A system of 100 kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/kgK to 0.4 kJ/kgK. At the same time, the entropy of the surroundings decreases from 80 kJ/K to 75 kJ/K.
Irreversible
Understanding the nature of a thermodynamic process (reversible, irreversible, or impossible) requires analyzing the total change in entropy of the system and its surroundings. The second law of thermodynamics provides the criteria for this analysis.
Calculating the System's Entropy Change
The system is the 100 kg mass. Its specific entropy changes from 0.3 kJ/kgK to 0.4 kJ/kgK.
The change in specific entropy ($\Delta s$) is the final specific entropy minus the initial specific entropy:
$$ \Delta s = s_{final} - s_{initial} $$
$$ \Delta s = 0.4 \text{ kJ/kgK} - 0.3 \text{ kJ/kgK} = 0.1 \text{ kJ/kgK} $$
The total change in the system's entropy ($\Delta S_{system}$) is the mass ($m$) multiplied by the change in specific entropy ($\Delta s$):
$$ \Delta S_{system} = m \times \Delta s $$
$$ \Delta S_{system} = 100 \text{ kg} \times 0.1 \text{ kJ/kgK} = 10 \text{ kJ/K} $$
Calculating the Surroundings' Entropy Change
The entropy of the surroundings changes from 80 kJ/K to 75 kJ/K.
The change in surroundings entropy ($\Delta S_{surroundings}$) is the final entropy minus the initial entropy:
$$ \Delta S_{surroundings} = S_{surroundings, final} - S_{surroundings, initial} $$
$$ \Delta S_{surroundings} = 75 \text{ kJ/K} - 80 \text{ kJ/K} = -5 \text{ kJ/K} $$
Calculating the Total Entropy Change
The total entropy change for the entire universe (system + surroundings) is the sum of the system's entropy change and the surroundings' entropy change:
$$ \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} $$
$$ \Delta S_{total} = 10 \text{ kJ/K} + (-5 \text{ kJ/K}) = 5 \text{ kJ/K} $$
Applying the Second Law of Thermodynamics
The second law of thermodynamics states the following regarding the total entropy change for a process:
- If $ \Delta S_{total} > 0 $, the process is irreversible and possible.
- If $ \Delta S_{total} = 0 $, the process is reversible and possible.
- If $ \Delta S_{total} < 0 $, the process is impossible as it violates the second law.
In this case, we calculated $ \Delta S_{total} = 5 \text{ kJ/K} $. Since $ 5 \text{ kJ/K} > 0 $, the total entropy of the universe increases.
Therefore, the process is irreversible.
Conclusion on the Process Type
Based on the positive total entropy change ($ \Delta S_{total} > 0 $), the process described is irreversible.
Important Questions from Second Law of Thermodynamics and Entropy
- Entropy of the universe is:
- Which statement correctly describes the total entropy change of the universe during an irreversible process?
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