Question

suppose 1.00 mol of ice at -30.0c is heated at ... Question Suppose 1.00 mol of ice at -30.0C is heated at atmospheric pressure until it is converted to steam at 140.C. Calculated Q, W, ΔH, and ΔU for this process. For ice, water, and steam, Cpm = 38.0, 75.0, and 36.0 J/mol K, respectively, and can be taken to be approximately independent of temperature. ΔHfus for ice is 6.007 kJ mol , and ΔHvap for water is 40.66 kJ mol . Use the ideal gas law for steam, and assume that the volume of ice or water is negligible relative to that of 1 mol of steam.

Answer 1

for constant pressure process Q=delH

delH includes converting ice

1. at -30 deg.c to ice at 0 deg.c = moles of ice* specific heat* temperature difference= 1*38.0*(0--(30)= 38*30 joules=1140 joules=1.140 joules

2. converting ice at 0 deg.c to water at 0 deg.c = moles of ice* latent heat of fusion= 1*6.07 Kj =6.007 Kj

3. for converting water at 0 deg.c to water at 100 deg.c = moles of ice * specific heat* temperature difference= 1*75*100=7500J= 7.5 KJ

4. For converting liqud water at 100 deg.c water vapor at 100 deg.c = moles of ice* latent heat of vaporization= 1*40.66 Kj= 40.66 Kj

5. For converting water at 100 deg.c to water vapor at 140 deg.c = 1*36*40 =1440 joules =1.440 Kj

Total delH= 1+2+3+4+5= 1.140+6.007+7.5+40.66+1.440 kj=56.747 kj

there is no work done on the system.

up to conditions of water vapor delU= 1.140+6.007+7.5+40.66=55.30 KJ

Change in internal energy of vapor =moles*specific heat at constant volume* temperature difference= 1*(36-8.314)*40 = 1140 Joules= 1.140 Kj

Change in internal energy= 55.30+1.140=56.44 Kj

work done by steam in onveting from 100 deg.c to steam at 140 deg.c (Expansion) = -nR*(T2-T1)= 1*8.314*(140-100)=-332.56 J

Work done for converting solid ice to water vapor at 100 deg.c is negligible.