in questions 31 and 32, how do you account for two mixtures? i understand how that you need to find the final temperature, but i am a little confused in the steps needed to be taken in finding delta s. HELP!

-- Anonymous, February 23, 2000

Answers

31. To find dS you need to find the dS for the 1 mol of cool water and 3 mol of hot water seperately using:

dS = n*Cp*ln(T2/T1)

Then use dStot = dSheat + dScool

to find the total change in entropy.

You can use the same method for 32 make sure you take in to account fusion and the different Cp of water and ice. Remember that dS = dH/T.

Try the promblems and let me know if you need more help.

-- Anonymous, February 24, 2000

i have a question!!! How do you calculate the final temperature. Why do you have to calculate the final temperature. Also, I know how to calculate dScool. How do you calculate dSheat? Please answer in a timely manner.

-- Anonymous, February 24, 2000

To calculate final temperature:

Heat loss(3mol) = heat gain(1mol)

n*Cp*dT = n*Cp*dT

(3mol)*(73.5J/molK)*(Tf-0C) = (1mol)*(73.5J/molK)*(100C-Tf)

solve for Tf and you get 298K

For #32 do the same thing but remember for heat gian by the ice you need to do it in parts.

Heat gain(ice) = n*Cp*dT(ice)+dHfusion+n*Cp*dT(water)

=(1mol)*(37.5J/molK)*(10C-0C)+(6010J)+(1mol)*(73.5J/molK)*(Tf-0C)

you then set that eqaul to heat lost by water and solve for Tf and you get 53.8C.

You find dSheat and dScool using the same formula dS=n*Cp*ln(T2/T1) for dSheat you use n=mol and for dScool n=1mol.

For #32 it's the same for dSwater but for dSice you must remember the heat of fusion so your equation is:

dSice=n*Cp*ln(T2/T1)+(dH/T)

I hope that helps let me know if you need more and what part is still confusing to you.

-- Anonymous, February 24, 2000