# Help with Entrophy

1. Feb 9, 2008

### tdusffx

One kg of ice at 0°C is added to one kg of boiling water. The mixture comes to
equilibrium. What is the change in entropy in cal/K of the system? (Lf = 80cal/g)

a. 300
b. 100
c. 200
d. 50
e. 25

Qcold = -Qhot

Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv

All has Mw in common, therefore it should cancel out

Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv

(1cal/g)(Tf - 0) + 80cal/g = -(1cal/g)(Tf-100) + 540cal/g;
I converted the latent heat of vaporation (2.26x10^6J/kg)*(1/4186cal*kg/J*g) ~
540cal/g

after equating the two sides, I got 280C for Tf

My answer for Tf doesn't seem reasonable because the Final Temp should be in between
0-100C

Anyways, My other problem is getting the dQ

I have no idea how to get it...or maybe I do..when it doubt I just ask people out!

dS = dQ/T

if my T is right then just finding dQ problem

2. Feb 9, 2008

### HallsofIvy

Staff Emeritus
Why didn't you cancel the "Mw" out in each term?
Dividing each term in your original equation by mW gives
Cw*dT+ Lf= -Cw*dT+ Lv

3. Feb 9, 2008

### tdusffx

I did cancel! ...

4. Feb 9, 2008

### HallsofIvy

Staff Emeritus
Then you must mean something different by "cancel" than I do.