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Entropy of Cube-Water System Problem

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Homework Statement



An 6.0 g ice cube at -12.0˚C is put into a Thermos flask containing 100 cm3 of water at 19.0˚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



Q = mcΔT
ΔS = Q/T

The Attempt at a Solution



-Qlost = Qgain -----> -Qw = Qice + Lm
Qw = (.1kg)(4187 J/kg*K)(Tf - 292K) = 418.7Tf - 122,260
Qice = (.006kg)(2200 J/kg*K)(Tf - 261K) = 13.2Tf - 3445
Lm = (333*10^3 J/kg)(.006 kg) = 1998 J

-418.7Tf + 122,260 = 13.2Tf - 3445 + 1998
-431.9Tf = -123,707
Tf = 286.425 K = 13.425 C

Is this the right process so far? If so I'm stuck on what to do next. What comes next in solving this?
 

Answers and Replies

  • #2
Andrew Mason
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[...
Q = mcΔT
ΔS = Q/T

...

If so I'm stuck on what to do next. What comes next in solving this?
You do not calculate the heat flow by itself. You have to calculate [itex]\Delta S = \int dS = \int dQ/T[/itex]

The heat gained by the ice from -12C to 0C is at gradually increasing temperature. So calculate the change in entropy over the reversible path from -12C to 0C by calculating this integral.

Just substitute the expression for dQ (the incremental heat gain - hint: it is a function of dT) and integrate.

That gives you the change in entropy of the ice up to the melting point. Then you have to find the change in entropy of the ice during melting. Finally, you have to find the change in entropy of the melted ice as its temperature goes from 0C to Tf.

To find the change in entropy of the surroundings you have to do the same process for the water in the thermos except it is much easier since there is just the heat flow out of the water as T decreases from 19C to Tf. The total entropy change is the sum of all these changes in entropy.

AM
 
  • #3
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You do not calculate the heat flow by itself. You have to calculate [itex]\Delta S = \int dS = \int dQ/T[/itex]

The heat gained by the ice from -12C to 0C is at gradually increasing temperature. So calculate the change in entropy over the reversible path from -12C to 0C by calculating this integral.

Just substitute the expression for dQ (the incremental heat gain - hint: it is a function of dT) and integrate.

That gives you the change in entropy of the ice up to the melting point. Then you have to find the change in entropy of the ice during melting. Finally, you have to find the change in entropy of the melted ice as its temperature goes from 0C to Tf.

To find the change in entropy of the surroundings you have to do the same process for the water in the thermos except it is much easier since there is just the heat flow out of the water as T decreases from 19C to Tf. The total entropy change is the sum of all these changes in entropy.

AM
I'm really confused at what you stated there to be honest. Are you saying all the work I did is not necessary?

Here's what I'm understanding from this problem. I have to find Qice and Qwater in the ice-water system. So I did that. I found Lm as well for melting the ice cube. Aren't those the three main things I needed to find? But then I get confused because now I need to utilize ΔS = Q/T. I found Q for each part of the process. I'm not sure what T values to use in this equation. I'm just really lost at this point. I understand that I need to find the entropy change for each Q and then add them all up and that's the final answer. But did I even find all the Q's for this process? There should only be 3 total right? -Qw = Qice + Lm. Divide each Q by the temperature at the process. So Qw/Tf = ΔS ; Qice/Tf = ΔS ; Lm/273 K = ΔS. Is any of what I said close to what I am supposed to be doing?
 
  • #4
Andrew Mason
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I'm really confused at what you stated there to be honest. Are you saying all the work I did is not necessary?

Here's what I'm understanding from this problem. I have to find Qice and Qwater in the ice-water system. So I did that. I found Lm as well for melting the ice cube. Aren't those the three main things I needed to find? But then I get confused because now I need to utilize ΔS = Q/T. I found Q for each part of the process. I'm not sure what T values to use in this equation. I'm just really lost at this point. I understand that I need to find the entropy change for each Q and then add them all up and that's the final answer. But did I even find all the Q's for this process? There should only be 3 total right? -Qw = Qice + Lm. Divide each Q by the temperature at the process. So Qw/Tf = ΔS ; Qice/Tf = ΔS ; Lm/273 K = ΔS. Is any of what I said close to what I am supposed to be doing?
The first thing you have to do is find the final temperature. You have tried to do that. But I am not clear on the process you are using. You did not seem to use the specific heat of 4186 J/Kg K for the 6 grams of melted ice going from 0C to Tf.

To determine the final temperature you have to first determine whether all the ice melts. If it doesn't, then the final temperature is 0C. If it does, then you have to work it out as follows:

The heat gained by the 6 grams that starts out as ice is:

[tex]Q_{ice} = mC_{ice}(T_{melt}-T_{i-ice}) + mL + mC_{water}(T_f-T_{melt})[/tex]

The heat lost by the water in the thermos is:

[tex]Q_{water} = mC_{water}({T_f - T_{i-water})[/tex]

Equating the magnitude of those two heat flows (Qgained = - Qlost) will give you a single expression for Tf that you should be able to work out.

Once you get Tf you can work out the change in entropy.

To do that you have to calculate the change in entropy of the ice/melted ice and then calculate the change in entropy of the 100g of water in the thermos.

The change in entropy is the integral of dQ/T over the reversible path between the beginning and end states. For the ice you have to break up the process into three parts.

1. First you have to calculate the integral of dQ/T in bringing the ice from -12C to 0C. In doing this, you have to express dQ as a function of temperature. (hint: the answer involves the natural logarithm).

2. Then you have to calculate the integral of dQ/T in melting the ice at 0C. That is easy because T is constant so it is just the heat flow divided by T.

3. Then you have to calculate the integral of dQ/T in bringing the melted ice (ie. water at 0C) to Tf. Again, you have to express dQ as a function of T (hint: it is not the same as in 1. because it is now liquid water).

4. Then you have to add all these entropy changes to find the total change in entropy of the ice.

Next you have to calculate the change in entropy of the water in the thermos. There is only one part to this since that water does not undergo a change of state. Just calculate the integral of dQ/T for 100g of water going from 19C to Tf. Again, you have to express dQ as a function of T.

That will give you the change in entropy of the water.

To find the total change in entropy of the system, you have to add the change in entropy of the water to the change in entropy of the ice.

AM
 
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  • #5
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I had to re-do the problem because it said my answer was wrong the first time. So it gives me new values to do the problem over again (this program seriously sucks). Now this time, I'm absolutely positive my answer is correct. Yet it says I got it wrong. I don't know what is going on now. I even followed the tutorials answers and compared with my own and they matched exactly until the last step where it says "Now that you know all this, find the total change in entropy of the ice-water system." Here's the new values in the question:

An 12 g ice cube at -23.0˚C is put into a Thermos flask containing 105 cm3 of water at 22.0˚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.

My Work:

mass of ice = 12 g = .012 Kg
Initial Ice Temperature = -23 C = 250 K
Initial Water Temperature = 22 C = 295 K
Volume of Water = 105 cm^3 * (1 m/100 cm)^3 = 1.05E-4 m^3
D = m/V -----> m = DV = (1000 Kg/m^3)(1.05E-4 m^3) = .105 Kg

-Qlost = Qgain -----> -Qwater = Qice

Qice = mCice(Tmelt - Ti,ice) + mLf + mCw(Tf - Tmelt)
Qice = (.012 Kg)(2200 J/Kg*K)(273 K - 250 K) + (.012 KG)(333E3 J/Kg) + (.012 Kg)(4187 J/KG*K)(Tf - 273 K) = 607.2 + 3996 + 50.244Tf - 13,716.612 = 50.244Tf - 9113.412

Qw = mCw(Tf - Ti,w)
Qw = (.105 Kg)(4187 J/Kg*K)(Tf - 295 K) = 439.635Tf - 129,692.325

-(439.635Tf - 129,692.325) = 50.244Tf - 9113.412
-439.635Tf + 129,692.325 = 50.244Tf - 9113.412
-489.569Tf = -138,805.737
Tf = 283.5264 K -----> 10.5264 C

ΔS = Q/T
ΔS = mcln(Tf/Ti)

Ice Entropy - Phase 1 (Raising Temperature to Freezing Point)
ΔS,ice-1 = mcln(Tf/Ti) = (.012 Kg)(2200 J/Kg*K)ln(273 K/250 K) = 2.3235 J/K

Ice Entropy - Phase 2 (Melting)
ΔS,ice-2 = Q/T = 3996 J/273 K = 14.6374 J/K

Ice Entropy - Phase 3 (Raising Temperature to Tf)
ΔS,ice-3 = mcln(Tf/Ti) = (.012 Kg)(4187 J/Kg*K)ln(283.5264 K/273 K) = 1.9009 J/K

ΔS,ice = ΔS,ice-1 + ΔS,ice-2 + ΔS,ice-3 = 2.3235 J/K + 14.6374 J/K + 1.9009 J/K = 18.8612 J/K

Water Entropy (Lowering Temperature to Tf)
ΔS,w = mcln(Tf/Ti) = (.105 Kg)(4187 J/Kg*K)ln(283.5264 K/295 K) = -17.4404 J/K

Total ΔS = ΔS,ice + ΔS,w = 18.8618 J/K + (-17.4404 J/K) = 1.4214 J/K

Did I seriously make a mistake in my work here? I don't see how that's possible since my answers for the Tf value and entropy changes equal the ones in the tutorial.
 
  • #6
Andrew Mason
Science Advisor
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Looks very well done to me. I haven't checked your arithmetic.

AM
 

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