Question

A mass of 5 kg of water at 293 K is turned completely to ice at 273 K. The entropy change in the process is

1. 6.13479 kJ/K
2. -1.4847 kJ/K
3. -7.6195 kJ/K
4. 8.3195 kJ/K

Answer 1

Correct Answer - Option 3 : -7.6195 kJ/K

Concept:

Entropy change for a process (without phase change):

\(Δ S=\int \frac{dQ}{T}=\int_{T_1}^{T_2}\frac{mcdT}{T}=mc \ln\frac{T_2}{T_1}\)

where c = specific heat of water (4.183 kJ/kg-K)

Entropy change (with phase change):

\(Δ S=\frac{Q}{T}=\frac{mL}{T}\)

where L = latent heat of fusion (334 kJ/K)

Calculation:

Given:

m = 5 kg, T₁ = 293 K, T₂ = 273 K

When water is changed from 293 K of water to 273 K of water no phase change is involved.

\(Δ S=mc \ln\frac{T_2}{T_1}\)

\(Δ S=5 \times 4.183 \times \ln\frac{273}{293}=-1.478\;\rm{kJ/K}\)

When water is changed from 273 K water to 273 K ice phase change is involved.

\(Δ S=\frac{mL}{T}\)

\(Δ S=\frac{5\;\times\;334}{273}=-6.117\;\rm{kJ/K}\)

∴ the entropy change of the process will be 

ΔS_process = -1.478 - 6.117 ⇒ 7.595 kJ/K ≈ 7.6195 kJ/K.