What is the temperature of equilibrium when mixing ice and vapor?

[Robin04]

Homework Statement

We mix 1 kg of ice at -10 C with 5 kg of vapor at 110 C. Calculate the temperature of equilibrium.

m_i = 1 kg
T_i = -10 C
m_v = 5 kg
T_v = 110 C
L_melting = 330 000 J/kg
L_condesation = 2 260 000 J/kg
c_water = 4200 J/kg

Homework Equations

Q_given = Q_got
Q = L*m = c*m*dT

The Attempt at a Solution

I do not know that at what state will be the water at the temperature of equilibrium. If it was water, if would be at 520 C. It can't be vapor because if the vapor gave energy to the ice to melt, then boil, the vapor would already start the condensation.

 

[SteamKing]

Homework Statement

We mix 1 kg of ice at -10 C with 5 kg of vapor at 110 C. Calculate the temperature of equilibrium.

m_i = 1 kg
T_i = -10 C
m_v = 5 kg
T_v = 110 C
L_melting = 330 000 J/kg
L_condesation = 2 260 000 J/kg
c_water = 4200 J/kg

Homework Equations

Q_given = Q_got
Q = L*m = c*m*dT

The Attempt at a Solution

I do not know that at what state will be the water at the temperature of equilibrium. If it was water, if would be at 520 C. It can't be vapor because if the vapor gave energy to the ice to melt, then boil, the vapor would already start the condensation.

How does ice and vapor at 110 C form a mixture whose temperature is 520 C? Have you found a new source of energy?

You should show your calculations for this.

 

[gneill]

You need to work out the energy required or produced by the various state transitions and compare them to see what transitions, in whole or in part, can take place. You can't necessarily know ahead of time what the final state will be --- you have to "sneak up on it" by moving the available energy around a step at a time.

 

[Robin04]

Thank you guys. I rethought, and i realized that I can have water and vapor at the same time, so I got that in the end from the 5 kg of vapor I have only 1,31 in the state of gas, and all the ice boiled to vapor. m_vapor_total = 1,31+1 = 2,31 kg
m_water_total = 5-1,31 = 3,69 kg
Am I right?

 

[gneill]

Thank you guys. I rethought, and i realized that I can have water and vapor at the same time, so I got that in the end from the 5 kg of vapor I have only 1,31 in the state of gas, and all the ice boiled to vapor. m_vapor_total = 1,31+1 = 2,31 kg
m_water_total = 5-1,31 = 3,69 kg
Am I right?

That doesn't look right. You've both boiled all the ice to vapor and condensed some vapor to liquid?

You'll have to present your work in detail for us to see where it's right or wrong.

 

[Robin04]

Finding the final state:
To boil the ice I need:
2100*10+330000+4200*100+2 260 000 = 3 031 000 J of energy.
I got it from the vapor.
For the vapor to condense I need:
1380*5*10+2 260 000*5= 11 360 000 J.
This is much more that is needed to boil the ice and at the same time the energy to cool the vapor to 100 C is smaller than the ice is needed to boil. So I am sure that in the end I will have a mixture of 100 C water and 100 C vapor.

Q_given = Q_got
1380*5*10+2 260 000 * m = 2100*10+330000+4200*100+2 260 000
Here m is the mass of the vapor that is required to condense which is 1,31 kg. And here was a mistake I made (I switched the meaning of m). So in the end I have 5-1.31+ 1 = 4,69 kg of vapor and 1,31 kg of water.
I hope I didnt make any mistake. :-)

 

[gneill]

I'm dubious about your value "1380". You haven't defined what it is anywhere, and I suppose it's meant to be the specific heat of water vapor (steam). I'd think it should be closer to 1900 J/(kg C ) for temperatures near 100C.

And why are you vaporizing the water you converted to liquid from ice? You've written:

1380*5*10+2 260 000 * m = 2100*10+330000+4200*100+2 260 000

Where that final term looks suspiciously like a heat of vaporization. If you're going to end up with a mix of liquid water and steam at 100C, shouldn't the ice be brought to 100C and left there?

However, that said, the good news is that any water that's vaporized will be exactly traded for an equal amount of steam condensed, so the oversight disappears in a puff of mathematics. Your final masses for the liquid and vapor look okay.

 

[Robin04]

Yes, 1380 is the specific heat of steam, it is given in my book, I forgot to write it down for you, sorry.

I think now I got it. I took out the boiling from the ice and now the equation is
1380*5*10+2 260 000 * m = 2100*10+330000+4200*100
where m is still the mass of vapor that is needed to condense to get the ice to 100C.
m=0,31 kg
So Total Vapor = 5-0,31 = 4,69 kg
Total Water = 1 + 0,31 = 1,31 kg
Interesting, the same result. :O

 

[gneill]

Yes, 1380 is the specific heat of steam, it is given in my book

Hmm. Seems rather low to me. I'd really expect something closer to 1900 or 2000 kJ/(kg K). See the table here for example.

 

[Robin04]

Hmm. Seems rather low to me. I'd really expect something closer to 1900 or 2000 kJ/(kg K). See the table here for example.

Yes, you're right, but 1380 is given in my book, so I have to calculate with this. By the way, next to the 1380 there is written "for isochoric prossesses". I don't really understand this comment indeed.