## We need help on 32!!!!!!!!!!!!!greenspun.com : LUSENET : UR General Chemistry : One Thread |

A few of us are up here in the management library trying DESPERATELY to figure out #32 in chapter 10. We have read all the postings about it, but still cannot figure out why 3 steps must be used instead of one for the ice when calculating the final temp. It would be great if someone could do the WHOLE problem out in detail and explain! Thanks!

-- Anonymous, March 01, 2000

Ok this is what I did (and I think it is correct): 1)Find joules of heat needed to melt the ice = nCdT + heat of fusion = (1)(37.5)(10)+(6010)=6385J 2)Find the temp change for the water because it lost 6385J heat. (6385J)/nC=28.26 So the water went from 373 to 344.74K. 3)Find the temperature of the mixture by averaging the temp of the water and the melted ice: [(1mol)(273K)+(3mol)(344.74K)]/4mol =326.81K 4)Now find the entropy change FOR THE ICE. Use dS=nCln(T2/T1) =(1mol)(37.5)ln(326.81/273) =6.74J which rounds to 7joules I hope this is helpful and correct. If I am wrong please tell me.

-- Anonymous, March 01, 2000

When a substance changes states you must take in consideration each part seperately, becuase of different specfic heat value for liquids, solids and gases. Also you must account for enthalpy of vaporization and/or fusion.18.02g H2O ice = 1 mol ice; 54.05g H2O water = 3 mols water;

Heat gain ice = heat lost by water

heat gain ice = nCpdT(from -10C to 0C)+n*6010J(dHfusion(icemelting)) +nCpdT(Tf-0C): in this part you are figuring the enthalpy for the ice warming up to 0C, melting and increasing to the final temp.

heat gain ice = (1mol)*(37.5J/Kmol)*(10)+(1mol)*(6010J)+(1mol)* (75.3J/Kmol)*(Tf-0C)

heat loss water = nCpdT = (3mol)*(75.3J/Kmol)*(100-Tf); you do it this way to get a positive value for dT

you then set: heat gain ice = heat lost water; and solve for Tf

6385+75.3Tf=22590-225.9Tf

301.2Tf = 16205

Tf=53.8C

You now calculate dS. You must do it in parts ice + water = total

dSice= nCpln(T2/T1)(-10to0) + dHfus/T + nCpln(T2/T1)(0to53.8) dSice= 1mol)*(37.5J/Kmol)*ln(273K/263K)+(6010J/273K)+(1mol)* (75.3J/Kmol)*ln(326.8K/273K)= 36.9J/K

dSwater= nCpln(T2/T1)(100to53.8) = (3mol)*(75.3J/Kmol)*ln(326.8K/373K) = -29.9J/K

dStotal= dSice + dSwater= (36.9J/K) +(-29.9J/K)= 7 J/K

-- Anonymous, March 01, 2000

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