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Skip to 0 minutes and 0 secondsNow, let's calculate the actual entropy changes for the reversible and irreversible process. To be noted here is that if a process is adiabatic and reversible, the entropy change for that process is zero, thus it is a isoentropic process. Let's go back to the actual calculation of entropy. Consider the two freezing processes as an example for reversible and irreversible changes. The first freezing happens at melting point temperature zero degree celcius. The second freezing is the freezing of supercooled water at -10 degree celcius. So the freezing occurs not at a melting point, but -10 degree celcius. It's the first freezing at zero degree celcius. The water transforms to ice and the the enthalpy of freezing is -334 J/g.

Skip to 0 minutes and 58 secondsThe same transformation can be written chemically like this. Liquid H2O changes into solid H2O. If we take the water as our system, the entropy change of the system is reversible heat over temperature. Here the enthalpy at constant pressure process is the same with the heat. So the entropy change of the system is -1.22 J/g.k. The entropy change of surrounding is actual heat over temperature. The actual heat transferred to surrounding is -delta H freezing, so +334J/g. The entropy change of surrounding is thus +1.22 J/g.k. So the total entropy change of the universe is zero. It means the freezing at the melting point temperature is the reversible process. The reverse transformation of freezing is melting.

Skip to 1 minute and 55 secondsIn the melting process, the solid H2O, the ice changes into liquid, and the enthalpy of melting is -enthalpy of freezing +334J/g. The entropy change of the system, the water, is now +1.22 J/g.K. The entropy change of the surrounding is -1.22 J/g.K since the actual heat transferred to surrounding is -delta Hm and it is -334 J/g. The total entropy change of the universe is zero now. So the melting at zero degree Celsius is also a reversible process. So the conclusion here is that freezing or melting at melting point temperature proceeds reversibly.

Entropy of reversible process

We know that the freezing of water at meting point is a reversible process.

Calculation of entropy for the system (the water) and surrounding for this freezing process is simply heat divided by melting temperature. The total entropy change of this process is thus zero, demonstrating the statement of the second law of thermodynamics.

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Thermodynamics in Energy Engineering

Hanyang University

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