At a pressure of 1 atm, the equilibrium melting temperature of lead is 600 K, and at this temperature, the latent heat of melting of lead is 4810 J/mole.
Calculate the entropy produced when 1 mole of supercooled liquid lead spontaneously freezes at 590 K and 1 atm pressure.
Given : The constant-pressure molar heat capacity of liquid lead, as a function of temperature, at 1 atm pressure is given by
and the corresponding expression for solid lead is
c_{p_{(s)} }-23.56+9.75\times 10^{-3}T J/KThe entropy produced during the irreversible freezing of the lead equals the difference between the change in the entropy of the lead and the change in the entropy of the constant-temperature heat reservoir (at 590 K) caused by the process.
First, calculate the difference between the entropy of 1 mole of solid lead at 590 K and 1 mole of liquid lead at 590 K. Consider the processes illustrated in Figure 3.11.
1. Step a → b : 1 mole of supercooled liquid lead is heated from 590 to 600 K at 1 atm pressure.
2. Step b → c : 1 mole of liquid lead is solidified reversibly at 600 K (the equilibrium melting or freezing temperature is the only temperature at which the melting or freezing process can be conducted).
3. Step c → d : 1 mole of solid lead is cooled from 600 to 590 K at 1 atm pressure.
Since entropy is a state function,
\Delta S_{(a\rightarrow d)} =\Delta S_{(a\rightarrow b)}+\Delta S_{(b\rightarrow c)}+\Delta S_{(c\rightarrow d)}Step a → b
\Delta S_{(a\rightarrow b)}=\int_{a}^{b}{\frac{\delta q_{rev} }{T} } \int_{a}^{b}{\frac{\delta q_p}{T} } =\int_{590K}^{600K}{\frac{c_{p,Pb(l)}dT }{T} }=\int_{590K}^{600K}\left\lgroup\frac{32.4}{T}-3.1\times 10^{-3} \right\rgroup dT
=32.4\ln \frac{600}{590}-3.1\times 10^{-3}\times (600-590) =+0.514 J/K
Step b → c
\Delta S_{(b\rightarrow c)} =\frac{q_{rev} }{T} =\frac{q_p}{T} =\frac{latent heat of freezing}{equilibrium freezing temperature}=-\frac{4810}{600} =-8.017 J/K
Step c → d
\Delta S_{(c\rightarrow d)} =\int_{c}^{d}{\frac{\delta q_{rev} }{T} } =\int_{c}^{d}{\frac{\delta q_{p} }{T} }=\int_{600K}^{590K}{\frac{c_{p,Pb(s)}dT }{T} }=\int_{600K}^{590K}{\left\lgroup\frac{23.6}{T} +9.75\times 10^{-3} \right\rgroup } dT
=23.6\ln \frac{590}{600} +9.75\times 10^{-3} (500-600)=-0.494 J/K
Thus,
\Delta S_{(a\rightarrow d)} =+0.514-8.017-0.494=-7.997 J/KConsider the thermal energy entering the constant-temperature heat reservoir at 590 K. As the thermal energy is transferred at constant pressure, q_p = Δ H , where Δ H is the difference between the enthalpies of states d and a . As H is a state function,
\Delta H_{(a\rightarrow d)}= \Delta H_{(a\rightarrow b)}+\Delta H_{(b\rightarrow c)}+\Delta H_{(c\rightarrow d)}\Delta H_{(a\rightarrow b)}=\int_{a}^{b}{c_{p,Pb(l)} }dT=\int_{590K}^{600K}{(32.4-3.1\times 10^{-3} )} dT
=32.4\times (600-590)-\frac{3.1\times 10^{-3} }{2} (600^2-590^2)=306 J
\Delta H_{(b\rightarrow c)} =-4810 J
\Delta H_{(c\rightarrow d)} =\int_{c}^{d}{c_{p,Pb(s)} }dT=\int_{600K}^{590K}{(23.6+9.75\times 10^{-3} )}dT
=23.6(590-600)+\frac{9.75\times 10^{-3} }{2}(590^2-600^2) =-294 J
Thus,
\Delta H_{(a\rightarrow d)} =-4799 Jand so, the heat reservoir absorbs 4799 J of thermal energy at 590 K. Consequently,
\Delta s_{heat reservoir}=\frac{4799}{590}=8.134 J/KThus, the entropy created is
\Delta S_{irr} =-7.994+8.134=0.137 J/KAn examination shows that the lower the temperature of the irreversible freezing of the supercooled liquid, the more irreversible the process and the larger the value of \Delta S_{irr} .
