Equilbrium temp of ice and steam mixture

[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.

m_s=mass of steam
m_i=mass of ice
L_HV,s=Heat of vaporization of steam=2257J/g
L_HF,i=Heat of fusion of ice=334J/g
T_i,s=100°C=373K
T_i,i=0.00°C=273K
c_w=4.19J/(g*K)

Homework Equations

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=Q_gained+Q_lost=Q_gained+(-Q_lost)

Q_lost=Q_gained

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

Q_lost=m_sc_w(T_i,s-T_eq)+m_sL_HV

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

Q_gained=m_ic_w(T_eq-T_i,i)+m_iL_HF

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

Q_lost=Q_gained

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

m_sc_w(T_i,s-T_eq)+m_sL_HV=m_ic_w(T_eq-T_i,i)+m_iL_HF

m_sc_wT_i,s-m_sc_wT_eq+m_sL_HV=m_ic_wT_eq-m_ic_wT_i,i+m_iL_HF

Moving all terms containing T_eq to one side

m_sc_wT_i,s+m_ic_wT_i,i+m_sL_HV-m_iL_HF=m_ic_wT_eq+m_sc_wT_eq

Factor out T_eq

m_sc_wT_i,s+m_ic_wT_i,i+m_sL_HV-m_iL_HF=(m_ic_w+m_sc_w)T_eq

and finally solving for T_eq results in

T_eq=m_sc_wT_i,s+m_ic_wT_i,i+m_sL_HV-m_iL_HF/(m_ic_w+m_sc_w)

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

T_eq=mc_wT_i,s+mc_wT_i,i+mL_HV-mL_HF/(mc_w+mc_w)

T_eq=m(c_wT_i,s+c_wT_i,i+L_HV-L_HF)/m(c_w+c_w)

T_eq=(c_wT_i,s+c_wT_i,i+L_HV-L_HF)/2c_w


Part 3)
Plug in known values

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

T_eq=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

 

[SteamKing]

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.

 

[castrodisastro]

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.

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.

 

[flatmaster]

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.

 

[SteamKing]

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.

 

[castrodisastro]

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 T_eq.

T_eq=[c_wT_i,s+c_wT_i,i+L_HV-L_HF]/[2c_w]

Plugging the same values in gives me the same answer

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

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

Q_lost=Q_gained

mL_HV+mc_w(T_i,s-T_eq)=mL_HF+mc_w(T_eq-T_i,i)

Now from here I factor and cancel the masses

m[L_HV+c_w(T_i,s-T_eq)]=m[L_HF+c_w(T_eq-T_i,i)]

Then becomes

L_HV+c_w(T_i,s-T_eq)=L_HF+c_w(T_eq-T_i,i)

Distributing the temperature differences...

L_HV+c_wT_i,s-c_wT_eq=L_HF+c_wT_eq-c_wT_i,i

Moving all the terms containing the T_eq to one side

L_HV+c_wT_i,s-L_HF+c_wT_i,i=c_wT_eq+c_wT_eq

Factoring out the T_eq

L_HV+c_wT_i,s-L_HF+c_wT_i,i=(c_w+c_w)T_eq

Now solving for T_eq

So I have

T_eq=[L_HV+c_wT_i,s-L_HF+c_wT_i,i]/(2c_w)

OR if it's clearer this way...

In the numerator we have
L_HV+c_wT_i,s-L_HF+c_wT_i,i

Divided by (denominator)
(2c_w)

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

 

[castrodisastro]

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

When I plug in values for the term c_wT_i,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"

 

[Chestermiller]

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

 

[castrodisastro]

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 m_sL_HV

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.

αm_sL_HV=m_iL_HF

Knowing that the masses are the same will leave me with
αL_HV=L_HF

α=L_HF/L_HV

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.

αL_HV=m_ic_w(T₂-T_i,i)

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

Lastly we calculate

m_sc_w(T_i,s-T_eq)=m_ic_w(T_eq-T₂)

m_sc_wT_i,s-m_sc_wT_eq=m_ic_wT_eq-m_ic_wT₂

Solving for T_eq

m_sc_wT_i,s+m_ic_wT₂=m_ic_wT_eq+m_sc_wT_eq

T_eq(m_ic_w+m_sc_w)=m_sc_wT_i,s+m_ic_wT₂

Finally

T_eq=(m_sc_wT_i,s+m_ic_wT₂)/(m_ic_w+m_sc_w)

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?

 

[Chestermiller]

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 m_sL_HV

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.

αm_sL_HV=m_iL_HF

Knowing that the masses are the same will leave me with
αL_HV=L_HF

α=L_HF/L_HV

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.

αL_HV=m_ic_w(T₂-T_i,i)

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

Lastly we calculate

m_sc_w(T_i,s-T_eq)=m_ic_w(T_eq-T₂)

m_sc_wT_i,s-m_sc_wT_eq=m_ic_wT_eq-m_ic_wT₂

Solving for T_eq

m_sc_wT_i,s+m_ic_wT₂=m_ic_wT_eq+m_sc_wT_eq

T_eq(m_ic_w+m_sc_w)=m_sc_wT_i,s+m_ic_wT₂

Finally

T_eq=(m_sc_wT_i,s+m_ic_wT₂)/(m_ic_w+m_sc_w)

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.