Calculate Change in Entropy of Water & Ice System

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Homework Help Overview

The problem involves calculating the change in entropy of a system consisting of one kg of ice at 0°C mixed with one kg of boiling water. The participants are exploring the thermodynamic principles involved in reaching thermal equilibrium and the associated entropy changes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy balance between the ice and boiling water, questioning how to properly set up the equations for heat transfer. There are attempts to derive the final temperature (Tf) and calculate the change in entropy (dS) using the formula dS = dQ/T. Some participants express uncertainty about their calculations and the reasonableness of their results.

Discussion Status

The discussion is ongoing, with participants clarifying their mathematical steps and questioning the validity of their assumptions regarding the final temperature. There is no explicit consensus yet, but guidance is being offered on how to approach the problem and check calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring the final temperature falls within the expected range of 0-100°C.

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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

Thanks in advance folks.
 
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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

Thanks in advance folks.
 
I did cancel! ...
 
Then you must mean something different by "cancel" than I do.

You start with
"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?
 

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