Consider a process in which 100 kj of energy is transferred reversibly and isothermally as heat to a large block of copper. calculate the change in entropy of the block if the process takes place at (i) 0 °c, (ii) 50 °c.

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

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