Irreversible adiabatic reaction

[liquidicy]

Hey everyone =) This is my first post here but I've been lurking for a while.

Anyway, so i am taking this physical chemistry class and I can't seem to solve this problem. I think it is still a physics problem.

Anyway, it goes like this:

small amount of moles of steam (.83 moles) @ 100 celsius is added to a larger amount of water (13.9 moles) @ 25 Celsius. The pressure standard is constant and walls are adiabatic. They want me to find the total entropy change for the process.

Basically this problem would be really easy if it was easy to find the final temperature of the system. I can't figure it out.

The problem gives data such as enthalpy of vaporization of water which is 40.656 kJ/mol @ 373.15 kelvin. They also give me molar heat capacities of liquid water @ constant pressure: (Cp) (75.25 J/(k*mole)) and steam (33.256J/(K*mole))

Basically i thought total enthalpy change would be zero. I am not even sure of that but assuming it is, I got this equation:

(moles of gas)*(enthalpy of vaporization) + (moles of vapor turned water)*(Cp of water)*([tex]\Delta[/tex] T of gas turned water) = (moles of the water @25)*(Cp of water)+([tex]\Delta[/tex] in T of water)

After finding the final temperature, i realized that it is impossible. I got a 25 degree change in temperature of the water after adding basically 15 grams of water to 250 grams of water. It is like adding a tablespoon of boiling water into a pretty large cup of water. It wouldn't change the temperature by 25 degrees. So that means that the formula that I had was wrong =(

Can anyone suggest an alternate solution? Thnx =)

 

[Chestermiller]

The first step in this problem is to establish the final state o the system. At constant pressure, there is no change in the enthalpy of the system.

Taking 25 C at 1 atm as the reference state for enthalpy, the enthalpy per mole of the liquid water in the initial state is 0. The enthalpy per mole of the steam in the initial state of the steam is 75.25(100-25)+40656 = 46300 J/mole. So the enthalpy in the initial state of the system is:
$$H_i=(13.9)(0)+(0.83)(46300)=38430\ J$$

Assuming that, in the final state, we have all liquid water at temperature T, the enthalpy of the system in the final state is $$H_f=(14.73)(75.25)(T-25)$$
This must be equal to the initial enthalpy, so we must have:
$$(14.73)(75.25)(T-25)=38430$$
Solving for T then gives a final temperature of T = 59.7 C

Taking the same reference state for entropy as for enthalpy, the initial entropy of the system was:
$$S_i=(0.83)(75.25\ln{(373/298)}+\frac{40656}{373})=104.5\ J/mole$$
The final entropy of the system is
$$S_f=(14.73)(75.25)\ln{(357.7/298)}=202.4\ J/mole$$ So the change in entropy between the initial and final states of the system is $$\Delta S = 202.4-104.5=97.9\ J/mole$$