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