Question

Question: Calculate the change in entropy when 50g of ice at 0C is dropped into 100.0g of water at 75C in an insulated vessel. (The enthalpy of fusion of ice is 6.01 kJ/mol and Cp for water is 75.3 J/K mol, and the molecular weight for H20 =18g/mol)

Answer 1

ice at 0 deg.c first gains heat of fusion to convert itself into liquid water at 0 deg.c.

heat of fusion = 6.01Kj/mole and moles of ice = mass of ice/ molar mass = 50/18=2.78

heat to be added to convert the ice into liquid water= molles of ice* heat of fusion= 2.78*6.01 Kj=16.71 Kj.

this much heat has to come from cooling of 100 gm of water at 75 deg.c Let us assume that it also reaches a temperature of 0 deg.c. This assumption is correct if heat gained by ice = heat losy by water at 75 deg,c

heat lost = mole of water* specific heat* change in temperature = (100/18)*75.3*(75-0)=31375 Kj. this value is much much larger than heat of fusion. So this suggests that liquid water at 0 deg.c gains additional heat until ice and water will be at thermal equilibrium at temperature T.

So heat added to ice= 16.71*1000 ( heat of fusion)+ sensible heat of liquid water from 0 deg.c to T= sensible heat of water from 75 deg.c to final temperature T

16.71*1000+2.77*75.3*(T-0)= (100/18)*75.3*(75-T), T is the final temperature of the system of ice and water

2.77*75.3*T +T*418.3 = 31375-16710, T= 23.39 deg.c

so both ice at 0 deg.c melts at 0 deg.c and then changes its temperature to 24.06 deg.c while water changes its temperature from 75 deg.c to 23.39 deg.c

entroy change of ice during melting= Q/T= 16.31*1000/273 J/K=59.74

entropy change from 0 deg.c to 24.06 deg.c= moles* speciifc heat* ln(T2/T1)= 2.77* 75.3*ln{(23.39+273)/(0+273)}=17.14 J/K, entropy change of ice from 0 deg.c to 24.06 deg.c= 59.74+17.14= 76.78J/K

entropy change of water from 75 deg,c to 23.39deg.c= (100/18)*75.3* ln {(23.39+273)/(75+273)}=-67.15 J/K

entropy change of combined system= 76.78-67.15= 9.63 J/K