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