i am still having trouble fuguring out Tf for #32. this is what i have and it does not turn out to be 53.8 degrees.

1 mol*37.5*10 + 6010 +1 mol*75.3*(Tf-0)=3 mol *75.3* (Tf-100)

thank you

-- Anonymous, March 01, 2000

Answers

In order not to run in to trouble with a sign problem you need to arrange your problem so dT is always positive. Therefore on the right hand side of your problem it should be (100C-Tf).

I hope that helps.

-- Anonymous, March 01, 2000

I'm very confused about #32. Why do you have to find Tfinal? What do you use it for?

I used dS(total)=nCpln(T2/T1) + (dHfus/T) + nCpln(T2/T1) dS(total) = 1 mol*37.5*ln(273/263) + (6010/273) + 3 mol*75.3*ln(373/273)

What am I doing wrong??

-- Anonymous, March 01, 2000

You are not using the right T's. You need fine Tf to know how to solve those equations. Go back up on the board and look at the other questions about 32. There is a full explainition with one of them.

let me know if you need more help.

-- Anonymous, March 01, 2000

I looked at the other responses but I'm still confused. If you have dS=nCp(T2/T1) + (dHfus/T) + nCp(T2/T1), do you use Tfinal for all the T2's? Then what do you use for the T1's? 273? 263? 313? When do you take into account the 0C--10C and 100C-0C? And what value do you use for the T in the (Hfus/T) equation?

-- Anonymous, March 01, 2000

You need to thin about what part you are doing. For the ice you go from -10C to 0C. So your T1=263K and T2=273K. Then you have fusion and you use 273K because that is the temperature that ice freezes at. Then the water goes from 0C to 53.8C(Tf). So for that part T1=273K and T2=327.8K.

Ask if you need more help.

-- Anonymous, March 01, 2000