An amount of ice (mass m) at temperature T_1 is added to an equal mass of water

Question

An amount of ice (mass m) at temperature T_1 is added to an equal mass of water at temperature T_2 and the mixture is allowed to reach equilibrium. Show that the change of entropy of the system is Delta S = m [c_1 ln (T_m/T_1) + Delta_fus H/T_m + c_W ln (T^2_3/T_m T_2)] where c_1 and c_w are the constant pressure specific heat capacities (e.g., c_P) of ice and water, respectively, Delta_fuss H the heat of fusion of ice at 273 K. The melting point of ice is T_m and the temperature T_3 is defined by 2c_W T_3 = T_m (c_W - c_1) + T_2 c_W + T_1 c_1 - Delta_fus H.

Explanation / Answer

from T1 to 0 deg.c, ice gains sensible heat which is = mass* specific heat* Temperature difference= m*Ci*(Tm-T1)

ice gains sensible heat to become liquid at 0 deg.c which is given by = mass * heat of fusion = m*deltaH fusion

total heat to be added to convert ice into liquid at 0 deg.c= mc*(Tm-T1)+m*deltaHm

then sensible heat has to be supplied to reach an equilibrium temperature T3.

Hence mCi(T1-Tm)+m*deltaHfu+m*Cw*(T3-Tm)= m*Cw*(T2-T3),

CITm*(Tm-T1)+deltaHfus+Cw*(T3-Tm)= Cw*(T2-T3)

2CwT3= Tm*(Cw-Ci) +T2Cw+T1C1-deltaHfu

entropy change from T1 to Tm = mCi* ln (Tm/T1)

entropy change at the melting temperature = m*deltaH/Tm

entropy of liquid water = m*CW*ln (T3/Tm)

entropy of water at T2= m*Cw* ln (T3/T2)

hence entropy change = mCi*ln(Tm/T1)+ m*deltaH/Tm + mCw*( ln T3*T3/T2Tm)

=m*( Ci* ln (Tm/T1) + deltaH/Tm+ Cw* ln (T32/TmT2) ( the answer)

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