Answer:
a) at 0°C the entropy change is ΔS=0.366 kJ/K
b) at 50°C the entropy change is ΔS= 0.309 kJ/K
Explanation:
from the second law of thermodynamics
ΔS ≥ ∫ dQ /T
where ΔS = change in entropy , Q= heat exchanged with the surroundings
and T= absolute temperature
- For an isothermal process ∫ dQ /T= 1/T* ∫ dQ = Q/T
- For an reversible process the ΔS = ∫ dQ /T
thus
ΔS= Q/T
replacing values
a) 0°C= 273 K
ΔS= Q/T = 100 kJ /273K = 0.366 kJ/K
ΔS=0.366 kJ/K
b) 50°C= 50 + 273K = 323K
ΔS= Q/T = 100 kJ /323K = 0.309 kJ/K
ΔS= 0.309 kJ/K
