# Equilbrium temp of ice and steam mixture

• castrodisastro
In summary, the final temperature of the mixture will be 334J/g*K for ice and 373K*4.19J/g*K for steam.
castrodisastro

## Homework Statement

You are asked to mix equal masses of of ice and steam at 0.00°C and 100.00°C respectively, What will the final temperature of the mixture be? The mixture is in a perfect insulator.

ms=mass of steam
mi=mass of ice
LHV,s=Heat of vaporization of steam=2257J/g
LHF,i=Heat of fusion of ice=334J/g
Ti,s=100°C=373K
Ti,i=0.00°C=273K
cw=4.19J/(g*K)

Q=mL
Q=mcΔT

## The Attempt at a Solution

I am fairly confident that my thought process is correct. I just keep making algebra mistakes(I assume) but I can't find what my error is. Please let me know where my process breaks down please.

I did this in 3 parts

Part 1)
So first I consider the situation.
Since the mixture is in a perfect insulator, that means no heat is lost to the surroundings.

Q=0

That means that the heat lost by the steam is gained by the ice. Q for the steam will be negative since it is losing heat.

Q=0=Qgained+Qlost=Qgained+(-Qlost)

Qlost=Qgained

From here, the Qlost is given by the sum of energy used to turn steam into water, then to bring the steam-turned-water to thermal equilibrium

Qlost=mscw(Ti,s-Teq)+msLHV

and Qgained is given by the sum of energy used to melt the ice, then to bring the ice-turned-water to thermal equilibrium.

Qgained=micw(Teq-Ti,i)+miLHF

Again, since there is no heat lost to the surroundings, and the mixture is allowed to reach thermal equilibrium then...

Qlost=Qgained

Part 2)
Now we plug into be able to solve for Teq

mscw(Ti,s-Teq)+msLHV=micw(Teq-Ti,i)+miLHF

mscwTi,s-mscwTeq+msLHV=micwTeq-micwTi,i+miLHF

Moving all terms containing Teq to one side

mscwTi,s+micwTi,i+msLHV-miLHF=micwTeq+mscwTeq

Factor out Teq

mscwTi,s+micwTi,i+msLHV-miLHF=(micw+mscw)Teq

and finally solving for Teq results in

Teq=mscwTi,s+micwTi,i+msLHV-miLHF/(micw+mscw)

We can do one more thing, since the masses are equal, we can factor out the mass.

Teq=mcwTi,s+mcwTi,i+mLHV-mLHF/(mcw+mcw)

Teq=m(cwTi,s+cwTi,i+LHV-LHF)/m(cw+cw)

Teq=(cwTi,s+cwTi,i+LHV-LHF)/2cw

Part 3)
Plug in known values

Teq=[(4.19J/(g*K)(373K)+(4.19J/(g*K))(273K)+(2257J/g)-(334J/g)]/[(2*4.19J/(g*K))]

Teq=552K Obviously incorrect

Please let me know where I am making my mistake. I separated my work into 3 parts to make it easier to point out where i made an error. Thank you I have a test on Friday I NEEEEEED to understand how to do this properly. Thank you

First, there is no need to bring absolute temperature into this problem. A 1 degree centigrade change in temperature is equal to a 1 degree Kelvin change in temperature.

Since you have equal masses of steam and ice, you know the equilibrium temperature of the mixture must lie somewhere between 0 C and 100 C.

In problems of this type, it's sometimes easier to pretend you are doing bookkeeping with heat instead of money. Instead of writing out some long equation where it is easy to make a mistake, set up a table. You know how much heat steam @ 100 C contains per unit mass, and you know how much heat ice @ 0 C contains per unit mass. After accounting for the latent heats of fusion and evaporation, you should be able to write a simple equation to determine equilibrium temperature of the liquid formed after the phase change is complete.

SteamKing said:
First, there is no need to bring absolute temperature into this problem. A 1 degree centigrade change in temperature is equal to a 1 degree Kelvin change in temperature.

The only reason I did was so that whenever I multiplied any terms with the initial temperature of the ice it didn't zero out.

SteamKing said:
In problems of this type, it's sometimes easier to pretend you are doing bookkeeping with heat instead of money. Instead of writing out some long equation where it is easy to make a mistake, set up a table. You know how much heat steam @ 100 C contains per unit mass, and you know how much heat ice @ 0 C contains per unit mass. After accounting for the latent heats of fusion and evaporation, you should be able to write a simple equation to determine equilibrium temperature of the liquid formed after the phase change is complete.

I understand this, this would be easier. My goal is to be able to do this problem with all masses, temperatures, inside of containers of different material. I NEED to do this algebraically since our exam will consists of concept questions and solving for different terms where the values aren't given.

Personally, I would prefer to work through the algebra rather than doing SteamKing's bookkeeping technique.

You missed one major simplification that you could have done very early in your algebra. You noted that you can factor out an m from everything. However, you kept m around for a while. You can divide out m in your first algebra step and cut your pencil mileage in half.

If you do that, you should be able to find your algebra mistake. I found something, but I think you corrected it. WHen you solved for Teq, you divided by (micw+mscw), but you are missing parentheses around the numerator in the next step.

castrodisastro said:
The only reason I did was so that whenever I multiplied any terms with the initial temperature of the ice it didn't zero out.

I don't understand this statement. What terms 'zeroed' out?

I understand this, this would be easier. My goal is to be able to do this problem with all masses, temperatures, inside of containers of different material. I NEED to do this algebraically since our exam will consists of concept questions and solving for different terms where the values aren't given.

Still, if you are going algebra only, at least try for simple algebra. Writing one equation which covers half a page is hard to understand. At least find out how much heat exchange occurs to get the ice and steam both to liquid phase first. There's no requirement everything be done in one step.

flatmaster said:
You missed one major simplification that you could have done very early in your algebra. You noted that you can factor out an m from everything. However, you kept m around for a while. You can divide out m in your first algebra step and cut your pencil mileage in half.

If you do that, you should be able to find your algebra mistake. I found something, but I think you corrected it. WHen you solved for Teq, you divided by (micw+mscw), but you are missing parentheses around the numerator in the next step.

Yes I chose the long way to force myself to keep track of everything no matter how complicated but I re-did it. This time I canceled out the masses from the beginning and I came up with the same equation for Teq.

Teq=[cwTi,s+cwTi,i+LHV-LHF]/[2cw]

Plugging the same values in gives me the same answer

Teq=[(1562.87J/g)+(1143.87J/g)+(2260J/g)-(334J/g)]/[8.38J/(g*K)]
Teq=(4632.74J/g)/8.38J/(g*K)
Teq=552.83 K

So I'll work out my algebra in detail here...here goes

Qlost=Qgained

mLHV+mcw(Ti,s-Teq)=mLHF+mcw(Teq-Ti,i)

Now from here I factor and cancel the masses

m[LHV+cw(Ti,s-Teq)]=m[LHF+cw(Teq-Ti,i)]

Then becomes

LHV+cw(Ti,s-Teq)=LHF+cw(Teq-Ti,i)

Distributing the temperature differences...

LHV+cwTi,s-cwTeq=LHF+cwTeq-cwTi,i

Moving all the terms containing the Teq to one side

LHV+cwTi,s-LHF+cwTi,i=cwTeq+cwTeq

Factoring out the Teq

LHV+cwTi,s-LHF+cwTi,i=(cw+cw)Teq

Now solving for Teq

So I have

Teq=[LHV+cwTi,s-LHF+cwTi,i]/(2cw)

OR if it's clearer this way...

In the numerator we have
LHV+cwTi,s-LHF+cwTi,i

Divided by (denominator)
(2cw)

Which is still the exact same thing. Did i make the same mistake?

SteamKing said:
I don't understand this statement. What terms 'zeroed' out?
When I plug in values for the term cwTi,i, or anything multiplied by the initial temperature of the freshly melted ice, multiplying anything by 0.00°C will make that term 0. That term has been "zeroed out"

You can see what's happening from your final equation. The heat of condensation of the steam is so high that the ice doesn't have enough oomph to even condense all the steam. Try the problem assuming that only a fraction of the steam condenses. Of course, the final temperature will have to be 100C. You will be solving for the fraction of the steam that condenses.

Chet

Chestermiller said:
You can see what's happening from your final equation. The heat of condensation of the steam is so high that the ice doesn't have enough oomph to even condense all the steam. Try the problem assuming that only a fraction of the steam condenses. Of course, the final temperature will have to be 100C. You will be solving for the fraction of the steam that condenses.

Chet

ah got it! you are correct. Bear with me as I am trying to do this without any notes or equations since I want to be able to do this properly during the exam.

So let's say that the energy required to turn the steam into liquid is given by msLHV

if I know that the melting of ice will contribute to the steam turning into water. So I could say that some fraction(α) of steam will have turned to water when the ice fully melts.

αmsLHV=miLHF

Knowing that the masses are the same will leave me with
αLHV=LHF

α=LHF/LHV

That's a proportion. Is that right so far?

So when the steam melts all the ice we have α% left.
Next we have to calculate the rest of the steam turning into water.

αLHV=micw(T2-Ti,i)

The T2 is the temperature that the water is once the steam has fully converted to water.

Lastly we calculate

mscw(Ti,s-Teq)=micw(Teq-T2)

mscwTi,s-mscwTeq=micwTeq-micwT2

Solving for Teq

mscwTi,s+micwT2=micwTeq+mscwTeq

Teq(micw+mscw)=mscwTi,s+micwT2

Finally

Teq=(mscwTi,s+micwT2)/(micw+mscw)

I haven't done the algebra by hand for it that's why I didn't cancel out the masses but was that the correct process?

castrodisastro said:
ah got it! you are correct. Bear with me as I am trying to do this without any notes or equations since I want to be able to do this properly during the exam.

So let's say that the energy required to turn the steam into liquid is given by msLHV

if I know that the melting of ice will contribute to the steam turning into water. So I could say that some fraction(α) of steam will have turned to water when the ice fully melts.

αmsLHV=miLHF

Knowing that the masses are the same will leave me with
αLHV=LHF

α=LHF/LHV

That's a proportion. Is that right so far?

So when the steam melts all the ice we have α% left.
Next we have to calculate the rest of the steam turning into water.

αLHV=micw(T2-Ti,i)

The T2 is the temperature that the water is once the steam has fully converted to water.

Lastly we calculate

mscw(Ti,s-Teq)=micw(Teq-T2)

mscwTi,s-mscwTeq=micwTeq-micwT2

Solving for Teq

mscwTi,s+micwT2=micwTeq+mscwTeq

Teq(micw+mscw)=mscwTi,s+micwT2

Finally

Teq=(mscwTi,s+micwT2)/(micw+mscw)

I haven't done the algebra by hand for it that's why I didn't cancel out the masses but was that the correct process?
Unfortunately, no. If there is any steam left in the end, the final temperature is going to be 100C. You need to calculate the amount of heat it takes to melt all the ice and raise its temperature to 100 C. That will tell you how much steam can be condensed at 100 C. That will all become water at 100 C. The rest of the steam will remain steam at 100C.

## What is the equilibrium temperature of an ice and steam mixture?

The equilibrium temperature of an ice and steam mixture is 0 degrees Celsius, also known as the melting point of ice. This is the temperature at which both ice and liquid water can coexist in equilibrium.

## How is the equilibrium temperature of an ice and steam mixture calculated?

The equilibrium temperature can be calculated using the principle of energy conservation, which states that the total energy of a closed system remains constant. In this case, the energy gained by the ice as it melts is equal to the energy lost by the steam as it condenses.

## Can the equilibrium temperature of an ice and steam mixture vary?

Yes, the equilibrium temperature can vary depending on external factors such as pressure, altitude, and the purity of the ice and steam. For example, at higher altitudes, the pressure is lower, and the equilibrium temperature may be slightly lower as well.

## Why is the equilibrium temperature of an ice and steam mixture important?

The equilibrium temperature is important because it helps to determine the state of a substance. In the case of an ice and steam mixture, the equilibrium temperature indicates whether the substance is in a solid, liquid, or gas state. It also plays a crucial role in various industrial and scientific processes.

## What happens if the equilibrium temperature of an ice and steam mixture is not reached?

If the equilibrium temperature is not reached, the ice and steam will continue to exchange energy until they reach a state of equilibrium. This could result in the ice melting or the steam condensing until they reach the same temperature. If the temperature is significantly different from the equilibrium temperature, the process may take a longer time.

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