Question

Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, it should be applied to the bar in combination with its surroundings (the lake). Assume that the entropy change of the bar is -1238 J/K . What is the change in total entropy ΔStotal?

Answer 1

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Step-by-step explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings