Entropy Change of Ice-Water System

[majormuss]

Homework Statement

An 11 g ice cube at -16.0˚C is put into a Thermos flask containing 120 cm3 of water at 24.0˚C. By how much has the entropy of the cube-water system changed when a final equilibrium state is reached? The specific heat of ice is 2200 J/kg K and that of liquid water is 4187 J/kg K. The heat of fusion of water is 333 × 103 J/kg.

Homework Equations

The Attempt at a Solution

I desperately need help.. I keep solving for almost one hour and my answers is surely wrong. I approached it by solving Q1 for the ice to heat up from -16C to 0, then melts, then comes to a final temperature, then i can use that to find entropy change.. but my final temperature camee out to be 36C- that is impossible.. pls help
these are my numbers to find the final temperature
FOR ICE ------------------------- FOR WATER
Q1= 387.2 J -------------------------468.944(T-final -24)
Q2= 3663 J
Q3=46.1(T- final)
my resulting answer is nonsensical! was my approach wrong? what is the best procedure?

 

[Andrew Mason]

Homework Statement

An 11 g ice cube at -16.0˚C is put into a Thermos flask containing 120 cm3 of water at 24.0˚C. By how much has the entropy of the cube-water system changed when a final equilibrium state is reached? The specific heat of ice is 2200 J/kg K and that of liquid water is 4187 J/kg K. The heat of fusion of water is 333 × 103 J/kg.

Homework Equations

The Attempt at a Solution

I desperately need help.. I keep solving for almost one hour and my answers is surely wrong. I approached it by solving Q1 for the ice to heat up from -16C to 0, then melts, then comes to a final temperature, then i can use that to find entropy change.. but my final temperature camee out to be 36C- that is impossible.. pls help
these are my numbers to find the final temperature
FOR ICE ------------------------- FOR WATER
Q1= 387.2 J -------------------------468.944(T-final -24)
Q2= 3663 J
Q3=46.1(T- final)
my resulting answer is nonsensical! was my approach wrong? what is the best procedure?

Does the ice all melt? Set out the algebraic expressions for: the heat flow from the water to the ice to get the ice to 0C.; the heat flow to melt the ice at 0C; if there is still a temperature difference, the heat flow to warm the ice to the final equilibrium temperature.

You do not appear to be using the heat of fusion in your answer. You have to use it.

AM

 

[majormuss]

I did use the heat of fusion of ice..mL(ice)= .011kg x333000=3663 J
I don't think I need the heat of fusion of water in this calculation because there is not melting of water in this case.
can you please write down a solution so that i can follow and safe my self more time to finish the remainder of the homework.. its due this night!

 

[Andrew Mason]

I did use the heat of fusion of ice..mL(ice)= .011kg x333000=3663 J
I don't think I need the heat of fusion of water in this calculation because there is not melting of water in this case.

There is ice. Ice is frozen water. Why do you say there no melting of ice?

AM

 

[majormuss]

There is ice. Ice is frozen water. Why do you say there no melting of ice?

AM

That is exactly what I already stated.. ok let me show you my entire work so you can correct my mistake..
Q1= cmT (all for ice)
2200 x .011 x (0 +16)
= 387.2 J
Q2= mLf = .011x 333000= 3663 J

Q3= cmT ----------at this point i assumed that all the ice is now liquid so I sued specific heat specific heat of water- 4187
Q3= 4187x .011x(Tf)= 46.1 xTf

FOR WATER
we assume the water will only decrease in temperature. therefore
Q4= 4187 x .112 x (Tf - 24)
=468.94(Tf - 24)IN TOTAL
adding all the numbers from the ice on left and water on the right..
387.2+ 3663 + 46.1 Tf= 468.9( Tf - 24)
final answer= Tf= 36.12 C
this is clearly wrong because 36 is bigger than what we started with.. where did i go wrong?

 

[majormuss]

are you online? pls help!

 

[Andrew Mason]

FOR WATER
we assume the water will only decrease in temperature. therefore
Q4= 4187 x .112 x (Tf - 24)
=468.94(Tf - 24)IN TOTAL
adding all the numbers from the ice on left and water on the right..
387.2+ 3663 + 46.1 Tf= 468.9( Tf - 24)
final answer= Tf= 36.12 C
this is clearly wrong because 36 is bigger than what we started with.. where did i go wrong?

Ok. I understand what you are doing now. You just have a sign problem.

Your equation should be: Qi = -Qw (1)

Qi = + 387.2 + 3663 + 46.1(Tf-0) > 0

Qw = mCwΔTw = 468.9(Tf-24) < 0

So (1) becomes:

Qi = 4050 + 46.1(Tf) = -Qw = -468.9(Tf-24) = - (468.9Tf - 11254)

Work that out and you should get an answer between 0 and 24 for Tf.

AM

 

[majormuss]

thanks very much, I got 13.9C as my final temperature and that makes more sense.. can you please also help me in determining the entropy change? This is my problem:
entropy change for water= Q/T , Q=cm x Ln(Tf/Ti)
=4187x.112x Ln(286.9/297)= -16.22 J/K --------Reasonable answer because the water got colder and so contracted.

my real problem is with ice.. what should I use to represent the Q for Ice? Should it be like; Q= Q1 + Q2 + Q3...??
or should it be just Q= m x c x Ln(Tf/Ti)= 2200x .011x Ln( 286.9/257)= 2.66 J/K

and thus finally Net Entropy Change= 2.66 + -16.22= -13.56 J/K -------which is slightly reasonable because the entire mixture had contracted.. Was this wrong? where did i go wrong again?

 

[Andrew Mason]

thanks very much, I got 13.9C as my final temperature and that makes more sense.. can you please also help me in determining the entropy change? This is my problem:
entropy change for water= Q/T , Q=cm x Ln(Tf/Ti)
=4187x.112x Ln(286.9/297)= -16.22 J/K --------Reasonable answer because the water got colder and so contracted.

my real problem is with ice.. what should I use to represent the Q for Ice? Should it be like; Q= Q1 + Q2 + Q3...??
or should it be just Q= m x c x Ln(Tf/Ti)= 2200x .011x Ln( 286.9/257)= 2.66 J/K

and thus finally Net Entropy Change= 2.66 + -16.22= -13.56 J/K -------which is slightly reasonable because the entire mixture had contracted.. Was this wrong? where did i go wrong again?

You seem to be using Q for ΔS

When temperature changes with heat flow: ΔS = ∫dQr/T = cm x ln(Tf/Ti)

So, for the ice going from -16C to 0C, ΔS = cm x ln(Tf/Ti) = 1.46 J/K

It is simple to find ΔS for the change of state since T is constant. Q/T = 3663/273 = 13.4 J/K

Add the ΔS for the heat flow from 0 to 13.9C and that is the change in entropy of the melted ice.

The ΔS for the bath going from 24C to 13.9C is .112 x 4186 x ln(286.9/297) = -16.22

The answer will be > 0.

AM

 

[majormuss]

I found an answer= .95 J/K ...somehow the homework software says I am wrong again.. This entire approach was my idea.. how would you have approached this particular problem

 

[Andrew Mason]

I found an answer= .95 J/K ...somehow the homework software says I am wrong again.. This entire approach was my idea.. how would you have approached this particular problem

If you followed my previous post, your approach is correct. You might have a significant figures issue. My answer is pretty close to yours:

ΔS_ice = 1.46 + 2.29 + 13.40 = 17.15 J/K
ΔS_bath = -16.22 J/K

ΔS_ice + ΔS_bath = 17.15-16.22 = .93 J/K

If you use three sig. figs. it would be 17.2 - 16.2 = 1.0 J/K

AM

 

[majormuss]

I entered your number and it was still rejected. That was my last try. The stupid software does not even have an answer! I just have to move on to a different question. Many thanks to you for helping out..:)