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