Finding the final temperature when steam and ice are mixed

[kiro484]

Homework Statement

Suppose that 20.0g of steam at 110°C is mixed with 25.0g of ice at -40°C What will the final temperature be?

Ice - Specific heat capacity 2.10 J/g°C
Water - Specific heat capacity 4.19 J/g°C
Water - Latent heat of fusion 334 J/g
Water - Latent heat of vaporization 2268 J/g
Steam - Specific heat capacity 2.08 J/g°C

Homework Equations

Q=mcΔt
Q=mH

The Attempt at a Solution

Heat loss (steam) = heat gain (ice)
mcΔt+mH_vap+mcΔt = mcΔt+mH_fus+mcΔt
(20.0g)(2.08J/g°C)(10°C)+(20.0g)(2268J/g)+(20.0g)(4.19J/g°C)Δt = (25.0g)(2.10J/g°C)(40°C)+(25.0g)(334J/g)+(25.0g)(4.19J/g°C)Δt
416J+45360J+83.8J/°C(Δt)=2100J+8350J+104.75J/°C(Δt)
45776J+83.8J/°C(Δt)=10450J+104.75J/°C(Δt)
35326J+83.8J/°C(Δt)=104.75J/°C
[(421.5513126)(1/°C)]Δt=1.25Δt
But then I get lost and confused as to what I need to do to isolate the variable. Any help or solutions would be greatly appreciated. Thanks in advance

 

[ehild]

Homework Statement

Suppose that 20.0g of steam at 110°C is mixed with 25.0g of ice at -40°C What will the final temperature be?

Ice - Specific heat capacity 2.10 J/g°C
Water - Specific heat capacity 4.19 J/g°C
Water - Latent heat of fusion 334 J/g
Water - Latent heat of vaporization 2268 J/g
Steam - Specific heat capacity 2.08 J/g°C

Homework Equations

Q=mcΔt
Q=mH

The Attempt at a Solution

Heat loss (steam) = heat gain (ice)
mcΔt+mH_vap+mcΔt = mcΔt+mH_fus+mcΔt
(20.0g)(2.08J/g°C)(10°C)+(20.0g)(2268J/g)+(20.0g)(4.19J/g°C)Δt = (25.0g)(2.10J/g°C)(40°C)+(25.0g)(334J/g)+(25.0g)(4.19J/g°C)Δt

Do not use the same notation for different things. What is Δt in case of the steam and what is it in case of ice?

ehild

 

[kiro484]

Well the first time /\t appears on the steam side of the equation it is equal to 10. And the first time it appears on the ice side of the equation it is equal to 40. But the second time it appears on both sides is the variable I'm trying to solve for because it is the final temperature of the mixture.

 

[ehild]

Well the first time /\t appears on the steam side of the equation it is equal to 10. And the first time it appears on the ice side of the equation it is equal to 40. But the second time it appears on both sides is the variable I'm trying to solve for because it is the final temperature of the mixture.

Δt is not the final temperature of the mixture.

ehild

 

[kiro484]

Then how would I find the final temperature?

 

[ehild]

We use the symbol Δ for difference, for change of some quantity. ΔT is change of temperature, ΔT=T(final)-T(initial). You added the heat released when the steam cools down to 100 °C to the heat released when it becomes water at the boiling point, that is at 100 °C. That water cools down to the final common temperature. So what is the ΔT you need to use for the steam-water?ehild

 

[kiro484]

I'm not sure I understand what you're saying. How are you supposed to come up with the the final temperature? That's the whole point of the question. The steam water will have the same temperature as the ice water but I don't know how to solve for that.

 

[haruspex]

I'm not sure I understand what you're saying. How are you supposed to come up with the the final temperature? That's the whole point of the question. The steam water will have the same temperature as the ice water but I don't know how to solve for that.

Create an unknown variable, T say, for the temperature you're trying to determine. Then you can replace each Δt with the appropriate difference between two temperatures. Some of those will involve T.

 

[kiro484]

So for the steam it would be (110-t) and the ice would be (t-40), correct? But the rest of the equation posted above would stay the same?

 

[haruspex]

So for the steam it would be (110-t) and the ice would be (t-40), correct? But the rest of the equation posted above would stay the same?

No, that's wrong in a couple of ways. I presume you meant (t-(-40)) = t+40, but that would be wrong too. The delta t's you are replacing in the equation by these terms are the changes in temperature of what, exactly?
Also, your equation does assume that the final state is all water. When you have solved it, how will you know whether that assumption was correct?

 

[kiro484]

Yes that is what I meant, with the t-(-40). These changes would be the final temperature. We know that the steam drops 10 degrees then condenses but we don't know the final temperature which would be the 100-t and we know the ice gains 40 degrees and melts and the final temperature would be t-(-40). Well the final temperature has to be the same, so when the equation is solved we will know the state because if the temperature is 0<t<100 it's water, and if t<0 then it's ice and if the t>100 then it's steam. But for sure it can't be steam, because the ice would have to gain a substantial amount of energy to become steam. Since we assume no heat was gained or lost to the environment, the only energy available is in the ice or steam.

 

[haruspex]

Yes that is what I meant, with the t-(-40). These changes would be the final temperature. We know that the steam drops 10 degrees then condenses but we don't know the final temperature which would be the 100-t

No, the final temperature is t. 100-t is the drop in temperature of the water after it condenses. Note that this is the right quantity to multiply by the s.h. of the water, not (110-t) as you wrote before. And for the melted ice, what temperature difference should you use there?

 

[ehild]

When the steam cools down to its boiling point Q1=c(steam)m(steam) (T(steam)-100) heat is released.
During condensation, Q2=m(steam)L(heat of vaporization) is released.

When the condensed steam cools down from 100 °C to the final temperature T, Q3=c(water)m(steam)(100-T) heat is released.

q1=c(ice)m(ice)(0-(-40)) is the heat needed to warm up the ice to 0°C.
q2=L(heat of fusion)m(ice) is needed to melt it and gain m(ice) water at 0°C.
q3=c(water)m(ice)(T-0) heat is needed to warm up the ice-water to the common final temperature T.

Q1+Q2+Q3=q1+q2+q3, if the final stage is water. But you can not be sure that all the ice is melt or all the steam condenses. So calculate the amounts of heat separately and see if the released heat of steam is is enough to warm up the ice, then melt it.

Q1=416 J.....q1=2100 J
Q2=45360 J...q2=8350 J

You see that Q1+Q2>q1+q2, so all the ice will be melt.

It can happen that the released heat is enough to warm the ice-water to 100°0. How much is the heat q4 needed to warm up the ice-water to the boiling point?

If the heat of the steam is enough to warm up the ice-water to 100°C then you have 20+25 g water at 100 °C and Q1+Q2-(q1+q2+q4) heat available to evaporate some water. Is it enough to evaporate all?


ehild

 

[kiro484]

Ok so I tried this again
Heat loss (steam) = Heat gain (ice)
mcΔt+mH+mcΔt = mcΔt+mH+mcΔt
(20g)(2.08J/g°C)(10°C)+(20g)(2268J/g)+(20g)(4.19J/g°C)(100-t) = (25g)(2.1T/g°C)(40°C)+(25g)(334J/g)+(25g)(4.19J/g°C)(t-0)
416J+45360J+83.8J/°C(100-t) = 2100J+8350J+104.75J/°C(t-0)
Combining like terms
45776J+83.8J/°C(100-t) = 10450J + 104.75J/°C(t-0)
35326J+83.8J/°C(100-t) = 104.75J/°C(t-0)
35326J+8380J/°C-83.8J/°Cx = 104.75J/°Cx
35326J+8380J/°C = 188.55J/°Cx
187.356139(1/°C)+44.444444444444444444444444444444 = x
231.8005834°C = x
232°C = x
But the final temperature can't be higher than the starting temperature of steam, because the steam is already the hottest substance.

 

[ehild]

Yes, so the final state is not all water. Try the second part of my post. Is the heat of the steam enough to warm up all the ice to 100 °C?

ehild

 

[kiro484]

It can happen that the released heat is enough to warm the ice-water to 100°0. How much is the heat q4 needed to warm up the ice-water to the boiling point?

If the heat of the steam is enough to warm up the ice-water to 100°C then you have 20+25 g water at 100 °C and Q1+Q2-(q1+q2+q4) heat available to evaporate some water. Is it enough to evaporate all?

Well this has already been answered by solving for the variable in the equation I posted before this post. That variable balances out the equation, because you can see that it was very unequal before. T is the final temperature which came out to 232 degrees. You will get the same answer trying to see if all the ice can become steam as you suggested. But if you think about this question logically, 20g of steam at 110 degrees isn't going to be able to change the temperature of a greater mass of ice by 140 degrees without condensing. Plus we know for a fact the the final temperature has to be between -40-110 degrees.

 

[ehild]

Well, what is the final temperature?

ehild

 

[kiro484]

Well I came up with 232 degrees which isn't the right answer

 

[ehild]

You supposed that the final state is all water which results in an impossible final temperature. So the final state is not just water. I tried to explain it in post #13.
The steam cools down to 100 °C and some condenses. The released heat is enough to warm up the ice to 0°C, melt it and warm up the ice-water from 0°C to 100 °C. How much heat is needed to that? How much steam need to condense to provide that heat?

ehild

 

[kiro484]

Well according to the equation, the steam releases enough heat to melt the ice and raise the temperature of the mixture to 232 degrees. Something about the equation itself has to change, but I don't know what

 

[ehild]

The water, reaching the boiling point, boils away, becomes steam. It never can be as hot as 232 degrees.
The assumption that the final state of the mixture is water is not true. So your equation is not valid.

Part of the steam will condense and the heat released during condensation raises the temperature of the ice, melts it, and rises the temperature of ice-water to the boiling point. Both water and steam coexist in the container: it can happen at the boiling point only.

ehild

 

[kiro484]

So how would I fix my equation to give me an answer like that?

 

[ehild]

You can not fix that equation. It is valid only if the final state is water. That is not true. The final state is water and steam. The water can not be hotter than 100 degrees. Calculate the available amount of heat and show that part of it is enough to warm up the ice to 100 degrees.

ehild

 

[kiro484]

I'm sorry but I just don't understand what I have to do. Thanks for the help, but this is due tomorrow and I will just hand in what I got.

 

[ehild]

If the whole steam condenses 35326 J heat is released, as you calculated already. The ice needs 10869 J to warm up to 0°C, and melt. It needs some more heat to warm up to 100 °C. Calculate it. You will see that all the heat needed to warm up the ice is much less than the heat released if all the steam condenses. The ice can not condense all steam. Steam and water coexist at the boiling point of water.

What is you do not understand?

ehild

 

[kiro484]

10475J is needed to warm the ice water to 100. That gives us a total of 21344J. This isn't a value I get in any way with the steam condensing

 

[ehild]

The steam cooled down to 100° but will not fully condense. What is unclear?

ehild