1.2 J/K

(e) -6.1 J/K

Answer 1

(e) -6.1 J/K

Step-by-step explanation:

Step 1: Given data

• Heat of fusion of ice (ΔH°fus): 333.5 J/g

• Mass of water (m): 5.0 g

Step 2: Calculate the heat (Qfreezing) required to freeze 5.0 g of water

We will use the following expression.

Qfreezing = -ΔH°fus × m

Qfreezing = -333.5 J/g × 5.0 g = -1.7 × 10³ J

Step 3: Calculate the entropy change (ΔS°) at 0 °C (273.15 K) and 1 atm

We will use the following expression.

ΔS° = Qfreezing/T

ΔS° = -1.7 × 10³ J/273.15 K = -6.1 J/K