Entropy of Liquid water Calculation.

[Joseph95]

Homework Statement

Calculate the ΔS° when 0.5 mole of liquid water at 0°C is mixed with 0.5 mole of liquid water at 100°C.Assume Cp=18cal/deg mole over the whole range of temperatures.

Homework Equations

ΔS=∫ Cp/T dT

 

[BvU]

Hello Joseph,

You erased part of the template. That's a nono in PF -- let's hope the spirits that watch over us don't delete your thread altogether. PF requires effort on your part, so work out the equation !

 

[Andrew Mason]

Welcome to PF Joseph!

The assumption is that the only heat flow into the cold water comes from the hot water, not from the surroundings. So there is no change in entropy of the surroundings.

Step 1: find the end state of each quantity of water (ie. the final temperature - easy).
Step 2: find a reversible path between the initial and final states for each quantity of water
Step 3: calculate $\Delta S = \int dQ/T$ over that path for each quantity of water
Step 4: add the two results to find the total change in entropy.

AM

 

[Joseph95]

Hello Joseph,

You erased part of the template. That's a nono in PF -- let's hope the spirits that watch over us don't delete your thread altogether. PF requires effort on your part, so work out the equation !

I am new in here.Forgive me.

 

[Joseph95]

Welcome to PF Joseph!

The assumption is that the only heat flow into the cold water comes from the hot water, not from the surroundings. So there is no change in entropy of the surroundings.

Step 1: find the end state of each quantity of water (ie. the final temperature - easy).
Step 2: find a reversible path between the initial and final states for each quantity of water
Step 3: calculate $\Delta S = \int dQ/T$ over that path for each quantity of water
Step 4: add the two results to find the total change in entropy.

AM

Step 1: Heat Loss=Heat Gain

m.Cwater.ΔT1=m.Cwater.ΔT2 (Moles are equal so the mass)
(100°C-T)=(T-0°C)
T=50°C

Step 2: Calculations for 0.5 mole liquid water at 100°C
Ti=100°C----> 100 +273.3=373.3°K
Tf=50°C------> 50+273.3=323.3°K

ΔS=∫dQ/T pressure is constant therefore ΔS=∫CpdT/T from 373.3°K to 323.3°K

ΔS=Cp.ln(323.3/373.3)=18 cal/deg (-0.144) = - 2.59 cal /deg

Calculations for 0.5 mole liquid water at 0°C
Ti=0°C------> 0 + 273.3 = 273.3°K
Tf=50°C-----> 50+ 273.3 = 323.3°K

ΔS=∫CpdT/T from 273.3°K to 323.3°K

ΔS=Cp.ln(323.3/273.3)= 18 cal/deg (0.168) =3.02 cal/deg

Step 3: ΔS=3.02 cal/deg - 2.59 cal/deg =0.43 cal/deg

Is it correct ?

 

[Andrew Mason]

If you have done the math right, that should be the right answer.

AM

 

[Joseph95]

What if we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C.How will the result change?
Because in this equation ΔS=∫dQ/T entropy doesn't depend on quantity of substance.So can we say entropy doesn't depend on mole of the substance for the whole concept?

 

[Chestermiller]

In post # 5, you should have multiplied by the number of moles of each liquid before adding the changes.

 

[Joseph95]

In post # 5, you should have multiplied by the number of moles of each liquid before adding the changes.

ΔS=-2.59 X 1/2 cal/deg mol =-1.295
ΔS= 3.02 X 1/2 cal/deg mol =1.51

ΔS°=1.51-1.295=0.215 cal/deg mol I think this is the correct answer now.Is there any mistake in the unit of ΔS° ?

 

[Chestermiller]

ΔS=-2.59 X 1/2 cal/deg mol =-1.295
ΔS= 3.02 X 1/2 cal/deg mol =1.51

ΔS°=1.51-1.295=0.215 cal/deg mol I think this is the correct answer now.Is there any mistake in the unit of ΔS° ?

Yes. It should be cal/deg.

Do you know what to do now when you have unequal amounts?

 

[Joseph95]

Yes. It should be cal/deg.

Do you know what to do now when you have unequal amounts?

We should multiply entropy by the number of moles of each liquid before finding the net entropy of the system.Thus,

If we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C so the result should be:

ΔS=1x3.02 cal/deg= 3.02 cal/deg
ΔS=2×(-2.59 cal/deg)= -5.18 cal/deg

ΔS°=3.02 cal/deg - 5.18 cal/deg
=-2.16 cal/deg (We can say that entropy of the system is decreasing )

 

[Chestermiller]

We should multiply entropy by the number of moles of each liquid before finding the net entropy of the system.Thus,

If we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C so the result should be:

ΔS=1x3.02 cal/deg= 3.02 cal/deg
ΔS=2×(-2.59 cal/deg)= -5.18 cal/deg

ΔS°=3.02 cal/deg - 5.18 cal/deg
=-2.16 cal/deg (We can say that entropy of the system is decreasing )

You forgot to calculate and use the new equilibrium temperature.

 

[Joseph95]

You forgot to calculate and use the new equilibrium temperature.

Ohh.You are right .I forget it.I am going to calculate again.

 

[Joseph95]

You forgot to calculate and use the new equilibrium temperature.

Qloss=Qgain

m1Water.Cwater.ΔT1=m2water.Cwater.ΔT2

2 mole × 18g/mole × 1 cal/deg(100°C-T)=1 mole × 18g/mole × 1 cal/deg (T-0°C)

2(100°C-T)=T , 200°C=3T T≅66.7°C

Calculation for 2 mole liquid water at 100°C

Ti=100°C ------> 100+273.3=373.3°K
Tf=66.7°C-------> 66.7 + 273.3=340°K

ΔS=∫CpdT/T from 373.3°K to 340°K

ΔS=Cp.ln(340/373.3)=18 cal/deg .(-0.093)=-1.68 x 2=-3.36 cal/deg

Calculations for 1 mole liquid water at 0°C

Ti=0 + 273.3=273.3 °K
Tf=66.7+ 273.3=340°K

ΔS=∫CpdT/T from 273.3°K to 340°K

ΔS=18 cal/deg x 0.021
= 3.93 x 1=3.93 cal/deg

ΔS°=3.93 cal/deg - 3.36 cal/deg=0.57 cal/deg

I hope it is correct.

 

[Chestermiller]

I'm not going to check your arithmetic, but your methodology is definitely correct now.

Chet

 

[Joseph95]

I'm not going to check your arithmetic, but your methodology is definitely correct now.

Chet

Thank you very much sir !