# Calculate Change in Entropy of Water & Ice System

• tdusffx
In summary: Did you multiply or divide only one side of the equation? In summary, the conversation discusses the addition of one kg of ice at 0°C to one kg of boiling water, resulting in equilibrium. The change in entropy of the system is being questioned. The equation used to calculate this is "Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv", which can be simplified to "Cw*dT + Lf = -Cw*dT + Lv". The Final Temperature (Tf) calculated from this equation is 280°C, but this does not seem reasonable as it should be between 0-100°C. The issue of finding the dQ is also brought up,
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

tdusffx said:
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
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

(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

I did cancel! ...

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

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

and conclude
"Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv"

What exactly did you do in order to "cancel"? Did you divide both sides of the equation by something? Did you add or subtract something from both sides?

## 1. What is the formula for calculating the change in entropy of a water and ice system?

The formula for calculating the change in entropy of a water and ice system is ΔS = mΔH/T, where ΔS is the change in entropy, m is the mass of the substance, ΔH is the change in enthalpy, and T is the temperature in Kelvin.

## 2. How do you determine the change in enthalpy of a water and ice system?

The change in enthalpy of a water and ice system can be determined by subtracting the enthalpy of the final state from the enthalpy of the initial state. This can be calculated using the equation ΔH = H(final state) - H(initial state).

## 3. What is the significance of calculating the change in entropy of a water and ice system?

Calculating the change in entropy of a water and ice system can provide valuable information about the system's thermodynamic properties and how it changes with temperature. It can also help determine the spontaneity of a process and the amount of energy that is available to do work.

## 4. What are the units of measurement for entropy and enthalpy?

The units of measurement for entropy are Joules per Kelvin (J/K) or Calories per Kelvin (cal/K). The units for enthalpy are Joules (J) or Calories (cal).

## 5. How does the change in entropy of a water and ice system differ from other thermodynamic processes?

The change in entropy of a water and ice system is unique because it involves a phase change from solid to liquid or vice versa. This means that the entropy change is not only affected by temperature, but also by the amount of energy required to change the phase of the substance.

• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
18
Views
5K
• Introductory Physics Homework Help
Replies
11
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Biology and Chemistry Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K