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### Topic: entropy  (Read 9473 times)

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#### cokuardi ##### entropy
« on: April 09, 2006, 10:53:10 AM »
Calculate the entropy change that occurs when a 23.5 gram piece of ice at 0deg C is placed into a styrofoam cup containing 129.0 grams of water at a temperature of 78.5deg C. Assume that there is no loss or gain of heat to or from the surroundings. Enter your answer in J/K as a number in scientific notation without units. The heat of fusion of water is 333 J/g. Assume that the specific heat of H2O(l) is constant at 4.184 JK-1g-1.

I know that m=Q/C*change in T
Change in H=M*heat of fusion

#### Borek ##### Re: entropy
« Reply #1 on: April 09, 2006, 02:56:34 PM »
First rule: you must show that you have tried.
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#### cokuardi ##### Re: entropy
« Reply #2 on: April 09, 2006, 07:19:26 PM »
I'm just not sure what to do with the T's.  I can't get passed that.

#### plu ##### Re: entropy
« Reply #3 on: April 09, 2006, 08:31:29 PM »
Assuming that you are only looking for the entropy change of the system, the equation you should use is dS = nCpln(T1/T2).

#### Hunt ##### Re: entropy
« Reply #4 on: April 09, 2006, 09:03:42 PM »
Quote
Assuming that you are only looking for the entropy change of the system, the equation you should use is dS = nCpln(T1/T2).

Since the rxn takes place in a styrofoam cup, then it's at const Pressure. I think you're right, but you inverted the temp's ...

dS = nCpLn(Tf / Ti)

Cokuardi , when ice is added to water, there will be heat transfer from water to ice. You have the Latent heat of fusion of water ( which is the same as the latent heat of melting of ice ) and the mass of ice, so you can calculate the heat absorbed q by ice to reach thermal equilibrium with water. Since no heat is lost or gained from the surrounding then the heat lost by water = heat gained by ice, or qice = -qwater

qice =mC(Ti - Tf ) where C = specific heat of water. You have Ti , you can determine Tf , then use the value in the equation above to compute the entropy change.

#### plu ##### Re: entropy
« Reply #5 on: April 10, 2006, 06:24:41 AM »

Since the rxn takes place in a styrofoam cup, then it's at const Pressure. I think you're right, but you inverted the temp's ...

dS = nCpLn(Tf / Ti)

Yes, you are right Vant_Hoff.  Thank you for the correction.

#### cokuardi ##### Re: entropy
« Reply #6 on: April 10, 2006, 08:29:44 PM »
I was told to calculate the final temperature by balancing the heat lost = heat gained The heat to melt the ice and to warm its cold water must equal the heat used to cool the hot water to the final temperature. Then calculate the entropy change when the ice melts (=q/m.pt) Then calculate the entropy change as the cold water warms up (=heatcap*ln(Tf/Ti)) Then calculate the entropy change when the hot water cools down and add up all three

#### cokuardi ##### Re: entropy
« Reply #7 on: April 10, 2006, 08:49:29 PM »

#### Hunt ##### Re: entropy
« Reply #8 on: April 10, 2006, 09:31:15 PM »
Quote

I'm not sure what that is, but this is how we got ?S = mCpLn(Tf / Ti) :

dSsys = dqrev / T

? dSsys= ? dqrev / T

?Ssys = mCp ? dT / T = mCpLn(Tf / Ti)

The problem is very straight forward, it's only a matter of computation.

The final temp should be about 64 oC

Quote
Yes, you are right Vant_Hoff.  Thank you for the correction.

No problemo #### Hunt ##### Re: entropy
« Reply #9 on: April 10, 2006, 09:46:18 PM »
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Then calculate the entropy change as the cold water warms up (=heatcap*ln(Tf/Ti)) Then calculate the entropy change when the hot water cools down and add up all three

It's the same process, water cooling down and ice warming up , they will eventually reach thermal equilibrium i.e. same final temp. You need to determine the final temperaure, and then compute the change in entropy for this process.

#### cokuardi ##### Re: entropy
« Reply #10 on: April 10, 2006, 10:41:13 PM »
your confusing me, first you use n in the equation and now your using m.

#### cokuardi ##### Re: entropy
« Reply #11 on: April 10, 2006, 10:52:42 PM »
I know how to find the dS for the cold water warming up and the hot water cooling down, I just don't know how to find the dS for the ice melting.

#### Hunt ##### Re: entropy
« Reply #12 on: April 11, 2006, 04:55:11 AM »
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your confusing me, first you use n in the equation and now your using m.

It depends on how C is given.

CM = q / n?T ----> Molar heat cap

C = q /m?T ----> Specific heat cap

But I thought you would already be aware of that ...

In the problem, "Assume that the specific heat of H2O(l) is constant at 4.184 J/K.g"

This means  C of water = q / m?T . So q =mC?T

?Ssys =  mCpLn(Tf / Ti)  or   ?Ssys =  nCpLn(Tf / Ti)

You should use one of them depending on the units of Cp. Since Cp is given in J/K.grams, then we use the 2nd form.

Quote
I know how to find the dS for the cold water warming up and the hot water cooling down, I just don't know how to find the dS for the ice melting.

You already know how much heat is abrosbed by ice, and at temp 273.15 K, so u can easily determine the change in entropy.
« Last Edit: April 11, 2006, 05:00:39 AM by Vant_Hoff »

#### Hunt ##### Re: entropy
« Reply #13 on: April 11, 2006, 05:10:24 AM »
?Swater =  mCpLn(Tf / Ti) , u have Ti of water and u can compute the final temp of water, so u determine the change in entropy for water.

?Sice =  mLiceLn(Tf / Ti) ( L : Latent heat of fusion or melting ) , again u have the initial temp of ice and final temp is the same as that of water.