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Entropy of irreversible process

Entropy of irreversible process
Now, let’s consider freezing of supercooled water. Somehow water exists as liquid at -10 degree celcius. It’s called supercooled water. The supercooled water will freeze at this temperature. We intuitively know that the freezing at this temperature is spontaneous and irreversible. To calculate the entropy change of the system under the irreversible process, we need to keep in mind that the entropy of the closed system is a state function. So it’s change only depends on the initial and final states. Here, we have a pair of initial and final states. There is a reversible path connecting the initial and final state. Also, there can be a irreversible path connecting the same initial and final state.
The entropy change is path independent, since it is a state function. So the entropy changes of these two processes are the same regardless of whether it is reversible or irreversible. So, to calculate the entropy change of the irreversible process such as freezing of supercooled liquid at -10 degree Celcius, we can take a reversible path for the same change. This is a reversible path. Instead of directly changing the supercooled liquid into ice -10 degree celcius, this step is made of three steps, kind of a detour for the original change.
This detour is a reversible path since the freezing at 0 degree, at the melting point temperature is reversible as we have seen before, and the heating and cooling can be done reversibly. So each constituent step is reversible and the whole detour path can be called a reversible path. So the entropy change of the system for the freezing of supercooled liquid is the summation of entropy changes of each step. Delta S1 is the entropy change of heating the supercooled liquid from -10 degree to zero degree celcius. For the heating of supercooled liquid, Cp of supercooled liquid is approximated as the Cp of the normal liquid since the temperature range delta 10 degree is not very big.
Delta S2 is the freezing at the melting point temperature and it is -1.22 J/g.k as before. Delta S3 is the cooling of ice from zero degree to -10 degree celcius, so it is the integration of Cp of ice over T dT. Refer to the data for Cp values. Cp of ice is about 0.5 cal/g.k and that of liquid water is 1 cal/g.K. Inserting these values into the integration yields the entropy change for the system as -1.14 J/g.k. To calculate the entropy change of surrounding for the freezing of supercooled liquid, we need the actual heat and actual transformation temperature T. The actual heat Q at constant pressure is the enthalpy of transformation.
The enthalpy of the freezing of supercooled liquid at -10 degree celcius is the detour path enthalpy since enthalpy is also a state function. So it is the enthalpy of reversible heating of supercooled liquid + enthalpy of reversible freezing at zero degree + enthalpy of reversible cooling of ice. The actual calculation with the Cp values results in the heat of transformation as -314.9 J/g. It is the heat goes into the system. So the heat goes into the surrounding is -Qp. The entropy change of surrounding is thus -Qp over the actual temperature of transformation. So it is +1.197 J/g.k. The total entropy change of the universe is the sum of entropy for the system and surrounding.
So the total entropy change is +0.057 J/g.K and it is positive. Since it is positive, it is a spontaneous process. So the freezing of supercooled liquid at -10 degree celcius is a spontaneous process and thus irreversible as we expected!

Intuitively, we know that the freezing of water below meting point, e.g -10 ℃ occurs spontaneously.

It means freezing of supercooled water is a irreversible process. Calculation of the entropy change for this process is by taking reversible detour path. Cp values of the materials involved and the latent heat give the entropy change and the enthalpy change of the system. The total entropy change is positive for this spontaneous process.

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Thermodynamics in Energy Engineering

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